This Is My New Favorite Number

i is the imaginary (complex) unit, but what is its square root?
We'll use some complex number theory and matching coefficients to solve a system of equations to figure this one out!
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/ @brithemathguy
Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.
#math #brithemathguy #i

[@BriTheMathGuy]

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[@jeff-1738]

That’s pretty cool; personally, I like number 4.

[@wockhardt9705]

lmfaooo

[@ktz1185]

LMAO

[@pasztesnik]

Perfection

[@qwertyflags]

welp he's definitely not asian

[@Bozitico]

SAME. Everyone always thinks it's weird, but it is important to solving math entirely. Well, every number has a role. But I like 4's role.😊😊😊

[@kingbeauregard]

The neat thing is, the fact that there isn't a meta-imaginary means that our number system is closed.

[@Kokurorokuko]

Yeah. It's interesting that humanity had to spent so much time to make numbers "complete". There are quaternions, though, but they don't have some properties which imaginary numbers have.

[@Noam_.Menashe]

Does been able tomae undefined numbers and singularities still keep it closed?

[@axoluna]

It is still possible to define further numbers; even natural numbers are “made up.” Check out the dual numbers, a separate number system where the new unit E^2 = 0

[@mertaliyigit3288]

@@Kokurorokuko imaginary numbers also lack some properties of real numbers. Inequalities dont work anymore

[@brandonklein1]

Sets are only closed under particular operations, so we need to specify which operations and sets we're talking about for that statement to be meaningful.

[@hassan55055]

i was actually wondering this the other day. how this recommendation was so spot on blows my mind

[@charliewallisch5581]

Literally the same for me

[@thedarkslayer9475]

I was studying complex numbers and this got recommended

[@wellen_good972]

@@thedarkslayer9475 same

[@benjaminkenney3706]

Apple is listening through our phones I swear

[@jacobw59]

Same for me. I think the video was recommended based on our search histories and overall interests.

[@0riginator]

Alternative method: Represent i in polar form Therefore, i = cos(π/2) + isin(π/2) By Euler's formula, e^(iθ) = cosθ + isinθ Therefore i = e^(iπ/2) Now √i = i^(1/2) = e^(iπ/4) Therefore √i = cos(π/4) + isin(π/4) √i = 1/√2 + i/√2 √i = (1+i)/√2

[@PunmasterSTP]

Yeah, that's what I thought about right when I saw the thumbnail.

[@flowingafterglow629]

@@PunmasterSTP Because it is so much easier

[@gitgudnoobs7917]

This is quite neat.

[@angeluomo]

Also my method. Had it in three steps.

[@matejnovosad9152]

So much uglier, but works

[@BriTheMathGuy]

Technically we define i by having the property i^2=-1. That is, sqrt(-1) is the principal root of i^2. It's just more convenient for this video say i=sqrt(-1).

[@createyourownfuture3840]

blackpenredpen found the √i using both this method and using polar coordinates.

[@themlaw9895]

Nice video but i is not equal to sqrt(-1) : suppose sqrt(-1) exist, so sqrt(-1)² = sqrt(-1)×sqrt(-1) = sqrt(-1×-1) = sqrt(1) = 1; wich is different of i², which is -1 So i is not a number that has a value, his only property is i²= -1, so your starting point is wrong, nice try though

[@kshitij7b286]

@@themlaw9895 it is wrong √(-1)×√(-1)= √[(-1)×(-1)=√1=1 The property √a×√b=√ab holds if and only atleast one of the number is positive that is either a≥0 or b≥0

[@createyourownfuture3840]

@@themlaw9895 The property of exponents you just used isn't true for complex numbers, but you enter that field when you talk about √-1. i is as real as real numbers, and nature works in complex numbers, not real ones.

[@MUJAHID96414]

I am excited to learn upper class math🤩, but I haven't books

[@Obotron7]

Just a small thing; 0:14 Imaginary numbers are a subset of Complex Numbers, not the same.

[@user-sb9ho5jz3e]

before it was official, people used imaginary numbers to refer to any numbers that have an imaginary part

[@juxx9628]

@@user-sb9ho5jz3e Official? Like what time? Cardano's time? Descartes time? Euler's time? Gauss time? Cauchy time? What is "official complex numbers"?

[@PunmasterSTP]

Another awesome video as always! I was curious about another method. Instead of using either Euler or De Moivre's formula, when you have the equation 0 + 1i = (a^2 - b^2) + 2ab*i, you could try setting a = b. If you do, then 1 = 2b^2 and so b = +/- sqrt(2)/2 = a, and sqrt(i) = +/- (sqrt(2)/2 + i*sqrt(2)/2). Knowing there are only two square roots of a complex number, you would be done. I also tried setting a = -b, and I got complex values for a and b, but plugging them into (a + bi) I got the same thing!

