i^1=i, i^2=1, i^3=-i, i^4=1, i^5=i, ...

[@driptcg]

Every time he says an angle in degrees and then says "but we are adults now" and writes it in radians i lose my mind😂😂😂

[@greece8785]

Absolutely!!!!

[@aashsyed1277]

not really!!!!!!!!!!!!!!!!!!!!!!!!!! i mean you are wrong!!!!!!!!!!!!!!

[@Ayush-yj5qv]

Those were the times in when we wrote in degrees in high school radian is way more important

[@martinp3621]

@@Ayush-yj5qv o

[@TheWingAnthony]

As the beard grows, knowledge shall grow exponentially.

[@guillaume5313]

To be honest I think it really isn't pretty lmao

[@guillaume5313]

It's super thin it's weird

[@vukpayn8711]

i think it's cool

[@angelmendez-rivera351]

Exponentially? You're mistaken. It grows super-exponentially. It grows faster than exp[x·log(x)].

[@JSSTyger]

Wax on. Wax off. Up. Down.

[@blackpenredpen]

I didn't take log base 1 this time! : )

[@quercus_opuntia]

@@realDewSplatGum 🍪

[@alfrednik07]

Think the infinite many solutions are produced in the euler’s formula; because exp(ix) =cos (x)+isin(x) but is also equal to cos(x+2*pi*n)+i sin(x+2*pi*n) with n being an integer

[@coc235]

@@alfrednik07 but these values are the same

[@ascaniuspotterhead2484]

@@coc235 That is pretty much what we want. Infinitely many solutions should come from e^ix being equal to e^i(x+2pi).

[@angelmendez-rivera351]

@@coc235 See my main comment to the comments section

[@patrickchapman8583]

One could say "and that's a good place to stop."

[@blackpenredpen]

Prof Penn!!!

[@orenfivel6247]

@@blackpenredpen nice 8:30 you may write x^i as x^i=e^(i⋅ln(x))=e^(i⋅lnx)= e^(i⋅lnx-2πni) so you solve e^(i⋅lnx-2πni)=e^(ln2) ⇒…x=e^(2πn)⋅[cos(ln2)-i⋅sin(ln2)] , n∈ℤ therefore exsist infinite solution for Q2 ⬛ and then you should say "🆗 Gr8, and that's a good place to stop🛑."

[@aashsyed1277]

@@blackpenredpen no there is a person namely "good place to stop"

[@gary.h.turner]

@@blackpenredpen Prof blackPenn-redPenn?

[@SlipperyTeeth]

Question 2 has infinitely many answers. You got rid of most of them when you took the i-th root of both sides at the start - similar to how taking the square root would get rid of one of the answers. Whenever you take a root, if you want to preserve all of the answers, you have to put the number in polar form (2e^(2n*i*pi))

[@hassanniaz7583]

I avoided the i-th root and still got the same single solution. x^i=2 i ln x = ln 2 ln x = (ln 2)/ i ln x = -i ln 2 ln x = ln(2^-i) x = 2^-i You see this is the same as bprp got. From here you can use Euler's identity to solve for x and see that it yields only one solution. So I don't think that by taking i-th root we reduce the solutions. Edit: But you can write e^(-i ln 2) as e^(-i(ln2 + 2*n*pi)) yielding infinite solutions.

[@SlipperyTeeth]

@@hassanniaz7583 You also just took the i-th root

[@leagueofleogends1192]

Hey guys, can't I do x^i = e^(i lnx) = e^ln2 => i lnx = ln2 => lnx = ln2/i => e^lnx = e^(ln2/i) => x=ln(2)/i to get one solution?

[@leagueofleogends1192]

I just realise it's the same as -i ln(2) lol

[@stranger0152]

@@hassanniaz7583 But this is just e^-iln2 * 1^n.

[@giovannipelissero1886]

Can someone make a *"I don't want to be on the bottom"* compilation? Love this man.

[@fotnite_]

I notice for the second one that even though there aren't multiple answers, there are infinitely many expressions in the same form as the original that lead to the singular answer at the end, since we're doing Euler's formula forwards instead of backwards.

