sqrt(i)

We will find both the square roots of i, i.e. sqrt(i). We will first write sqrt(i) as a complex number a+bi and then square both sides. Then we will solve for a and b by setting a system of equations! This is the algebra way to find the square root of the imaginary unit i.
Check out these related videos:
polar way: • sqrt(i) in polar form
sqrt(i+sqrt(i+sqrt(i+...))): • sqrt(i+sqrt(i+sqrt(i+....
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[@blackpenredpen]

sqrt(a+bi)=? Answer here: kzbin.info/www/bejne/eZa5lZtrga1sZpo

[@user-RN123]

👍

[@onradioactivewaves]

White Chalk Red Chalk, nice 😊

[@allozovsky]

Steve: 0:00 As we all know, √(-1) = 𝒊 Also Steve: 8:50 √𝒊 = ±(1 + 𝒊)/√2 - *two answers* That's inconsistent!

[@onradioactivewaves]

@@allozovsky it's not inconsistent. There's one in Q I and another in Q III separated by p u/2 rotation Much like using an inverse trig function to return an answer in a limited domain where the actual solution may be outside of that domain. Recall that -1 is a real number, I is imaginary, and thr square roots are complex. Complex numbers don't behave quite the same as real and purely imaginary numbers.

[@allozovsky]

@@onradioactivewaves If √𝒊 = ±(1 + 𝒊)/√2 gives two complex square roots, then √(−1) also should return two complex square roots, that is √(−1) = ±𝒊, isn't it? Otherwise it is inconsistent. That's pretty strange that Steve often evaluates in his videos multivalued complex functions alright, but at the same time uses only single 𝒊 for the square root of −1.

[@Electric_Bagpipes]

There ya go psychologists; the root of imagination.

[@jologer20]

This is so underrated

[@SpiderMan-in9rl]

This deserves more likes...

[@kaustubhi5910]

Bro what 30 likes 😂

[@BritishBeachcomber]

The blackboard was designed by a guy named Hilbert.

[@NStripleseven]

Woo

[@jonhues5552]

I happened to watch this on my break from studying before my leaving cert maths exam. Square root of a complex number was on the exam and I got the right answer using this method. What a crazy lucky coincidence.

[@dolumdoli7621]

Congrats man

[@artemis_furrson]

Yeah I remember the exam. I used a different method tho.

[@randomzedotp9579]

@@artemis_furrson What was it? 🧐

[@Someone-wj1lf]

@Mosinlogan Being interested in what you study is a blessing many students would strive to have.

[@arian6565]

@@Someone-wj1lf there's no way people are like that? Lol I didn't know that, that's strange

[@Abram--S]

What I miss when I zone out for 30 seconds in class:

[@rubensf7780]

Underrated comment

[@kdog3908]

ADHD moment. Felt it.

[@nayjer2576]

There is a simpler way without the polar form. We know a² - b² = 0 and 2ab = 1, therefore a² = b² a = b or a = - b. But because of the second equation we can cancel out a = - b as a possibility. That means a = b and if we plug this in the second equation we get: 2a² = 1 a = +/ - (1/sqrt(2)). Thats it. :D

[@Arcangel0723]

yeah I instantly saw that, its cool how there are so many different ways to arrive at the same answer

[@gamerdude7800]

No , just cause a² = b² , Doesn't mean a = b , cause (-1)² = 1² but obviously -1 ≠ 1

[@nayjer2576]

@@gamerdude7800 read my comment again

[@priyanshusharma1812]

​@@gamerdude7800 he mentioned a=±b

[@musty2468]

This is how I did it too, nice

[@BlackIGO]

Plot twist: He has endless layers of boards.

[@muhammadhafiz1337]

Hahhaa

[@stewie3k_]

Truuue

[@franciskang]

blackboardwhiteboard

[@christopherjacobo1669]

😂😂😂

[@BlackIGO]

@BLVGaming / Y1000 Couldn't not agree

[@PG6813]

Me: trying to go to sleep KZbin: BuT wHaTs ThE sQuArE rOoT oF i???

[@laerciocivali]

-1

[@alexting827]

@@laerciocivali no

[@galoisalexander3833]

lol

[@RubyPiec]

@@laerciocivali thats i squared

[@AymanSussy]

Same here XD

[@neobaud513]

You can also convert to e^(i*π/2). Then sqrt(e^(i*π/2))=e^(i*π/4). Then convert back to get 1/sqrt(2)+i/sqrt(2)

[@at_one]

And losing second solution )

[@CorvusSapien]

You can get the second solution by generally writing sqrt(i)=exp(i*(pi/2+k*pi)) for k any integer

[@at_one]

@@CorvusSapien It's not a solution. You write just answer. Initial post suppose to write i, but not sqrt(i) in exponential form. And than use powering properties. One could write i = exp(i*π/2 + 2πk) i**0.5 = exp(i*π/4 + πk) But why we didn't lose something else in this solution? Using powering properties is totally wrong way of thinking in this case.

[@peted2783]

@@at_one doing square root gives 2 solutions sqrt(i)= ±sqrt(e^πi/2)= ±e^πi/4 = ±(cos(π/4)+isin(π/4)) giving the 2 solutions: sqrt(i)=1/sqrt(2) + i/sqrt(2) or sqrt(i)=-1/sqrt(2) -i/sqrt(2)

[@at_one]

@@peted2783 this is not a question. The question is about using powering properties. My opinion is that your's approach is wrong. On this case it gives correct answer, but in general case not. One must use this formula while square rooting: en.m.wikipedia.org/wiki/De_Moivre%27s_formula But you shouldn't use this formula: (e^z)^0.5 ≠ e^(0.5z) to find all roots. Initial post is about powering. And my comment is about it. I know how to square rooting in complex field 😂

[@lifeatitsfinest3019]

bro has infinite blackboards

[@eastonrocket3273]

🎅🎅🐡🥕

[@thexavier666]

I, as a non native english speaker, watched your video at 2x speed. Got everything you said. Keep it up.