[@BriTheMathGuy]

Cool!

[@pNsB]

Nice! I always thought of it as: i = e^(i pi/2) i^(1/2) = +- e^( i pi/4) and just leave it at that, because I prefer polar form for some reason

[@data50090]

Imagine a 6th grader watching this video, couldn't be me right?

How's 8th grade going?

[@data50090]

pretty well so far, math score really good and I am going to be one of the 5 people in my school to do the mathematics national competition in my country (I hope that goes well)

[@FlyFlux]

​@@data50090Gl

[@kaauchpohtato]

dang howd that go?

[@data50090]

@@kaauchpohtato it did not go well

[@JakubS]

you can also just know that a 45 degree, 90 degree, 45 degree triangle with hypotenuse of 1 has sides of 1/√2

[@v44n7]

this, its a beautiful way to see it

[@lathasri3483]

Amazing

[@reevsy3]

I was given this problem in an introductory maths module for physics and thought of a simple geometric interpretation. I figured √(i) as being half the rotation around the origin in an argand diagram from 1 to i, since multiplying by i has the effect of "rotating" a number 90 degrees counter clockwise. Therefore the coordinates for √(i) would be the points on the unit circle that were half way between 1 and i either way. This would gives two points on the unit circle, one going 45 degrees counter clockwise from 1, and another going 135 degrees clockwise from 1. You can quickly see these are both 45 degree angles from the axis, thus a^2 + a^2 = 1 --> 2a^2 = 1 --> a^2 = 1/2 --> a = +1/√(2) or -1/√(2) for the respective top right and bottom left quadrant solutions. Therefore √(i) = +1/√(2) * (1 + i) or -1/√(2) * (1 + i)

[@sujaiy6646]

Nice 😊👍👍👍

[@kaan8964]

What is the concept of rotating imaginary numbers called?

[@andrewkarsten5268]

@@kaan8964 it’s really more of the general interpretation of multiplication in general. It’s a great interpretation. I recommend 3blue1brown’s video on complex numbers, he gives a wonderful way to rethink what addition and multiplication actually mean. The way to think of complex numbers falls right out of that.

[@Diaming787]

@@kaan8964 complex numbers are in a form of a+bi but so is r*e^(i*theta), and theta is the rotation part. Example: i is equal to e^(i*pi/2) because it is rotated pi/2 radians from the +Real axis. -1 = e^(i*pi) because it is rotated pi radians, which is why when you multiply by negative numbers, you 'flip' to it's negative side.

[@Firefly256]

Shouldn’t the result be the “positive result” only because we’re talking about the principal root?

[@simong1666]

Think of the case of sqrt(-i)... then the roots are (1-i) and (-1+i) which one qualifies as positive?

[@Firefly256]

@@simong1666 When dealing with b⁴ = 1 you only get 1 as answer, not -1 Because sqrt(36) = x x² = 36 Now we have created an extraneous solution x = -6 Because of principal root, sqrt(36) only equals 6. If talking about all roots, sqrt(36) = 6, -6 So because we squared the square root, when we do the inverse of x² = 36, we just do the step x = sqrt(36), and not x = ±sqrt(36) In the video he did ± which I think is wrong if we're talking about principal root, which we usually are

[@qwerty687687]

the whole point of complex numbers is to calculate all roots, not just the principal root. As Simon G pointed out, the concept of a principal root doesn't even make sense for roots of complex numbers (with a non-zero imaginary part). So I don't think we atre talking just about principal roots here.

[@JordanMetroidManiac]

No, there are not two values for sqrt(i). There is only one. There are two values for z in the equation z^2=i. You get that equation by making the invalid step of squaring both sides of z=sqrt(i). It’s like starting from 1=-1 and then squaring both sides to find that 1=1, and then using that fact to “prove” that 1=-1. Obviously that logic doesn’t work there, so don’t try to use that logic to prove that sqrt(i) has two values. Also, you wouldn’t say that sqrt(1)=-1, would you? But you would obviously say that x=1 and x=-1 in the equation x^2=1.

[@佐藤広-c4p]

Your comment is √ i={cos(π/2)+isin(π/2)}^(1/2)=cos(π/4)+isin(π/4)=1/√2 +i1/√2, and -√ i =-{cos(π/2)+isin(π/2)}^(1/2)=-{cos(π/4)+isin(π/4)}=-(1/√2 +i1/√2)=-1/√2 -i1/√2 ? I certainly think it's more reasonable to think so. However, complex numbers that are different from real numbers are difficult for ordinary people to understand. If you have an explanation or material that is easy to understand, please let me know.