[@pawel413]

There are multiple solutions if you replace the 2 with 2e^(2πni) you end up with e^2πn (cos(ln(2))-i sin(ln(2))) (It's not a mistake the i in the exponent cancels out)

[@fotnite_]

@@pawel413 You could do this with any complex expression at any time, no matter what kind of problem you're doing, since e^2πni = 1. So I'm not sure how this is relevant.

[@micheledejana7849]

8:25 I think it could have multiple solutions: remember that 2^-i is like saying (2^-i)*1 and 1=e^(2*pi*i*n), so the correct answer to the problem might be [cos(ln2)-i*sin(ln2)]*e^(2*pi*i*n)

[@hassanniaz7583]

Genius!

[@hanzhoutang9235]

But they are still the same.

[@gamerdio2503]

It's the same numerical value though

[@AnonymousAAL]

I would like to se a 1 hour video maybe where u just factor terms with every method that exist...

[@JoseEduardo-rw2rh]

100 factorizations Time!!!

[@ProCoderIO]

My ability to do complex analysis has gone through the roof thanks to these vids.

[@tanelkagan]

Just for clarity, at 7:50 there's a slip of the tongue with "because sine is an even function". Sine is of course an odd function and I'm sure that's what was meant, based on what was written down.

[@ahmedamin1557]

at 5:04 we could write " 2 " in the following way " 2e^(2*n*pi*i) " to get all the solutions.

[@TrimutiusToo]

Problem is, they will loop on itself because sin and cos at the end have period of 2*pi so it cancels out

[@quantumsoul3495]

@@TrimutiusToo the sin and cos won't be on the 2π after the ^(-1). See my other comment

[@TrimutiusToo]

@@quantumsoul3495 why wouldn't they be? it will just change from positive to negative, but as n can be any integer sign doesn't matter.

[@quantumsoul3495]

@@TrimutiusToo if you take 2e^(i2πn) to the -i th power you get 2^(-i)*e^(2πn). The i and -i cancel out so you have no more i. If you have e^x without i, you don't have cos and sin.

[@TrimutiusToo]

@@quantumsoul3495you cannot split powers like that when exponentiating a complex number, because it doesn't work, these fancy operations only work for real numbers...

[@StevenHodder]

The second problem also has infinitely many solutions as well: exp(-j*ln(2)) could actually be exp(-j*(ln(2)+2*n*pi))

[@scirium7199]

That still leads to a single answer though, as when you use Euler's Formula and get cos (ln(2) + 2πn) - i * sin (ln(2) + 2πn), both trig functions will still output the same answer regardless of what n is equal to sin(π) = sin(3π) = sin(5π) = 0 There are infinitely many ways to express the answer, but it is still just one (unless I'm missing something else).

[@Misteribel]

@@scirium7199 which means there are infinitely many answers. That’s the thing with the polar form (and the cos/sin equivalent), they’re cyclic over 2π.

[@sanyalox01]

saying "it still leads to the same answer" is not a good idea here, as multiple equations exist, where there are multiple solutions and yet still one answer, such as x²=1 (x=1; x=-1). The general answer to the second question in the video would be x=(cos(ln2)-i*sin(ln2))*e^(2πk), where k is an integer, which is same as e^(2πk-iln2). Put that to the original equation: e^((2πk-iln2)*i)=e^(ln2+2πki)=2. It's the same result, but the solutions are infinite, and the fact that the result is the same doesn't exclude the other solutions.

[@antonyqueen6512]

Second equation has indeed multiple solutions: x = [e(2*n*Pi)] * [cos(ln(2)) - i*sin(ln(2))] and could have been solved using log on both sides: x^i=2 To consider all solutions we have: x^i=2*e(i*2*n*Pi), n € Z Now we can build log on both sides: ln(x^i)=ln(2) + ln[e(i*2*n*Pi) i*ln(x)=ln(2) + i*2*n*Pi Multiply by -i on both sides: ln(x)= -i*ln(2) + 2*n*Pi x=e(2*n*Pi - i*ln(2)) x = [e(2*n*Pi)] * [cos(ln(2)) - i*sin(ln(2))] Multiple solutions, n € Z e(2*n*Pi) is the i-root of 1, i.e. [e(2*n*Pi)]^i=1