[@fattymuffinbbx]

You should be an English teacher

[@trobin]

Cough cough the pinned comment cough cough

[@JannisAdmek]

me too, I guess as a non english native speaker you are used to a bigger variety of accents

[@nicholasleclerc1583]

OMG EXACTLY ME!!!!!!!!!!

[@guiguspi]

Me too. I think improving the accent would be nice still, though.

[@n484l3iehugtil]

I don't know why people keep complaining about this guy's solution to the problem, and why they offer geometric proofs instead. I really like this guy's answer because it uses only the simplest arithmetic/algebra and the simplest definition of a complex number: a + bi. Also, he did it in an incredibly concise manner. (My only complaint was at the end where he could have used a² - b² = (a+b)(a-b) = 0 --> a = ±b. But, typing this out, I suddenly realise how clever he is that even *factorization* doesn't need to be used in his answer.) This answer is teachable on someone's very first lesson on complex numbers; even the average 15-year-old will comprehend it very well, and he's earned my amazement.

[@blackpenredpen]

Billy Ma-gusta thank you Billy. I think people just got too excited. It's like once they see derivatives, they want to show algebra students to take the derivative to find the vertex of a parabola.

[@alcesmir]

Billy Ma-gusta You want concise? Polar form: R*exp(iθ) = sqrt(i) Square: R^2*exp(2iθ) = i = exp(i(π/2 + 2πn)), n integer Identify: R=1, θ=π/4+nπ Done. Or if you prefer rectangular form the unique representations are: ±(1+i)/sqrt(2) Tl;dr: Rectangular is great for addition and subtraction, polar for multiplication, powers and roots. Right tool for the job.

[@n484l3iehugtil]

My point: Your working is as opaque as it gets. For all the average high schooler or KZbinr can see, you're just writing a whole bunch of Greek because nothing is explained. The polar form of complex numbers isn't explained at all. The polar form of √i specifically also isn't explained, neither is i. Converting between the polar form and the rectangular form also isn't explained. You cannot call this concise when you leave so much unexplained; someone who doesn't know the required background knowledge would call this gibberish. (Like how I call bullshit when I am told "using string theory, we can show that 11 physical dimensions exist". Just an example, don't digress pls.) And the greater point: the polar form does not NEED to be explained in order to prove this result, as bprp has shown. Simpler is better in maths, and a short proof is not necessarily a simple proof.

[@alcesmir]

I would happily agree that I assume too much if I for some video on an integral complain about someone not using residuals and Cauchy's integral formula to solve an integral on the real line since it would be trivial that way. But assuming knowledge of the polar form of a complex number, which is as fundamental as the rectangular form, is not a stretch imo. As for the specific polar forms. I never use the polar form of sqrt(i), only the polar form of i, to keep things simpler. I could happily have shown a justification of the polar form of i, but it's a bit clunky without being able to show any figures. Basically you have an angle of π/2 to i and any 2π increment of that still leaves you at i. If I had shown this on a blackboard the justification is explained in a few seconds. The conversion to rectangular form is really not part of my solution, I'm happy to stay in polar form. My point is that this result is trivial in the polar world. It also gives some really nice insights about how the root behaves in general, which is totally lost in rectangular form. Using polar form the result also trivially generalizes to any real power of i. Just like addition/subtraction is trivial in rectangular but a mess in polar form. Sure you could technically use the strategy as shown in this video to describe the 5th or 50th root of i as well, but it would be very cumbersome and very brute force.

[@grrrlag]

you actually don't need factorization to get that a = b. 1. a² - b² = 0 -> a² = b², therefore a and b have the same magnitude. 2. 2ab = 1, ab = 1/2, therefore a and b have the same sign (if the product of 2 real numbers is positive, then 2 real numbers have the same sign) 3. a = b, because a and b have the same magnitude (1) and the same sign (2) (definition of equality for real numbers) 4.a*a = 1/2 = a², from (3) and (2) 5. a = ± sqrt(1/2), sqrt both sides of (4) 6. b = ± sqrt(1/2), from (3) I actually prefer this method of solving the system of equations over the video because it takes less steps and is more intuitive. the videos method involves a lot of seemingly arbitrary moving of symbols around while each of my 6 steps have much clearer purpose.

[@CannedMan]

My professor taught me something valuable when writing my master’s: Never start an argument/discussion/presentation with ‘as we all know’; you never know who doesn’t know, and thus risk pushing away potentially interested readers.

[@finnboltz]

I completely agree with your professor about that.

[@adityellectual4563]

Another way can be e^(i*π/2)= i for r=1, square root both sides and it will be e^(i*π/4)=√i, which will give √i= (1+i)√2

[@waxyacrobat]

6:05 when he got the answer he started moving closer to the speed of light

[@SusEGP]

Lmao underrated

[@francescolorenzelli8912]

When you beat a boss in Kingdom Hearts

[@Star-rd9eg]

@@francescolorenzelli8912 didnt think id find a kingdom hearts reference here ..... i still dont get the joke tho edit: Ahh i see the vision is blury...

[@lqwe5627]

lmaoo

[@joey7107]

I thought the same thing haha

[@stalebread9833]

I litterally read it as "Squ(i)rt"

[@o_5553]

I'll remember that forever

[@user-dz5im1vt6h]

ㅋㅋㅋㅋㅋㅋㅋㅋ

[@jlitodelcid326]

@@o_5553 same

[@dinamosflams]

Time to stop waxing your carrot

[@kommandantgalileo]

Me too

[@AniketTurkel]

Just a heads up, writing it in Euler form is way faster. That is, i = e^[iπ/2] => √i = e^[iπ/4] = 1/√2 + i.1/√2. Edit: forgot a - for the second root.

[@NotYourAverageNothing]

Wanna know what's behind my board? It's another board!

[@_carrbgamingjr]

bruh this teacher is so prepared. he uses 2 different colors of chalk to distinguish between terms and grouping symbols. Good Job!

[@_carrbgamingjr]

Wanna know what's behind the second board? A third board!

[@vishwaajithn.k3266]

@@_carrbgamingjr yep

[@stevens5541]

I'm board.