[@sagnikasen2402]

Hello Brian! Your videos are captivating as usual. Really enjoyed it. I was hoping if you could kindly make a video on tensors. It would be really helpful

[@theuserings]

Find i^i Know: e^(iπ) = -1 Raise both sides to 1/2 th power e^(iπ)^(1/2) = (-1)^(1/2) e^(iπ)^(1/2) = i Raise both sides to the i th power e^(iπ)^(1/2)^(i) = i^i By properties of exponents, we have: e^(i²π1/2) = i^i i² = -1 by definition, in the end we have: e^(-π/2) = i^i

[@KazuyaMLBB]

Some dude that ask: Hey what's your favorite number? Me: √i

[@elchile336]

that same guy: ...in real values? Me: 4throot(-1)

[@matejnovosad9152]

This was a problem on my first Further Maths test in AS-Levels. Pretty easy but very fun!

[@nuquicharlesyaelq.8584]

This is my new favorite number-🤓

[@Anonymous4045]

This mightve just been my algebra teacher being picky, but arent you not supposed to leave roots in the denominator? I mean, i understand why 1/sqrt(2) looks better than sqrt(2)/2, but i believe the latter is easier to work with

[@flaht3964]

Draw a unit circle. Bcoz |sqrti|=1 r=1 do positive 45 degree angle the height is imaginary part the base is real part

[@pinedelgado4743]

Love this stuff!! Complex numbers are among my favorite mathematical objects!! Thanks lots for posting!! :) :)

[@BriTheMathGuy]

Glad you enjoyed it!

[@ZachWackAttack]

yo I know im prob wrong with my logic since mathemeticians didn't use something even close to as simple as this but what if you think of it as the square root of I = the square root of the square root of -1. Then, put it = x. Square to cancel out the big square root, the square root of -1 = x^2. Then, square that to get -1 = x^4. Then, take the fourth root of both to get i + or - 1 = x. So, the square root of i = i + or - 1. If this somehow does work then ig it doesn't matter cuz its not a real number like your solution

[@TamjidANoor-sh9bv]

Actually for any root of complex we can use a formula r=√(x^2+y^2) It's modulas Z=x+yi √z=1/√2{√(r+x) +i√(r-x)}

[@axbs4863]

I wanna know what i to the i’th tetration is

[@aweebthatlovesmath4220]

Wait is i-th tetration definited? Because i did it by my own and i got ln(0) in eqution which is undefined (in limit -∞ )

[@axbs4863]

@@aweebthatlovesmath4220 rip

[@EvilSandwich]

Its cool that you used solid math to completely prove something that someone with a small amount of trig skill could have just intuitively deduced. If -1 is a 180 degree rotation on the complex plane and the square root of -1 is i (which is a 90 degree rotation on the complex plane), it stands to reason that the square root of i would be (1/root2) + (1/root2)i. Which is a 45 degree rotation on the complex plane. I wonder if that means that the 4th root of i is a 22.5 degree rotation? And so on. Proofs can be so elegant sometimes. Edit: YEP! The pattern holds! 4th root of i = cos(pi/8) + sin(pi/8)i 8th root of i = cos(pi/16) + sin(pi/16)i and so on. So you could generalize this to: Nth-Root of -1 = cos(pi/n) + sin(pi/n)i Neat!

[@WaterMeetsLava]

That's basically De Moivre's theorem

[@EvilSandwich]

@@WaterMeetsLava Oh! Good to know! I'll look it up!

[@JojoJere]

Why is 0^2-1^2-=0, I thought it'\ is -1? Also, why is 2*0*1*i=1, 2*0*1=0 so 0*i=0? WHYYY

[@genepirate7114]

It is not entirely true... the principle root, or square root means the one with the biggest real part. That is, sqrt(4) is always 2 and not -2 even though (-2)^2=4. Using that logic, sqrt(i) can only be the positive one

[@ViniciusTeixeira1]

i = e^(i(pi/2 + 2kpi)) => i^(1/2) = e^(i(pi/4 + kpi)) = cos(pi/4 + kpi) + i*sin(pi/4 + kpi) = + or - (sqrt(2)/2 + i*sqrt(2)/2), k being an integer. :)

[@maximilianarold]

It actually makes sense. If i^1 is a rotation by 90° or π/2 then i^(1/2) or √i is a rotation by 45° or π/4

[@comradezach]

I'm so happy that I managed to get the real and imaginary parts being cosπ/4 and sinπ/4 just by looking at the thumbnail, feels nice to have something make Intuit sense. A neat way to think about this is rotation, multiplication by i gives a 90° or π/2 movement, so √i would be half that

[@tcthebeast3804]

Just took Complex Analysis last semester. Good vid

[@YOUSIFPOTATOYT0]

I mean just use the polar form

[@BriTheMathGuy]

also works of course!