[@jimschneider799]

At about 4:40, instead of multiplying the numerator by -1, one can multiply the denominator by -1, which, upon realizing that n can be any integer, gives a denominator of (4 n - 1) pi, so that x = 2 i ln(2)/((4 n - 1) pi)

[@RexxSchneider]

7:47 "because sine is an even function" - I don't think so. 8:13 "this one has only one answer" - maybe try cos(ln2+2n.pi) - i.sin(ln2+2n.pi)?

[@alexandrucalitescu5822]

Yes, but is unnecessary because sin and cos are periodic. It's the same solution, for every n

[@RexxSchneider]

@@alexandrucalitescu5822 It's just as "unnecessary" as the solution to the first part. Complex log is just as periodic as cos and sin, and for exactly the same reason.

[@KasabianFan44]

@@RexxSchneider Except that in the first problem the n is *outside* of the logarithm, which means that every integer n will give you a DIFFERENT solution. In the second problem, adding a bunch of n-s inside trigonometric functions would just give you one distinct solution.

[@RexxSchneider]

@@KasabianFan44 Indeed that's true for the most part; but degenerate solutions are still distinct solutions. Cos(x+2n.pi) is *not* the same thing as cos(x), even if it has the same value.

[@KasabianFan44]

@@RexxSchneider I don’t think you know what the word “distinct” means mate. Two solutions (say, a and b) to an equation (f(a) = f(b)) are said to be distinct if a ≠ b. But if a := cos(x) and b := cos(x+2π), then we get that a = b, which means they are NOT distinct.

[@ThePiMan0903]

Nice video blackpenredpen !

[@Kumar-oe9jm]

7:50 i think u meant sine is an odd function not an even function

[@victorpaesplinio2865]

Wolfram Alpha gives a interesting solution for x^i = 2. If you take the ln in both sides you can imagine taking ln(1 * 2) and 1 = e^(2 * i * pi * n). You'll get: i * ln(x) = 2 * pi * n + ln(2) (divide i on both sides) ln(x) = 2 * pi * n - i * ln(2) WA assumed the principal logarithm with n ∈ Z. Then the final answer was 2^(-i) * e^(2 * pi * n) with n ∈ Z For me it makes sense, but I think that it is some kind of redundancy.

[@quantumsoul3495]

I have the same result. However 2^(-1) still has to be calculated

[@pawel413]

The e^2*pi*n gives the equation infinite solutions because the exponent is only equal to 1 when n = 0

[@spudhead169]

You guys should read the comments before commenting. We only need one to explain how question 2 has infinite answers and literally half the comments are just saying the same damn thing.

[@antonyqueen6512]

You are right, people (Including myself) jumped directly formulating the answer before reading the comments.😁. Should pay attention next time!

[@Amoeby]

For you guys. We have x^i=2. So what is 2 in terms of complex numbers? Yes, it's 2*e^(i*2pi*n). Then we can raise our equation to the power (-i) and it'll give it to us x=2^(-i)*e^(2pi*n). Lastly, we can simplify it to x=e^(-i*ln(2)+2pi*n). And, of course, n is an integer.

[@gamerdio2503]

The +2πn is unnecessary, because it's the same value for all n. It's like how 1 = e^(2nπi). Sure, 1, e^2πi, e^4πi, etc all look different, but they're just different ways to write the same number. You still end up with only one solution

[@Amoeby]

@@gamerdio2503 nope. There is a different situation. You may see it more clearly if you put different integer n in the first form of the answer x = 2^(-i)*e^(2πn). So basically we have here complex number 2^(-i) multiplied by real number e^(2πn). I guess now you see why +2πn is a necessity.

[@gamerdio2503]

@@Amoeby Sure, but e^(2nπ) is just 1, so it's the same numerical value

[@Amoeby]

@@gamerdio2503 you should pay more attention because e^(2πn) is a real number which I mentioned before and it is equal to 1 only if n = 0. In other cases it is not because there is no i in the power.