[@_carrbgamingjr]

@@stevens5541 ok

[@ketofitforlife2917]

I appreciate the depth of this explanation, rather than memorizing forms, for the sake of speed and ease, you showed me understanding. I appreciate knowing why, over mechanical speed.

[@woophereigo9755]

You'll get mechanical speed with practice. Idiots just don't know to practice.

[@manperson6234]

@@woophereigo9755 Smart guy over here. Shut up.

[@woophereigo9755]

@@manperson6234 Bunch of morons. Get good.

[@k-fedd]

@@woophereigo9755 yikes bud 😬

[@sam_music555]

@@woophereigo9755 One thing is be fast while doing your own calculations, another is following another one being fast doing his calculations

[@kritishalli5073]

This showed up on my recommended, and I could feel myself getting smarter throughout the video because of your amazing teaching style. You have earned yourself a new subscriber, so thank you

[@a_voice_in_the_wilderness]

That was interesting to follow, I've forgotten so much math including basic algebra, this was very helpful and you did a great job of explaining it all.

[@TheDailyEgg]

Okay, that's easy. The real question is: How many blackboards does this guy have?

[@nogame-nolife8537]

lol the same here ;

[@lukeahandsa]

@The Daily Egg yes

[@pinklady7184]

He must be teaching in many different classrooms or renting classrooms.

[@nurphurecarnium]

He's like people in the math problem

[@slamwall9057]

i

[@badhbhchadh]

whitechalkredchalk

[@mmishra7214]

Lol

[@jumbledbee3603]

I had an aneurysm trying to figure out how u were changing the chalk color for like a solid 10 sec until I realized it was just 2 pieces of chalk

[@antasosam8486]

Like two pieces of sqrt(i) ;=)

[@Bikeswede7568]

Another way to see it is that multiplying by i makes the complex number rotate around origo by 90° (pi/2). Multiplying by i^(1/2) instead rotates 45° (pi/4). So, for example, 1 × i^(3/2) = -1/sqrt(2) + i/sqrt(2) since that is where a rotation of 135° from 0° takes us.

[@socrates727]

Thank you for sharing this.

[@megablademe4930]

My man’s too drippy for us, wearing supreme and teaching maths

[@blackpenredpen]

😆

[@samuelluce8286]

When the number is imaginary!

[@larrydonica9321]

Ù⁸⁸

[@boltthebot9625]

He’s on another level

[@junkgum]

Root_4 (-1)

[@chetanraikwal5766]

**He lifts up the black board** Me - what the hell is thisss?

[@SuperUghe]

*lifts second blackboard revealing a third* “Oh my god!”

[@chetanraikwal5766]

@@SuperUghe yeah I guess the future is here!

[@BenitezCarvalho]

What? This is pretty common xD

[@TimThomason]

@@SuperUghe It's blackboards all the way down.

[@nanashi_74_]

No it is -green board-

[@warwick802]

This is genuinely the first time I've watched math videos for entertainment. Props to bprp

[@sophisticatedplayer]

Euler's formula just makes everything easier

[@asforax8909]

There is an other easy way : We have i=e^(i(π/2)) So √i =[e^(i(π/2))]^((1/2)) So √i=e^(i(π/4))=√2/2 +i √2/2 It means that √i = √2/2 +i √2/2

[@ayaacademy7657]

Brilliant 👍

[@IconoclastX]

Damn bro nice flex lol

[@akasakasvault7597]

Aka 1+i

[@donebro4403]

i don't understand

[@donebro4403]

@@ericw2391 ok l will check it.Thanks+

[@GamingConsole]

Ok fine but, Why is he holding a grenade in his hands

[@kayjaad3349]

that's a microphone

[@user-yj5cm1jz7h]

@@kayjaad3349 $Thanks, I did not know that_

[@GamingConsole]

@@kayjaad3349 there's something called sarcasm sis......

[@GamingConsole]

@AFancySpoon you'll only get attention if you comment on the board that this guy is using XD *Bitter truth tho*

[@qubatistic4788]

His parents made him hold it, it'll go off the moment he makes a mistake.

[@lynxprime1778]

"sqrt" being sus

[@TRINI123A]

thanks. i was wondering about this when i woke up.

[@RobbyRatPoison]

Only thing confusing was saying 1/(2*(1/√2)) = 1/√2 When I looked I instead got √2/2 but if you multiply by √2/√2 you get 2/2√2 which gives you 1/√2 so you were right but that part was the only thing I found to be unclear

[@TheWannaramble]

glad you commented because that tripped me up as well

[@someoneunimportant3064]

@@TheWannaramble 2/sqrt(2) should be multiplied with sqrt(2)/sqrt(2) (which changes nothing since it's just 1). 2*sqrt(2)/sqrt(2)*sqrt(2) We know that square root of n multiplied with itself gives us n so: 2*sqrt(2)/2 Both sides divided by 2 Sqrt(2)

[@TheWannaramble]

@@someoneunimportant3064 very clear, thanks

[@someoneunimportant3064]

@@TheWannaramble you are very welcome, glad it helped

[@57energon]

@@TheWannaramble that is what I thought

[@jabir5768]

Very understandable even as a non native speaker

[@blackpenredpen]

thanks!!

[@anandviswanathan4625]

I agree. Your work is very good. Dont worry about trolls who complain and then fight.

[@YellowToad]

*egg*

[@Nylspider]

@@YellowToad egg

[@rubineronaldvallevivar]

Es verdad mientras voy en un bus lo miro, por el alto volumen de bus no puedo oír el vídeo, soy hispanohablante, aún así se entiende todo.

[@jawad9757]

Equating real and imaginary components in general is pretty useful

[@graquaza8180]

in a^2-b^2=1 you can simply add b^2 to both sides to get a^2=b^2 and then cut the roots then use this to get a=b then in 2ab=1 divide both sides by 2 to get ab=1/2 and then use a=b to get a^2=1/2 and then root both sides to get the answer to a and therefore b

[@brunoamezcua3112]

I don't understand how this can be SO FUKING PERFECT

[@stumpfightskills571]

Bruno Amezcua, because mathematics is a series of quantitative tautologies, where each system builds off the previous system.