[@johns.8246]

I never liked the definition of a square root, since two different numbers squared yields the same value. We define the square root of -1 as i, but -i works just as well. And how do we decide which square root of an imaginary number is THE square root?

[@Anonymous4045]

-i and i arent the same thing tho. One is -1 * sqrt(-1), the other is sqrt(-1) respectively. Yes, when multiplied by each other, they yield -1, but that doesnt mean they are the same

[@billmorrigan386]

@John S. The square root is exactly as you logically think it should be. I mean it returns two values, one positive, one negative. For simplification we often use principal roots and because of our laziness (mathematicians can also be lazy) we often don't even bother to mention it (whether we use principal roots or just roots), although folks might have huge problems because of that. Actually, i is defined as +√-1 because √-1 = +i, -i. Nowadays we often define a complex number as an ordered set of two real numbers with a specific rule for multiplication, which means we don't have to deal with complex numbers on a formal level. And that's how it is done in more rigorous math books. Bottom line: there's nothing wrong with definitions. It's just authors neglect finer points and some explanations. *Actually, it's a malpractice in my opinion. All you said, all your critique is 100% valid, while other people just pretend to understand stuff they have no idea about, unfortunately.*

[@Dragonaiis]

Anywhere you see a 1/√2 can be replaced with the following (according to unit circle): cos(π/4) = 1/√2 cos(7π/4) = 1/√2 sin(π/4) = 1/√2 sin(3π/4) = 1/√2 Alternatively you can get negatives: cos(5π/4) = -1/√2 cos(3π/4) = -1/√2 sin(5π/4) = -1/√2 sin(7π/4) = -1/√2 Perhaps these could all be used as substitutes? I tried replacing some of these with 1/√2 in the final equation at the end of the video and I graphed it but the equation looks wildly different. Thoughts?

[@PubicGore]

In the complex numbers, the nth root function takes on n values. Probably important to mention that.

[@angelmendez-rivera351]

No. Do you know what the definition of a function is?

[@PubicGore]

@@angelmendez-rivera351 Oh boy. Look what we have here. A person who seems to correct every little thing just to feel superior because they want to demonstrate that they know what they're talking about. All you really demonstrate is that you're insecure and probably terrible at that which you boast to practice with such skill. There is something called a multivalued function. This is clearly what I meant by "function." This association is hardly needed when you're talking to people who are somewhat competent. In complex analysis, the definition of function is extended. If you knew that, you wouldn't be trying to correct me. It remains a fact that the nth root is a multivalued function on C, and every nonzero complex number has n different complex nth roots.

[@angelmendez-rivera351]

@@PubicGore *Oh boy. Look what we have here. A person who seems to correct every little thing just to feel superior because they want to demonstrate that they know what they're talking about. All you really demonstrate is that you're insecure and probably terrible at that which you boast to practice with such skill.* Being said by someone who is being outwardly and intentionally condescending, and putting their train of insults before any valid arguments, this is just meaningless bluster. Did you get it out of your system? Did that stroke your ego enough? *There is something called a multivalued function. This is clearly what I meant by "function."* I know what it is you are talking about, but your misuse of terminology was worth pointing out. "Multivalued function" is a contradiction in terms. *In complex analysis, the definition of function is extended. If you knew that, you wouldn't be trying to correct me.* No. I know you are wrong, and I have more than sufficient education to know that. Your ignorant self being in a state of denial does not change this fact. I know the study of Riemann sheets is important in complex analysis and projective geometry, and they are also an an important foundation for studying presheafs. It changes absolutely nothing about the fact that you are wrong. No textbook in complex analysis defines function differently than in any other context. If you actually were literate on the subject, you would never talk about "multivalued functions." Almost no one educated on the subject does.

[@Gautam-tk8tf]

@@PubicGore woah chill, you wrecked that guy-

[@mujtabaalam5907]

You can think of this geometrically by drawing a unit circle on the complex plane.

[@RyanthePokemonTrainer]

sqrt(i) = a+bi has one solution, (a+bi)^2=i has 2 solutions because the square root function is defined as the positive answer to the square root of a number. Also, there is a faster way. 0 = a^2 - b^2 a^2 = b^2 a = b 1 = 2ab 1/2 = b^2 = a^2 a & b = 1/sqrt(2) sqrt(i) = 1/sqrt(2) + 1/sqrt(2)i Just wanted to point that out, also it doesn’t require any powers of 4 so it’s simpler for younger audiences.

[@TerryPlays]

Hi there.