[@gamerdio2503]

@@Amoeby SHIT you're right I thought there was an i

[@samarendra109]

ln(2) also can be written as 0.693.. + 2nπi , as 2 = e^(0.693... +2nπi) Once you introduce complex numbers log function no longer remains a 1 to 1 fn.

[@marielxenovia3062]

Inside cos & sin we can also add +2πn because they are periodic

[@jgsh8062]

There's a better way to do the number 1. Since e^i*pi = -1 = i^2, i = e^(i*pi / 2). therefore ln(i) = i*pi / 2. We can therefore say x = (2*ln(2))/(i*pi).

[@factsheet4930]

Both have infinitely many solutions, because for every natural n, e^(2*pi*i*n) = 1 so if 2^(-i) is a solution, so is 2^(-i)*e^(2*pi*n), because you raise to the i-th power!

[@angelmendez-rivera351]

That isn't at all a different solution, because e^(2nπi) = 1, so you get the same number. In order for the solutions to be different, you need to multiply by something that changes the result. For example, if sqrt(x) solves y^2 = x, I can get a different solution by multiplying by -1. Not multiplying by 1.

[@factsheet4930]

@@angelmendez-rivera351 No, the solutions are 2^(-i)*e^(2*pi*n) without the i, you then multiply by i because the equation is x^i

[@dimwittedhuman7798]

This guy bring back my love of maths in school!! I regret not doing a job in field of mathematics😭😭😭😭😭😭😭😭

[@antoniusnies-komponistpian2172]

I'm studying maths. Some months ago, I watched blackpenredpen to distract myself with something harder, today, I'm watching blackpenredpen to distract myself with something easier lol.

[@refrainrestrainresist-3rs49]

Hey bprp! I have a procedure for the first problem, I'm getting x-> infinity : i ^ x = 2 i = 2 ^ (1/x) i ^ 4 = 2 ^ (4/x) 1 = 2 ^ (4/x) ............... (i ^ 4 = 1) 2 ^ 0 = 2 ^ (4/x) 0 = 4/x ................. (if bases are same, powers can be compared) x -> infinity

[@LearnPlaySolve]

You have awesome videos. Great problems, and your explanations are perfect. Keep up the good work!

[@6754bettkitty]

now, which one is bigger O(i^n) or O(n^i)? (Or does complexity not apply for complex numbers)

[@quantumsoul3495]

complex numbers exponents can either diverge or end in a loop iirc. You should check mandelbrot set.

[@DontYouDareToCallMePolisz]

"i don't like to be on the bottom, i like to on the top" This should be a meme

[@greece8785]

I like how your beard are growing more and more at every video!!! 😊😊😊

[@geosalatast5715]

Βλέπει και η Ελλάδα blackpenredpen!

[@greece8785]

@@geosalatast5715 Ναι... Ο,τι καλύτερο... Σε ταξιδεύει σε έναν άλλο κόσμο!!!

[@tzonic8655]

Ελα Ελλάδαρα!!!

[@Marcox385]

Hey man are you greek?

[@greece8785]

@@Marcox385 just look at my name and my photo

[@nathanisbored]

i would guess you should expect a root to have as many answers as the magnitude of the index. so a square root has 2, a cube root has 3, etc. this would also mean that a 1st root only has 1, a (-1)st root only has 1, and similarly an i-th root only has 1, since its magnitude is 1. dont know if that's true, but it makes intuitive sense anyway.

[@quantumsoul3495]

if in x^n, n is a integer, the plot of the n solutions draw a regular polygon in the complex plane

[@nathanisbored]

@@quantumsoul3495 yeah but im talking about when n isnt an integer

[@quantumsoul3495]

@@nathanisbored Yes I know, I just wanted to say that. I wonder if it works with non integers too

[@nathanisbored]

@@quantumsoul3495 well i dont know what a 1/2-gon is or an i-gon for that matter... interesting to think about tho

[@quantumsoul3495]

@@nathanisbored It would be a 1-gon not i-gon, because the magnitude is a real number. For fractional-sided polygon, it would be awesome if it matched those: chalkdustmagazine.com/blog/fractional-polygons/

[@diedenarios8688]

on the second problem, although i am not sure I belive that you cant just cancle the i-th power with the i-th root, because that would rely on a rule of exponents which doesnt necessarilly hold true for complex exponents. I could also be wrong.