[@Infinite_Precision]

@@stumpfightskills571 Well said!!⚡🔥

[@michelmln]

He almost made a writing mistake at the end (= instead of or) but fixed it immediately...

[@simi_says]

Surprisingly complex yet also surprisingly simple

[@henkhu100]

But wrong! The answer is wrong! There are two square roots of i. A value x is a square root of i if x^2 = i. So what you find are the square roots of i. But one of those square roots is ⎷i. Just like 2 and -2 are the square roots of 4, but ⎷4 = 2. and not also -2 So only the principle root is ⎷i. the symbol ⎷ is used for the principle root. So his answer is wrong. For ⎷i see also 5.1 on en.wikipedia.org/wiki/Square_root#Principal_square_root_of_a_complex_number

[@ZotyLisu]

@@henkhu100 what do you mean? in complex numbers n-th root always has n answers, and he showed both

[@henkhu100]

@@ZotyLisu Indeed: there are two square roots(as I mentioned in my reaction). But only one of them can be written with the ⎷ symbol. Example: 4 has two square roots: 2 and -2. But ⎷4 is just one of those values: 2 (the non negative) and is called the principle square root. So when he gives two values for ⎷i he's wrong. In the case of complex numbers we have a similar situation. There are again two square roots, but only one of them can be written with the symbol ⎷ See the paragraph Algebraic Formula. on en.wikipedia.org/wiki/Square_root#Principal_square_root_of_a_complex_number For the principle root of x+iy. You will see that that principle square root (written as ⎷(x+iy). stands for just 1 value. And in the video he gives two values for ⎷i. and that's not correct. May be he does not know the differents between the square roots of a value x and the value ⎷x. The problem he solved is not "find ⎷i." but "find the square roots of i."

[@ZotyLisu]

@@henkhu100 yeah he could've wrote w0 and w1 or whatever, but I disagree that that's an only definition, in my course the value of an expression with ⎷ symbol was defined as a set of all answers - I'm still a bit confused on how that works tbh

[@henkhu100]

@@ZotyLisu If ⎷ is a set then what is the meaning of for instance (⎷5)/(⎷6) ? A quotient as a result of deviding one set by another? ⎷5 is not a set, it is just a value. Example: solve x^2=7. In your opinion the solution is x=⎷7 because ⎷7 stands for all answers . But the solution is x=⎷7. or x=-⎷7. The definition of the ⎷ symbol in your course was not correct. I am sure you know the solution of a standerd quadratic equation: x=(-b+⎷(b^2-4ac))/2a. or x=(-b-⎷(b^2-4ac))/2a Your definition of the ⎷ symbol would only give the first value because the ⎷ symbol already stands for the + end the - in the formula.

[@garylawrence7547]

The most amazing part here is that he can write on the board with two different colors in one hand!! Truly amazing.

[@Pengochan]

When doing roots on complex numbers (especially on the unit circle) it's really instructive to look at what happens in the complex plane and represent them in polar representation: i=exp(pi/2 * i), sqrt(i)=exp(pi/4 * i). The other thing that should be talked about here is mathematical conventions/notation when taking (principal) square roots, especially when first stating that sqrt(-1)=i unambiguously and then having two results for sqrt(i).

[@mohitdeb6728]

Teacher : What is the value of infinity? Me : The amount of boards that blackpenredpen has . Teacher : 🙏

[@tsukr28]

you know he can erase the chalk, right?

[@thetitanium2353]

🤣🤣🤣

[@astitva3072]

@Stev Da Great you're*

[@shubh.bapi_9423]

Teacher said namaste!!🙏

[@blackpenredpen]

Hi there! If you like this video, you will also like: (a+bi)^(c+di) kzbin.info/www/bejne/kGWzpH-qbsSer5o Be sure to subscribe for more interesting math videos! bprp

[@einsteingonzalez4336]

Where's the hashtag #YAY that we've been waiting for so long?

[@yoavcarmel1245]

almost reached 0.5M views!

[@blackpenredpen]

Yoav Carmel I know!!!! I am super excited!!

[@yasserouhamiche4418]

When it's power 4 we will find 4 solution i guess!!

[@kutuboxbayzan5967]

Another way1: İ=e^(i*pi/2) Sqrt(i)=e^(i*pi/4)= Cos (pi/4)+i*sin (pi/4)= (İ+1)/sqrt(2) Another way 2: Sqrt (i)=x X^4=-1 (X^2+sqrt (2)x+1)* (X^2-sqrt (2)x+1) Solutions (İ-1)/sqrt(2) (İ+1)/sqrt(2) (1-i)/sqrt(2) -(1+i)/sqrt(2) And later we found Sqrt (i)=(i+1)/sqrt(2)

[@HuntingKingYT]

3:20 from these equations you could have just done: a^2-b^2=0 => |a|-|b|=0 => |a|=|b| Then, knowing that you can go to the next equation: 2ab=1 => ab=1/2 => sign(a)=sign(b) (Either both positive or both negative, because their multiplication results in a positive number) Then, because their signs are equal *AND* their absolute values are equal, you can assume their both equal. So now you have 2 solutions that differ by sign: a=b=+-sqrt(1/2)

[@D.A.Jamal.]

Loved the way u explained!

[@TheMasterfulcreator]

Make sure you don't focus too much on improving your accent first of all. I could comprehend it just fine. Chinese pronunciation works just fine on English language as long as you have practiced. More good videos please.

[@DavidGarcia-nx2gj]

you are studying math you dont even know how to speak english to understand this. math is universa, i didnt even listen the audio to understand everything i jus tskipped it

[@liujanis636]

@TheMasterfulcreator I see what you did there.

[@wildchicken2049]

Nice I like what u did

[@zek9277]

Math is its own language

[@mattfrankman]

My calc class is taught by a Romanian woman, half the class is Chinese. Communication is not a problem. Math is the universal language, numbers unite us all.

[@iaagoarielschwoelklobo6342]

8:54 10 people got *TRIGGERED* because he didn't rationalize the denominator

[@blackpenredpen]

IAAGO ARIEL SCHWOELK LOBO lol!!!!