[@benbooth2783]

Imaginary numbers are not complex numbers. Both real and imaginary numbers are 1 dimensional numbers representing a point on either the real or imaginary number lines. A complex number is 2 dimensional representing a point on the complex plane.

[@佐藤広-c4p]

Your comment "Imaginary numbers are not complex numbers" is strange. If the imaginary number is not a complex number (as is the case with real numbers), it should not be represented in a complex plane. The complex number was mentioned in the dictionaries as follows: Encyclopedia MyPedia "Real x,y and the number that can be expressed as z=x+yi by imaginary units. y=0 is a real number, x=0 is a pure imaginary number. ・・". Britannica International Encyclopedia "Any complex number z is written in the form of z=a+bi with a and b as a real number, in this case, if b=0, z is real a, so the range of complex numbers includes real numbers. ・・". In other words, if you do not think about the complex number even if the real part and the imaginary part are 0, the explanation of this video will not be possible.

[@angeldude101]

Every real number is a complex number with the imaginary part equal to 0. Every imaginary number is a complex number with the real pay equal to 0. Every integer is a rational number with the denominator equal to 1.

[@wiggles7976]

sqrt(x) is a function of x so it can only have one result, not two.

[@mathisnotforthefaintofheart]

Yes, and actually using the square root symbol is abuse of notation when dealing with complex numbers. A better way of putting the problem is "Solve the quadratic equation x^2=i" and that equation has....TWO (complex) solutions of the form a+bi.

[@fewwiggle]

In mathematics, a square root of a number x is a number y such that y2 = x So, by definition, you have two solutions

[@wiggles7976]

@@mathisnotforthefaintofheart How is it an abuse of notation? You can always just take the root with the smallest angle, even if you have the 5th root of i or any root of i, or any root of any complex number.

[@mathisnotforthefaintofheart]

@@wiggles7976 Then that needs to be clearly mentioned in the question. Very often it isn't.

[@pedroivog.s.6870]

That's why my teacher doesn't like square roots out of non negative real numbers: since it gives you two different values, it cannot be used as a function

[@barakathayek8204]

It's more simple if you use the exponential form

[@ilovetiananmen]

Correction: Imaginary numbers don't have a real part, but complex numbers have both. So it's not confusing anybody.

[@SQRTime]

Hi Alex. If interested in math competitions, please consider kzbin.info/www/bejne/gaiadJaCmKiIm5Y and other videos in the Olympiad playlist. Hope you enjoy 😊

[@sauhardyahaldar4416]

Please increase the volume... It's too low..

[@anshumanagrawal346]

What a Coincidence! You and Professor Penn posted at the same time

[@BriTheMathGuy]

🧐

[@samyakmarathe3434]

u can also prove it using the fact that e^ix = i, where x = pi/2

[@charliebowditch7975]

This is the same as rotating 1/sqrt2 + (1/sqrt 2)i by 45 degrees to get i, which is neat this means the cube root will be 30 degrees. The third root of i is cos(30) + sin(30)i = sqrt3/2 + (1/2)i. The rule is therefore nth root of i = cos(pi/2n) + sin(pi/2n)i

[@userhomer]

3:22 you aren't supposed to keep root in the denominator

[@_PsyChoPaThiCX.___.]

i always love your videos. could you plz make a dedicated video on phi or golden ratio.?? how it came? how to prove golden ratio? i would be very happy if you make a video about this. love from bangladesh

[@marcusscience23]

Imagine a unit circle centered on the origin of the complex plane. Start from 1 and rotate a certain angle to get to another number on that circle, for example -1, and to take its square root, take the point that you would get with half that angle. You can convert the numbers with angles by using sin for the imaginary part and cos for the real part. Edit: You can rotate halfway both directions because square roots can be positive and negative.

[@angeldude101]

The question is then which direction did you rotate. You can go the long way around the circle, or the short way (-1 had both paths be the same length) the path you took to get the destination determines whether the answer has a negative sign. Half of 360° is 180°, but 360° is the same as 0° and half of 0° is still 0°, and 0° is not the same as 180°.

[@marcusscience23]

@@angeldude101 It doesn’t matter because square roots can be positive or negative. sqrt(1) can be +1 or -1.

[@_notch]

An imaginary number is also known as a complex number? I was under the impression a complex number was a number with two parts; a real part and a complex part, usually written an x + yi.