[@derwolf7810]

8:18: "I thought this one has more than one answer, because we have the complex exponent. But... i don't see a place that you can actually produce more answers" If i remember right, then you have used the real logarithm (in the step 2^-i = (e^ln(2))^i), but within complex exponentiation, you should have used the complex logarithm. Let be LN(x) the complex logarithm and ln(x) the real logarithm. x = 2^-i = (e^LN(2))^-i = e^(LN(2)*-i) = e^((ln(|2|)+i*arg(2))*-i) = e^((ln(2)+i*2*PI*n)*-i) = e^(-i*ln(2)+2*PI*n) = e^(-i*ln(2)) * e^(2*PI*n) = (cos(ln(2))-i*sin(ln(2))) * e^(2*PI*n)

[@ameliablight6888]

Your beard is so majestic. Never get rid of it! 8D

[@filipposchiabel]

Could we go even deeper in the second one? I thought we could elevate to 2 and to 1/2, to obtain the square root of -cos(ln2)sin(ln2). At this point we can easily simplify it in square root of -sin(2ln2). Our final resolution now is i√sin(2ln2)

[@andreimiga8101]

You can produce more answers if you use complex logs on real inputs. Write ln(2) as ln(2+0i)=ln(2*e^i(0+2npi))=ln2+2i*n*pi. The final ln2 is a real log, not a complex log, so you cannot apply this definition again.

[@gamerdio2503]

But the final answer is the same for all n. You didn't get more answers, you just got different ways to write the same answer

[@andreimiga8101]

@@gamerdio2503 Really? So you think that ln2+2i*pi is the same as ln2+8000i*pi ?

[@gamerdio2503]

@@andreimiga8101 Sorry, I misread your comment. I thought you were thinking of doing e^-i(ln2 + 2npi), since the ln2 is just an angle. However, doing this just gives the same value for all n. My bad

[@aminaazizi9877]

Very cool and interesting Video. Hope you'll make more Videos like that.

[@bobitza490]

he is starting to look like a real sensei

[@GOLDman4856]

I dont like to be on the bottom I like to be on the top. I will never not laugh

[@aashsyed1277]

Are you really jesus christ????

[@iansamir18]

1. We have x = ln 2 / ln i = -2i/pi ln2. 2. We have x = 2^(-i) = e^(-i ln2) = cos(ln2) - i sin(ln2)

[@pianistmanoj5780]

Why you have the pikeman ball in your hand buddy? Just asking!

[@Sam_on_YouTube]

I thnk the second one ends up having only one answer because the multiple answers remain inside the repeating sin and cos functions, and so only one result comes out. You could add in infinitely many solutions in Euler's function, but they all lead to the same answer in the end.

[@angelmendez-rivera351]

Well, your reasoning regarding periodicity is correct, but it does not prove there is only solution. In fact, the equation has infinitely many solutions, as some of us have already proven in the comments.

[@Sam_on_YouTube]

@@angelmendez-rivera351 Thanks. I'll take a look.

[@jmiesterlifeiscool4047]

There is infinite answers for question 2. When you converted 2 to e^ln2, you could have instead converted it into 2 × 1 = e^(ln2) × e^(2pi × n i)

[@burk314]

That is the same as replacing the cos(ln 2) with cos(ln2 +2n pi) and the same for sin, except those don't actually give different values, so you just end up with the same solution.

[@quantumsoul3495]

@@burk314 No

[@yoav613]

there are infinite answers for both q1 and q2.in q1 the answer is the answer in the video plus any even number because i power any even number is 1.in q2 the answer is the answer in the video multiple with exp(2pik) because exp(2pik) power i is 1

[@yoav613]

sorry in q1 the answer is the answer in the video plus any number that divisible by 4

[@awesome_757]

Did anybody else realize the title is wrong? (i^2 = 1)

[@alkamishra9974]

Imaginary log any one? We can use li for it. For eg: li(x) = ln(x)/ln(i).