[@Aramil4]

IAAGO ARIEL SCHWOELK LOBO No one rationalizes denominators in 2017 - that's what people did back in the day when there were no calculators and you'd prefer dividing the memorized decimal form of the square root by the rationalized denominator.

[@scitwi9164]

Another thing is that rationalizing the denominator often hides the geometric connections between quantities. It's a bit harder to see that `√2/2` is the inverse of `√2`, but it is obvious when you didn't rationalize it: `1/√2` (one over something is the inverse). It's even more hard to see it with some more complex expressions with radicals. That's why I usually leave it unrationalized, as an inverse, unless I really have to rationalize it.

[@Aramil4]

Sci Twi I wonder what modern books have answer keys that use rationalised denominators..

[@JM-lh8rl]

Sci Twi Look; I've got mixed feelings for the conventions of rationalizing denominators or not. You are completely right about maintaining the instant recognition for inverses, but then you'd be compromising the recognition of like terms. For example, 3/sqrt5 does not look like it could be added to 2sqrt5/2, but after rationalizing, you can see clearly that 3sqrt5/5 _can_. Now, on the issue of having the same answers as the teachers do (and trying to overlook the insult to many great math teachers I've met that I'm sure you were not trying to offend), you can't really be opposed to unification of measures or answers -at least to some extent you have to accept it. Of course it makes your life easier to save extra moments on a test or whatever, but using a more real life example with more important implications, the SIU (International System of Units)'s purpose is to ease scientists' endeavors at "sciencing", if you will, by having set standards as to what units are official, what they measure, and how much of that something they measure. This, of course, may mean little to a mathematician's job, but if you can apply this same smooth interchange of information through the answers and numbers you represent, what you try to state will be better conveyed and understood by the audience to whom you present the information to. Anyway, I know I can't force someone to think the way I do, and you have to use the methods that you know are better for your learning (very similar to the π/τ argument), but thanks for reading to the end.

[@gurjarriya9]

this was so informative! thank you.

[@toyfabrik2993]

Isn't it much easier to have a look at the situation in polar coordinates, using the complex e-function? That way you see right away that there are two solution, each one a vector in the complex plane with length 1, and theta either being pi over 4 or five pi over 4, which then can be rewritten in regular form using Euler's formula.

[@CharlesPanigeo]

You can also think about it in polar form. i is on the unit circle, so it's roots are also on the unit circle. The argument for the principal value must be π/4 (½ the argument of i). So if you have your unit circle memorized the principal root is clearly sqrt(2)/2 + i*sqrt(2)/2. The other root is opposite the principal root at -sqrt(2)/2 -i*sqrt(2)/2

[@benjamintete7335]

Damn Its so much simpler

[@maztrjayz007]

That's what I was think. You are rotating half way towards the imaginary number line from the real number line. That would be pi/4 rotation. Then figure out your polar coordinates and trig.

[@spiderjerusalem4009]

x = √i x² = i x⁴ = i² = -1 x⁴ + 1 = 0 x⁴ + 2x² + 1 = 0 + 2x² (x²)² + 2x² + 1 = 2x² (x² + 1)² = 2x² x² + 1 = x √2 x² - x√2 = -1 x² - 2(x)(√2 / 2) = -1 x² - 2(x)(1/√2) + ½ = -1 + ½ x² - 2(x)(1/√2) + (1/√2)² = -½ (x - 1/√2)² = -½ x - 1/√2 = ±√(-½) = ± √(-1) / √2 x = 1/√2 ± i/√2 √i = 1/√2 (1 ± i)

[@laurentwatteau8835]

I also solved the problem that way.

[@RichardJohnson_dydx]

Yep, that's the way I learned it.

[@blackpenredpen]

What if we have infinite nested square roots of i sqrt(i+sqrt(i+sqrt(i+...))): kzbin.info/www/bejne/ana9g4uNjJtjbcU

[@Star-fr8em]

no

[@asifalijatoi7965]

Can we also solve it by polar form ???

[@onassissalinas1195]

no

[@nathanblanchard2578]

Juste use the i=e**(i*pi*0.5)

[@vavkgp8029]

1.3 + 0.675i I think 😅

[@woodwardscreditcard7482]

This just shows that teachers needs to be smiling more when doing lectures. Just his smile alone made it more fun to watch

[@loghtsy8057]

6:31 isnt an easier way of doing that rearranging equation a² - b² = 0 to a² = b² in which case a = b then substitute a or b in the second equation so 2a² = 1 rearrange so a = +-√2/2

[@infinity8686]

same doubt

[@user-jz5sm4vv6v]

It can be a=-b

[@ampleman602]

If a² = b² a≠b Never make that assumption But it can be a = -b Or b = -a

[@nayjer2576]

@@ampleman602 a = -b or b = -a is the same. And he's right, it can only be a = b because of the second condition 2ab = 1, that only works if a = b. If a = -b you would get a negative output.

[@dcmurphy5157]

I’m not even doing this for school. I’m just interested.

[@austinlincoln3414]

same

[@aaronmisra735]

Lucky

[@aryanpuri6192]

True

[@boltthebot9625]

Same

[@joey7107]

Same, I'm really into this while i cant comprehend every step but i can feel his teaching energy pulling me into it.

[@bleesev2]

You went a really complicated way of solving these equations. In my head I did it like so a^2 - b^2 = 0; a^2 = b^2; a = +-b 2ab = 1; we know a and b should be of the same sign so we'll say a = b and get 2a^2 = 1 a^2 = 1/2 a = sqrt(1/2)

[@mienzillaz]

yup.. same here

[@apocalypticachmodio]

Exactly what I was thinking. But we would probably get less marks than him cause he did the longer method. 😂

[@rpgamer1002]

Thats why so many people don't get into maths. Maths can be wathever you want, depending on how you enter in. Many examples show very simple solution or very complex ones for the same question. Which one do you prefer ?