[@CDeeMondays]

Blackpenredpen made the same proof

[@deidara_8598]

2:00 that is a massively overcomplicated way of solving it. Just say a^2 = b^2, since 1 = 2ab you know they have the same sign so you know a=b, thus you can just say that a=sqrt(0.5) and you've solved it. Sqrt(i) = sqrt(0.5) + i*sqrt(0.5)

[@serulu3490]

Ikr 1=2a² 1/2 = a² a = b= 1/√2

[@TheSmegPod]

is it possible/meaningful to take the i-th root of some real number

[@fetch7312]

en.m.wikipedia.org/wiki/Euler%27s_formula this might be helpful to you as long as you can recall logarithms and what the reciprocal of the imaginary unit is

[@mysticstelios]

how can you have something inside the root that is negative? It is not possible...

[@Anonymous4045]

Thats why its called _imaginary_ lol

[@meeptothemax375]

Very useful video, I had previously thought that sqrt(i)=-i.

[@Ostup_Burtik]

No, because (-i)²=-1 like i²=-1

[@meeptothemax375]

@@Ostup_Burtik Which I realized after watching the video.

[@TheNormMan]

Nicer to see with the "unit circle" (is this the right vocabulary here?) and multiplications of numbers on it as adding the angles

[@78anurag]

I'm kind of proud I did it myself

[@bendesch98]

Shouldn't the only correct answer be when a and b are both positive? I feel like the reason you get +/- is because you started by squaring both sides, which added another solution

[@kchromaticpiano]

You will realize it is not extraneous if you plug it into the original equation though!

[@simong1666]

The sqrt(-i) is 1-i and -1+i As can be seen there is no only positive or only negative rule to be followed

[@0average_enjoyer044]

already i’m confused. the square root of the square root of -1? isn’t that just the quartic root of -1?

[@simong1666]

It is, -1^(1/4) has 4 complex roots

[@佐藤広-c4p]

Given that, it's natural to be confused. In the unit circle of the complex plane, consider where i comes. The " i " is located in the direction of 12 o'clock on a hand of the length 1 of the analog clock. Since the direction at 3 o'clock is 0 degree of initial line, when the hand turns 90 degrees counterclockwise, it becomes i. Multiplication of complex numbers is multiplied by absolute values and added by arguments. Since squared is multiplying the same thing, to come at 12 o'clock by multiplying, it would be good if the argument is 45 degree, or there is the hand in the middle of 1 and 2 o'clock (In this case, the absolute value (radius vector) is 1, so 1 *1 = 1, you do not have to think). This is cos45°+isin45°=1/√2+i1/√2. (1/√2+i1/√2)^2=i, so it's √ i. √ i*√ i=i. There is another thing that comes at 12 o'clock, multiplying the same thing. The argument is 225 degree, or there is the hand in the middle of 7 o'clock and 8 o'clock. cos225°+isin225°=-1/√2-i1/√2. (-1/√2-i1/√2)^2=i, so it's √ i. 5/8 laps + 5/8 laps = 5/4 laps, even if you add 90 degrees more to 360 degrees per lap, it will be i at 12 o'clock.

[@arunkrishna2669]

Correct solution..... But "legends" may make errors this way...... 😁..... Note: Legends should always solve it by complex exponential way.......

[@Rene_Christensen]

This turns up in for example micro acoustics where the problem has both diffusion and propagation like properties.

[@еуууу-с6о]

Ничего не понятно, но очень интересно

[@АзатМингалеев-в1к]

Also, you can think about sqrt(i) in context of rotation: -1 is turn on 180 degree and if we multiple any number by i twice, it's means we rotate by 90 degree two times and get negative number with respect to original one. So, in that case, we should to find "rotation" of real number by 45 degree. Using knowledge about trigonometry we find out length of imaginary projection and real one the same. So, the answer will be (1/sqrt(2)) * (1 + i)

[@n0mad385]

I remember back in high school, we had to put the square root of I in the form a + bi. Still makes my brain hurt to this day

[@kailashbani5835]

What about this? i^1/2 = i^4/8 = (i^4)^1/8 We know i^4 is 1 So (1)^1/8 equals to 1.

[@theuserings]

(1)^1/8 has 8 solutions, one of them is one shown in the video

[@theuserings]

Its called the fundamental theorem of algebra, every polynomial equation of degree n with complex number coefficients has n roots, or solutions, in the complex numbers

[@佐藤広-c4p]

Please read my reply comment in badlydrawnjolteon-san's comment below.