[@volodymyrgandzhuk361]

Is it practical?

[@alkamishra9974]

@@volodymyrgandzhuk361 No. But it's nice.

[@volodymyrgandzhuk361]

ln(i)=pi/2 i so li(x)=-2 ln (x)/pi i

[@saranyadas5522]

Really loved it,it was amazing,can you please make a video on Wilson's theorem and other number theory

[@shivansh668]

His interest is in integrals 😅

[@saranyadas5522]

@@shivansh668 yeah, I understood 😀

[@babajani3569]

Hi love ur vids, I had a request. Could you plz make some videos on the STEP Exam. It is the Cambridge entrance exam for High School students and the questions in it are absolutely brutal. There are 3 papers in total and they increase in difficulty from STEP 1 being the easiest and STEP 3 is the hardest. I think you will really enjoy some of the questions since in my opinion, having looked at both the JEE advance maths questions and STEP 3, I would say that STEP 3 is actually even harder than Jee so plz give it a go. Thanks

[@matengelo3812]

actually, question 2 has more than one answer, i found another one by solving for x: x = e^(ln2/i), it also works, and surely there are more

[@pconnell89]

cos(ln(2)+2πk)-i*sin(ln(2)+2πk) for any integer k could be infinite answers for the 2nd problem

[@gamerdio2503]

It's the same numerical answer for all k.

[@BlissNett]

Please did you make any video on Solution of Algebraic and Transcendental Equation? If so, please how do I find it?

[@aashsyed1277]

What is a transcendental equation?? Please tell me then I will answer you

[@aashsyed1277]

Are you from Pakistan BLISS NET????

[@calcuquack1206]

I swear blackpenredpen can reach mathematical immortality if he grows the beard more

[@dadoo6912]

ln(i) has infinite values

[@emanuelvendramini2045]

about x^i, maybe there is only one sol. because u can multiply both exponents by negative i and get a first degree polynomial. maybe, the number of solutions is related to the absolute value of the exponent at x (for integer values).

[@seroujghazarian6343]

Nope. There are (countably) infinitely many solutions.

[@emanuelvendramini2045]

@@seroujghazarian6343 how? for me, xî=2 has only one solution.

[@seroujghazarian6343]

@@emanuelvendramini2045all numbers of the form (cos(ln(2))-i×sin(ln(2)))×e^(2×n×pi) with n in /Z are solutions

[@emanuelvendramini2045]

@@seroujghazarian6343 holy crap, you're right!

[@jamesstrickland833]

In the second case,, Can't you always add 2pi to the theta to get inf solutions

[@mohamedibrahim1023]

Wow log base i

[@dwaipayandattaroy9801]

last day I did some bodmas and it exponent, this one i= 1+ 2^0 is 2 , 0^1+2 is 2

[@triacylglycerollipase5210]

I didn't know this man have a bear.

[@anishadj6157]

a video about de Rham's theorem and manifolds ?

[@Z_F-P_22]

surely any equation has solutions equal to the highest power, so this would have “i” solution, or because it’s just 1i does that mean it has 1 complex solution and if that’s the case would x^2i have 2 complex solutions- and furthermore how does this account for integer powers that have complex solutions!

[@deatheagleep8759]

I want to ask something: how can I calculate e!

[@MathElite]

I don't know how to calculate it but 2

[@deatheagleep8759]

@@MathElite i was thinking using π(x) or γ(x).... lol. I don't know how tough

[@MathElite]

I looked it up and there is really no way to calculate it by hand However, n! ~ n(n-1) as n->2 And, n! ~ n(n-1)(n-2) as n->3 So...... e! ~ 0.5[e(e-1)(e-2) + e(e-1)] ~ 4 On WolframAlpha: e! ~ 4.26

[@McNether]

If you mean "how do i calculate e?" the answer ist easy: e:= 1 + 1/2 + 1/3! + 1/4! + 1/5! + 1/6! +... Where 3!=3•2•1 and 4!= 4•3•2•1 . The new terms are getting smaller very fast. So you can approximate e as good as you want. If you really want to know what e! is, the answer is not that easy. The short answer would be: e! is not defined. The longer answer: use the Gamma-function and Gamma(n+1)=n! to Interpret e! as Gamma(e+1). For that you have to approximate an Integral... Maybe the intgral can even be solved easily but i am not sure about that.