[@Gehr96]

Even simpler: sqrt(i) = (e^(i pi/2))^(1/2) = e^(i pi/4) = cos(pi/4) + i sin(pi/4)

[@mohamedwalid6181]

Even if signs are same It doesn't make a=b

[@aurora6814]

I SWEAR the other day I was just thinking about this and here comes this video in my recommended. So enlightening, thank you

[@vannypin]

Thank you for your video. There are two square roots of the imaginary unit (i), but what above the notation sqrt(i)? As I know, the notation sqrt(i) indicates the principle square root. And the answer should be the complex number with positive real part. Please correct me! Sorry for my English. I've learnt a lot from you.

[@harrystuart7455]

All the people complaining he didn't use polar co-ordinates are completely missing the point. If you haven't already studied exactly why e^ix = i sin(x) + cos(x) then that would make this video completely pointless, the kind of people who want to know the answer to this problem most likely haven't come across that level of mathematics yet

[@blackpenredpen]

Harry Stuart thanks!

[@GaryKildall]

Harry Stuart, well assuming you have a calculator for inverse Tan or a book full with tables of precalculated values for not carefully selected examples. And dividing is multiplying with the power of -1 defined as (a-bi)/(a^2+b^2) but you are free to learn such tables like some people like studying phone book numbers 😋

[@suhailmall98]

I agree that it doesn't look nice when you first come across it, but polar representation is one of the reasons complex numbers are so useful in the sciences

[@Taricus]

densch123 you'll wind up being thankful for the exponential form when you hit differential equations :P

[@Taricus]

or theoretical physics... or having to find real and imaginary roots of numbers... and complex functions and variables... lol ;P

[@spitalhelles3380]

Everybody is like 'there' s an easy way:' and then has full paragraphs of calculations. Just think in polar coordinates and rotations and the answer is obvious

[@mrfiermath5036]

Right? It should just be "what rotation composed with itself brings you to where i sits (90 degrees)". Boom 45 degrees. Boom, express as cos(45d)+i sin(45d)

[@AlanCanon2222]

Yep, I got it that way in seconds. When I was a teenager I was into the Mandelbrot/Julia sets, and the complex plane became my main jam.

[@inigofield6264]

@@AlanCanon2222 r/iamverysmart

[@hispantrapmusic301]

MrFierMath what I did, like sqr of i is equal to sqrt -1, leave that sqrt and make -1 to polar, and then the moivre method and that is right?

[@randomness4989]

I didn't have good teachers so this is the first time I understood this out

[@montyhall2805]

You can just turn "i" (rectangualr) to e^pi/2i (polar) and then just do exponentiation - something we all know. Then convert e^pi/4i back into rectangular. Looks like sqrts divides the rotation angle by 2.

[@MrSeezero]

That is an awesome way to solve that. You can also put i in a phasor-angle format, and take the positive square root of 1 which is 1 and divide 90 degrees by 2 to get 45 degrees and add 180 degrees to 45 degrees to get 225 degrees for the other square root of i. i; (1/__90 degrees)^0.5; 1/__45 degrees, 1/__225 degrees, cos(45 degrees) + sin(45 degrees)i, cos(225 degrees) + sin(225 degrees)i; (1/2)^0.5 + [(1/2)^0.5] i, -(1/2)^0.5 + [-(1/2)^0.5] i

[@henryolsen6248]

I may be the only one who liked you accent. And could you do a video on differential equations?

[@blackpenredpen]

Pink Floyd is the Best Band of All Time. Hi there, thanks!! I do have diff eq videos here www.blackpenredpen.com/math/DiffEq.html

[@henryolsen6248]

blackpenredpen Thanks!

[@nholrknrjrk]

The funny thing is that 1/sqr(2)=cos(45)=sin(45)

[@kanekeylewer5704]

That's not really surprising, you get this immediately using Euler's identity. e^ix = cosx + isinx, so i = e^(i*pi/2). sqrt(i) = i^(1/2) = e^(i*pi/4)

[@Infinite_Precision]

@@kanekeylewer5704 Wow, what is that, I couldn't understand it much but I found it really interesting!

[@opposite342]

I think this method can only get one answer of the square root of i, but I'm not sure. If there is a way to get the other answer, then please reply. I want to know.

[@TomGalonska]

@@opposite342 There is multiple ways to express an angle as a multiple of pi. The angle pi/2 is the same angle as (5*pi)/2 (by just adding 2*pi = 4*pi/2). If you now say sqrt(i) = sqrt(e^(i*5*pi/2)) = e^(i*5*pi/4) = cos(5*pi/4) + sin(5*pi/4) * i = - 1/sqrt(2) - 1/sqrt(2) i

[@Abdul-pv1qt]

Opposite34 I know I’m a month late but as a person who doesn’t study maths at that level, you explained it really well and i could follow every step

[@petermartin5030]

To do it in your head notice that a complex number squared is required to give i. So its real part must cancel out. So it is something like 1+i all squared which is 1+2i-1, so that 1 squared cancels with i squared. So (1+i)/root(2) will do, as will minus the same thing.

[@leosheehan7574]

This is Incredible. Never would have thought of the system of equations created.

[@jamessaker270]

Really good explanation! Thanks! As an Englishman, I can completely understand your accent!

[@blackpenredpen]

James Saker thank you James

[@destinyovbiebo8988]

Yeah, bro. Everyone that understand English would understand this man. Some people are talking about his accent in poor because they don’t understand English. As a matter of fact, they don’t even know the origin of English, to go further, English comprises of different European languages.

[@davidadegboye773]

Or you could just use polar form and the tiniest bit of trig. Complex multiplication is a rotation and a stretch, because the magnitude of i is 1, this case is just a rotation. i Is a 90° rotation counter clockwise from 1 so you need to find a number that you rotate twice to get 90°. That's 45° So sqrt(i)=cos(45)+isin(45) which is what you got

[@egoreruamercy4056]

Nicely

[@Sjobban112a]

Well you missed out on the other solution there, which is rotation by 225º twice, i.e. √i = cos(225)+isin(225) = - [1/√2 + i/√2] (which is also in the video). All solutions to √i are on the form √i = cos(π*n + π/4) + isin(π*n + π/4), where n = 0, 1, 2 etc. Albeit they only result in two unique solutions: ± [1/√2 + i/√2].