[@Glitchnb]

Me : *watch an math video* Also me : *doesn't understand any single word of what the guy said*

[@ttxxxxxxxxxxxxxxt]

My solution: knowing that Log(z)= i arg(z) + ln(lzl) (complex log) we can substitute i for z and we get Log(i)=i pi/2 so e^i pi/2 = i now take the square root of both sides so sqrt(i)=sqrt(e^i pi/2) taking square root is the same as putting something to the 1/2 power so we get sqrt(i)=(e^i pi/2)^1/2 now we can multiply the powers (this rule does not always work in the world of complex numbers but it works most of the time) so sqrt(i)=e^i pi/4 now we can a) use euler's formula to evaluate the expresion so sqrt(i)=cos(pi/4) + i sin(pi/4) = sqrt(2)/2 + sqrt(2)/2 i or b) locate the points on the complex plane using the unit circle we know that the point with angle pi/4 and distance from the origin of one has coordinates sqrt(2)/2 , sqrt(2)/2 which on the complex plane will be sqrt(2)/2+sqrt(2)/2 i now we know that l sqrt(i) l = l e^ i pi/4 l wich is always one becouse rel(i pi/4) =0 so we need to multiply everything by one so we get sqrt(i)=sqrt(2)/2+sqrt(2)/2

[@wildfire_]

you could do this: i = sqrt(-1) sqrt(i)=sqrt(sqrt(-1)) sqrt(sqrt(-1)) = (-1^1/2)^1/2 sqrt(i)=-1^1/4 this is should be an extra complex number because there is no number that can be 4th powered to a result of -1. i^4 = (i^2)^2 = -1^2 = 1

[@manfredwitzany2233]

Sorry to correct you, but your first mentioned statement is wrong. i is not defined by i=V(-1) If i would be defined this way, you could calculate i² by the following way: i²=(V(-1))²=V(-1)*V(-1)=V(-1*-1)=V(1)=1 which obviousely is incorrect. Therefore i is defined by i²=-1

[@monse53]

Calculating it in polar coordinates is much simpler: i is 1 / 90 deg. Consequently sqrt(i) is 1 / 45 deg, or sqrt(0.5)+i sqrt(0.5) (Pytagoras)

[@benjaminkarazi968]

Hello, Although, ¡, √-1 is a fallacy and is an untruth value to be valid upon *agreements, √¡ = (√1÷2 ) + (√1÷2)*¡ and √-¡ = (√1÷2 ) - (√1÷2)*¡. What (number or digit, ±)² could probably possibly satisfy √-1? Math mimics physic; nevertheless, most of the time delivers humorous and untruthful consequences. Regards,

[@TheMemesofDestruction]

I see what you did there.

[@BriTheMathGuy]

😄

[@jerimyv3314]

Better insight is to use geometric polar coordinates and makes it easier to see why this works. i is a vector length 1 at 90 degrees. the square root is just same length 1 at (45 degrees) since square root of 1 for the length. Multiplying polar coordinates requires adding the angles. Therefore if I multiply vector 1 at 45 degrees by itself it results in 1 at 45+45 = 90 degrees. Therefore the square root of i is the vector 1 at 45 degrees. This vector in rectangular coordinates is now = Cos(45 deg) + iSin(45deg) = 1/sqr(2) + i/sqr(2) = 1/(sqr(2) (1+i). On the Cartesian plan it is a vector length 1 at 45 degrees. Square root of i. Likewise i^2 = 1(180 degrees) or a vector length 1 at 180 degrees since it is twice 90 degrees. You can take i to any power by just multiplying the exponent by the angle in polar coordinates and then convert to rectangular coordinates. The vector is always a length of 1. The math is easier.

[@martinstephens4633]

"one over two plus one over root two i, or MINUS ...." Please stop saying minus when you mean negative. If any child/pupil watches this video, they may think minus is appropriate language for a mathematician to use. I claim it is not.... Subtract or Take Away - the operation Negative - the directed number Minus - CONFUSION

[@davidlabovic268]

When I saw 1/sqrt(2). I was like it has to be racionalised, so it is sqrt(2)/2+(sqrt)2/2i. And instant remebered the eulers formula e^i0=cos0+i*sin0. So if theha is 45° this is answer. So is i^1/2 45°😂😂😂

[@jeffreypenkoff6178]

Since (a^2-b^2) = 0, that gives 2 cases: a=b or a = -b. Is there a reason that the a = -b case wasn't addressed? Is there some reason that I can't figure out why that case doesn't work or something? If that cse doesn't work, the video should explain why. If that case DOES in fact work, then your solution would be incomplete

[@g_vost]

dont reałly need to do much algebra imo. in the same way i is halfway rotated to -1, √i is halfway rotated to i. so in polar coordinates, its 1∠ π/4. but if you want it in normal coords or whatever, itd be (1+i)×√2/2 just going off the unit circle

[@darianleyer5777]

Also, by logical reasoning, if 2ab = 1, then ab = 1/2, and therefore, there are two valid options for the values of a and b. The only known factors of 2 are 1, the Square Root of 2, and 2 itself, as it is prime. Therefore, if it is given that a or b != Square Root of 2, then a = 1 and b = 1/2, or vice versa. Thus, the 2 and the 1/2 cancel out to 1, and 1^2 = 1.