[@deatheagleep8759]

@@McNether No, I mean e!. The Π(e), (integral of t^e×e^-t dt) is confusing me... lol

[@10baseballdmo]

Can part 2 just be generalized by saying: let 2 instead be n, which is any member of the integer set? Then at the end we just get cos(-ln(n)) +isin(-ln(n))?

[@thedigitaluniversity7428]

I don't know if you are aware of this but KZbin has polluted your carefully crafted videos. While one is following along and paying close attention **POW** something totally unrelated to the subject matter pops up on the screen and completely shatters the train of thought. I mean if KZbin will not respect your video work product how on Earth do you expect anyone else to????

[@SidneiMV]

So cool bro! Thanks a lot! x to the power of i.. Wow!

[@tabris1135]

my first instinct tells me that i^x forms the unit circle. Let's see if that is right or wrong

[@LilyKazami]

For real x greater than zero, both i^x and x^i go around the unit circle, the difference is one goes at a constant rate and the other at a logarithmic rate

[@tabris1135]

@@LilyKazami that's really cool to know. Thank you

[@angelmendez-rivera351]

You can also write i^x = exp[x·log(i)] = exp[x·i·(π/2 + 2·n·π)] = 2, and since 2 = exp[ln(2) + 2·m·π·i], this implies (π/2 + 2·n·π)·i·x = ln(2) + 2·m·π·i. This implies x = 2·m/(1/2 + 2·n) - i·ln(2)/(π/2 + 2·n·π) = 4·m/(4·n + 1) - i/π·2·ln(2)/(4·n + 1), for arbitrary integers n & m. Meanwhile, x^i = exp[i·log(x)] = exp[i·Log(x) - 2·n·π] = 2 = exp[ln(2) + 2·m·π·i], so i·Log(x) - 2·n·π = ln(2) + 2·m·π·i, implying i·Log(x) = ln(2) + 2·n·π + 2·m·π·i, implying Log(x) = 2·m·π - i·ln(2) - 2·n·π·i, implying x = exp[2·m·π - i·ln(2)]·exp(-2·n·π·i) = exp(2·m·π)·exp[-i·ln(2)] = [exp(π)]^(2·m)·(cos[ln(2)] - i·sin[ln(2)]), where m is any arbitrary integer. Both cases result in an infinite number of solutions.

[@ruathak1106]

How about x^i = i^x?

[@egilsandnes9637]

Well, at least i is an obvious solution.

[@quantumsoul3495]

x^i = i^x i lnx = x lni lnx/x = lni / i x^(-1) lnx = -i lni e^ln(x^-1) lnx = -i lni e^(-lnx) lnx = -i ln i -lnx e^(-lnx) = i ln i - lnx = W(i ln i) x = e^(-W(i ln i)) x = e^(-W(i (i(π/2+2πn)))) x = e^(-W(-π/2 - 2πn))

[@t7H2si0vß2]

The audio is really quiet on this video, I can barely hear it on my phone. Look into audio normalization.

[@teoviro]

If I do the second problem in this way: x^i = (e^lnx)^i = e^(i lnx) = cos(lnx) + i sin(lnx) = 2, then sin(lnx) = 0 and cos(lnx) =2, how is this possible?

[@Grundini91]

In the second equation couldn't θ= ln2 +2npi? This would give infinite answers, right?

[@felislapuz4637]

xin(i)=in2 x=in2/in(i)=in2/in1+i(π/2)

[@oenrn]

4:58 "Do we need to use logarithm? No." 6:14 uses logarithm...