[@rallynub956]

@@Sjobban112a wouldnt it be sqrt (i) = cos (pi*n + (pi/4)) + isin (n*pi + (pi/4))? cos pi*n = 1, sin pi*n = 0, where n € Z? Sorry if confusing btw :P

[@Kyulnjir]

Yeh, but it's for beginners so many people wouldn't know about those

[@Sjobban112a]

@@rallynub956 You are right, I was typing a bit fast. Correcting my comment... :)

[@londonalicante]

i has modulus 1 and argument pi/2 + 2n*pi. sqrt(i) has modulus 1 and argument pi/4 + n*pi from that you can work out what the answer is, then check that the rectangular coordinates algebra are actually giving the right answer.

[@viz8746]

Beautiful and elegant solution.

[@Gmod2012lo1]

English is also not my motherlanguage and i understood you justy fine, no complains, ignore these morons :D, they should be thankful you didnt speak your language and just put subtitles

[@JoaoVictor-gy3bk]

Gmod2012lo1 english is not my mother language either, so it got me thinking: what if non-native english speakers understand other accents better when compared to native speakers?

[@analuizafelixdesouza6336]

@@JoaoVictor-gy3bk We really do. That's why I prefer being a non native English speaker.

[@hayden6700]

@@JoaoVictor-gy3bk that doesn't make sense, nor is it true

[@Matlockization]

He was speaking bamboo English. Its when I get up in the morning and the memory is loading to the ram but it takes a short time so I'm incoherent in the mean time.

[@notimportant6340]

What is your mother tongue? I assume it is west Germanic because you made the typo "justy" which reminds me of German richtig. German? Dutch? Frisian? Afrikaans?

[@spencerdavis867]

my nigga wearing a supreme shirt real shit

[@raajjann]

Multiplication by i by I rotates by Pi/2. So, one can expect sort(i) to be i rotated right by Pi/4. Checking (1 + i) / sort(2) * (1+i)/sqrt(2) = 2i/2 = i. So, that is the solution. This also should be able to generalize the answer for any real root of i.

[@iHATEbigots666]

thanks for everything, over many years. you're the best

[@traso56]

i don't care about the haters and their circle formulas, using various methos is really useful and gives more options, i know that formula but using this for fun is really nice

[@blergblergblerg1343]

There is a method for determining square roots which uses a system of 3 equations on module, real part and imaginary part. It wasnt fully used here, and wasn't relevant in this case. This guy sucks at math and saying so doesn't make anyone a hater, just a skeptical person who knows a tiny bit about math

[@Xandawesome]

Clearly this guy sucks at math even though he got the right answer

[@ivanneto817]

So you just proceeded to state that there is such a method and didn't explain the method or at least name it, then you said he sucks at math even though he got the right answer in simple steps. Nice.

[@norpedholland5696]

I just realized, this answer has a magnitude of 1 on the complex number plane. If you just looked at a and b while disregarding the i, you could say that with Pythagorean’s Theorem and with a and b as the x and y coordinates, the hypotenuse is 1. If you draw a unit circle with radius 1 on the complex plane (which touches points 1, -1, i, -i), then you can draw angles based on the points plotted on this unit circle. The angle of this answer with the real number 1 is 45 degrees. The angle between i and 1 is 90 degrees. The angle between -1 and 1 is 180 degrees. The angle between 1 and 1 is 360 degrees. This answer squared is i, i squared is -1, -1 squared is 1.

[@alexcarpentier5698]

True, you can also solve multiplications using the complex plane: any two nubers mulitpilied will have an angle equal to the angle of the first and the angle of the second number summed up, and will have a distance from 0 equal to the distance of the first number multiplied by the distance of the second number I’m french sorry if this isn’t very clear

[@spiderjerusalem4009]

x = √i x² = i x⁴ = i² = -1 x⁴ + 1 = 0 x⁴ + 2x² + 1 = 0 + 2x² (x²)² + 2x² + 1 = 2x² (x² + 1)² = 2x² x² + 1 = x √2 x² - x√2 = -1 x² - 2(x)(√2 / 2) = -1 x² - 2(x)(1/√2) + ½ = -1 + ½ x² - 2(x)(1/√2) + (1/√2)² = -½ (x - 1/√2)² = -½ x - 1/√2 = ±√(-½) = ± √(-1) / √2 x = 1/√2 ± i/√2 √i = 1/√2 (1 ± i)

[@ConstantDerivative]

since it is a power of i, it lands on the unit circle on the complex plane

[@kyzer422]

@@alexcarpentier5698 That's really interesting, I never realized that before.

[@fart-man]

so -1, I, and sqrt(i) are all on the unit circle. I wonder if the fourth root of i is as well.

[@captainhero4116]

Thanks to your newer videos, I did it in less than 5 minutes on my own :D

[@butwhoasked1821]

I believe it's easier the equal the module of both such that you get a equation of the form a² + b² = k which is very easy to combine with a² - b² = q. One can then chose which pairs of a and b go together knowing the sign of 2ab...

[@hypercoder-gaming]

Another way to solve the fourth root of four being equal to four is this: Sqrt(Sqrt(4))=x Sqrt(4)=2, so Sqrt(2)=x

[@lc1777]

He didn't even need to do all that things for finding A, it just required the simple observation that a = b or -a = -b This would mean 2a² = 1 and a = ± 1/√2

[@user-cc3iu4mp7x]

The Best way for me is to think of complex numbers as vectors on a complex plane. And if you raise an imaginary nomber to some power you make a rotation. i^2 = i^(1+1) = -1 hense you make a 90 degree rotation anti-clockwise and went from the imaginary axis to the real one. i^0.5 = i^(1-0.5) means that you make a 45 degree rotation clockwise. Now you only need to decompose your vector. The real part is cos(45 deg), and the imaginary one is i*sin(45 deg).

[@beeble2003]

@@lc1777 Yes, substituting a=±b from the first equation saves about five minutes of algebra compared to substituting b=1/2a from the second.

[@pi_xi]

@@lc1777 I had the same approach. Much easier.