[@nicolastorres147]

2:46 Irrationalizing the denominator

[@ericgarcia3013]

Technically the answer is wrong. The square root of a number is always the positive root only. However, if written down in an equation, the negative version of that root will also satisfy the validity condition. His final answer is +-(sqrt. of i) where „sqrt. of i“ is ONLY the term in brackets but not negative 1 times the term in brackets. Also imaginary numbers are a subset of the complex numbers. If you’re gonna do math videos, at least be precise.

[@bloxrocks5179]

I think an interesting method to this is realizing e^(i π/2) = i then taking the square roots of both sides e^(i π/4) = i^1/2 From there apply eulers formula.

[@andrewkarsten5268]

You miss a solution that way. If you start with e^(2nπi+iπ/2)=i, then take the square root, (to account for periodicity), then you’ll see you actually get two distinct solutions. The other is e^(5πi/4), which when you put into a+bi form gives -√2/2−i√2/2, which indeed gives i when squared.

[@Russienede]

√2i is superior Because since a+b= √(a²+2ab+b²) If a=1 and b=i It's 1+i= √(1+2i-1) So 1+i= √2i

[@Nanook128]

Would the square root of I also be the 4th root of negative 1?

[@khongocten161]

the way 1^2 = (-1)^2 = 1 made me think that there should be 4 answers to x^2 = i (*) since i is an imaginary number and that other 2 solutions is just 2/4 answer to (*)

[@ricardoguzman5014]

Here's a quick solution: Multiply by 2/2 inside square root. √i = √(i x 2/2) = √(2i) / √2 = (1 +i)/√2. Done.

[@billmorrigan386]

Folks, just remember that i = +√-1, otherwise things won't make sense. That's an important point 99% of average books neglect to mention. I think it is a malpractice--actually a gross malpractice to omit such fundamental details. Things that are obvious for pro mathematicians (they don't even remember to mention them) can be a dead end for normal people. *And no, it's not more convenient, it is just lazy not to bother with important details and let people do things according to templates and rote-learn stuff.*

[@willemkoning2142]

2:00 “Pick your favorite way of solving this”. Doesn’t it just include lesser steps to conclude from a^2-b^2=0 that a=+/-b?

[@JDCullison1]

All you're doing is changing it from i = sin(90°) to sqrt(i) = +/- sin(45°). Neat.

[@v44n7]

if i= -1^1/2 --> the square root of i it's -1^(1/2)(1/2) which equals to -1^(1/4) right?

[@juandavidrodriguezcastillo9190]

Cool , another way to do it is by the eulers identity, e^i(a)= cos(a)+i*sin(a), if we replace a=pi/2 then cos=0 and sin=1 so e^i(pi/2)=i and we can extract the square root (elevate to 1/2) and by properties of the exponents we have that square root(i)=e^i(pi/4) sorry for my english, i'm learning english and math with this Chanel and is cool

[@BriTheMathGuy]

Nice! Thanks and thanks for watching!

[@scrpr3961]

Question. You wrote that "sqrt(i) = a + bi". Shouldnt it be that "i = a + bi", so "sqrt(i) = sqrt(a+bi)"?

[@nrgkickz4158]

The belligerent bacon contradictorily slap because accordion similarly cheer notwithstanding a raspy wish. chemical, languid paint

[@misterlau5246]

NO i is just a pi/2 ROTATION you can't just mix it up, when in quantum we use i, it's PHASE

[@priyanshrawat442]

Euler's formula makes it easier

[@BriTheMathGuy]

True!

[@raianaratti9943]

1:58 Love the video but the fact that you chose the second equation here instead of just taking (b^2) to the other side thereby getting (a^2 = b^2) and thus (a = b) really bothered me

[@ramahawai7056]

Complex analysis is my favorite math branch.

[@BriTheMathGuy]

One of mine too!

[@magicmulder]

Fun fact: The German term is the misleading “Funktionentheorie”, “theory of functions”. To me it always was too rigid - every continuously differentiable function is infinitely continuously differentiable, where’s the fun in that? Look at all those real function that are differentiable but not even continuously differentiable.

[@PXO005]

@@magicmulder wait, don't get me wrong I'm just a high schooler but I thought only continuous functions can be differentiable throughout their domain?

[@Elias___]

I don't know how this guy successfully clickbaited math

[@HenrikhWolf]

0:13 imaginary is not complex Imaginary numbers are a subset of complex numbers

[@Pure_Imagination_728]

There’s an easier way to do this. I solved it in a few seconds in my head using polar/exponential form.