[@lc1777]

@@beeble2003 yup

[@AstroHolden]

i may be complex, but 1 is still the loneliest number.

[@nationalstudyacademykim5030]

Just 2 punny!!!

[@fanimeproductionst.v.3735]

This shit got me sad asf.

[@nabeelkhan4910]

But can't debate bt being lonely he is also the glory coprime

[@yashuppot3214]

I isnt complex

[@darthvader2994]

at least he's positive about it, poor -1, lonely and negative :(

[@user-FMWM]

By posing i = e^(pi/2), it is quickier to answer. At one moment, you divide by "2a", you have supposed that 2a 0.

[@whiteeyedshadow8423]

the fact that this man made something so utterly incomprehensible make sense is incredible to me

[@TheAwesomeMister]

Very understandable, couldn't be more detailed. And nice that you already did a video on the polar way ;). Keep up the good work.

[@mueezadam8438]

I like how you hold your microphone the whole time.

[@blackpenredpen]

thanks!

[@Xrelent]

And two pens in the other hand. That blew my mind.

[@kaegansloan3794]

honestly though, it just truly added to the enjoyment of the video. He shows so much excitement when explaining and it altogether created a really good video

[@drewisbadden]

I have no idea what brought me to watch this, but I love it

[@WhiteDragon103]

Using a geometric interpretation, since i is unit length, its square root would be a unit vector (real, img) with half the angle. If i is at 90 degrees, sqrt(i) is at 45 degrees: 0.707, 0.707.

[@harrygenderson6847]

5:06 I noticed that not only are +1 and -1 solutions to a^4 = 1, but i and -i are too. More generally for a^n = 1, all nth roots of unity are solutions. Not that ignoring these solutions matters in this case, as they essentially just switch a and i*b but give the same final answers.

[@sebm2334]

I dont think we need to consider a or b being i or -i since theyre respectively the real and imaginary parts of the solution were looking for, therefore a and b are real numbers

[@woophereigo9755]

Then shut up

[@ayushman1940]

a and b are real they can't be i or -i

[@ayushman1940]

@@sebm2334 exactly

[@bensrandomshows1482]

does anyone else notice how oddly motion blurred he get when he's on the edge of the screen, and only the edge

[@kzushii]

the camera's not in focus?

[@mariabelleazemar7831]

yesss I've noticed

[@vizender]

@@kzushii it does not look like because it just seems out of temporal sinc, not spatial distortion

[@quandarkumtanglehairs4743]

Dude that's so clean and badass dude lmao Glad I watched and paid attention to the end lol fin perfect

[@medsalem4490]

You can use the complex exponential function sqrt(exp(i(pi/2))=sqrt(i)=exp(i(pi/4))=cos(pi/4)+isin(pi/4)=1/sqrt(2) + i/sqrt(2) 😅

[@adventure.assistant]

your energy is so uplifting. We love you. Keep doing what you're doing, with that little smile of yours you'll get anywhere :)

[@frankcabanski9409]

He has no energy. SJW's love his gimmick of holding a weird mic.

[@noahnaugler7611]

I like the more unit circle method, where squaring imaginary numbers doubles the angle to the positive real line, so square root must half. i is 90°, thus it's square root must be 45°, the coordinates of which are ((√2)/2,(√2)/2), or (1/√2)+(1/√2)i

[@alialknane8859]

هذا المثال في المنهج الدراسي في دولة العراق وهو من أبسط الامثلة في المرحلة الاخيرة (السادس الاعدادي) ❤

[@RCHobbyist463]

I just realized you could divide the angle by 2 since multiplying a complex number by itself is the same as doubling its angle. This explanation is pretty nice too.

[@jamesahibbard]

Yeah, pretty easy to tell that the square root of cos(pi/2) + i sin(pi/2) should either be: cos(pi/4) + i sin(pi/4) OR cos(5 * pi/4) + i sin(5 * pi/4)

[@Ocklepod]

Actually it can be done way simpler, if we know on the complex plane multiplication means that the absolute value of complex numbers multiply and the angle to the Real axis add up, So i is 90° with an absolute value of 1 sqrt(i) is 45 ° with absolute value 1, use some cos and sin to get the actual values if necessary

[@MrRoyalChicken]

it can also be 225° since you're going for 2*α mod 360° = 90°

[@lukapopovic5802]

MrRoyalChicken Actually, there is whole spectrum of solutions. They are e^(i*pi/4 + k*pi), where k is a whole number (positive or negative, or zero)

[@lukapopovic5802]

niklas schüller So there are infinitely many solutions ;)

[@MrRoyalChicken]

Luka Popovic out of all of those infinitely many solutions only two lay between 0° and 360° all other solutions are only a different way to reach those two points.

[@jmfriedman7]

coming from the other standard form of a complex number: z = a + i * b = r*e^(i * phi + 2*pi*k), with r = sqrt(a^2 + b^2), phi = atan2(b, a). For the complex number "z = i", r=1, phi=(pi/2), so i=e^(i*(pi/2+2*pi*k)), i^1/2 = the result above.

[@quickg0ld446]

i love how happy he is when talking about it

[@kingderper928]

There is also a restriction put into place when you remove the denominator of 4a^2, 4a^2 ≠ 0 => a ≠ 0 Since none of the answers given were 0, it didn’t matter in this case, but you can’t simply forget the restriction.

[@MythicalHandler]

I showed this to my physics teacher, even he can't keep up with you. Good job

[@norielsylvire4097]

Did anyone realise he is holding an Ood translator sphere?

[@blackpenredpen]

Dem Rottensoul u realized too late. kzbin.info/www/bejne/iYW0gah-eM6Fbrs

[@bubbao6209]

2:47 is legendary

[@RuwansaDewsandi]

I was unclear about this one and after i watch this video it was perfectly understood .Thank you so much❤❤❤

[@ommakadia3815]

u can also write in polar form sqrt (i)= i ^(1/2)= (e^(pi/2*i))^1/2= e^(i*pi/4) now, write is as cos(pi/4) + i*sin(pi/4) and u get the answer ...✌