[@eng-5011]

i think solution of x^i=2 is cos(log 2 +2pi n)-i sin(log 2 +2pi n) n is integer

[@shimotakanaki]

cos(log 2 +2pi n) is equal to cos(log 2). You can add how many 2pi you want, the value will be the same

[@eng-5011]

@@shimotakanaki i know i want to say this equation has an infinity of solutions for example, solve cos x =1 the answer is x=2pi n (because there is not a domain of x)

[@maitland1007]

I have a question.. you are doing all these great examples to show that by including complex numbers, we can get solutions to things that have 'no solution' in the real world. Does including complex numbers mean everything will have a solution? (except i guess dividing by zero.. or can we work around that too?) can you talk about that?

[@angelmendez-rivera351]

No, complex numbers can provide solutions to many equations that have no solutions, but not every equation can be solved with complex numbers. Trivial examples are 0·x = 1 & x = x + 1. Also, the equation exp(x) = 0 has no complex solutions, and thus the equation 0^x = 2, which was on a pinned comment in the previous video on this subject, also has no solutions. To be specific, the equations that have complex solutions that had no real solutions are those that involve polynomials and some transformation of the exponential function, along with products and sums thereof. This is because polynomials and the exponential function are known as examples of what mathematicians call *entire functions.* Entire functions are functions C -> C which are differentiable everywhere, a.k.a are holomorphic everywhere. There exists a theorem in complex analysis, called Picard's little theorem, that says every entire function that is not a constant is a surjection to either C or C\{s}, where s is a single complex number. Therefore, if I have an entire function z |-> f(z), then the equation f(z) = c for arbitrary c, unless c = s, will always have at least one complex solution. Furthermore, if such an s exists such that f(z) = s has no solutions, then for every complex c not equal to s, f(z) = c *must* have infinitely many solutions. This is so powerful, because products and compositions of entire functions yield entire functions, products of entire functions yield entire functions, as do sums of entire functions. The conjugate of an entire function is entire, and even more surprisingly, derivatives and integrals of entire functions are themselves entire. So combining entire functions in nice ways gives entire functions. As such, equations with only entire functions *almost always* have complex solutions. All of these exponentiation problems have no real solutions, but they have complex solutions, because exp(•) is an entire function. Polynomials are entire functions, and unless they are constant, they always have complex solutions. However, exp(z) is a surjection to C/{0}, not to C, so exp(z) = 0 has no solutions. In other words, exp(z) = c has infinitely many solutions *except* for c = 0. Then it has none. If you have an equation with non-entire functions or constant functions, in it, then this can constitute a problem when solving equations, even if complex numbers are involved. For example, |z| = -1 has no complex solutions, but this is because z |-> |z| is *not* an entire function. It is defined for every complex number, but this function is not complex differentiable everywhere. As it happens, |z| > 0 or |z| = 0 is a restriction that this function obeys on the complex numbers. Similarly, 0·z = 1 has no complex solutions, but this is because 0·z is in reality a constant function. It is a constant function, because the multiplicative inverse of 0 does not exist in the *field* of complex numbers. The equation z = z + 1 has no complex solutions, but this is because this too secretly involves constant functions, which is evident if you simplify the equation to 0 = 1. So in short, not every equation is solvable in the field of complex numbers. *Almost all* equations that use only entire functions are solvable. However, *almost no* complex functions are entire, unfortunately. It is very difficult for a function to be complex differentiable.

[@angelmendez-rivera351]

Now, regarding division by 0... You totally can divide by 0, as long as you work in structure called *wheel,* which is a generalization of a *ring,* which is what we usually work with. The set of integers can form a *ring.* The set of rational numbers can form a *ring.* The set of real numbers can form a *ring.* The set of complex numbers forms a *ring.* The set of function f : R -> R forms a *ring.* The set of algebraic numbers forms a *ring.* The set of computable numbers forms a *ring.* The ring of periods form a *ring,* as the name suggests. The sets we work with in modular arithmetic form *rings* too! The p-adic numbers form a *ring.* Rings are a very nice mathematical structure, and it is the preferred mathematical structure of mathematical studies, because the operations in those structures just behave so nicely and it makes them very practical, yet also very interesting for their own sake. Unfortunately, rings have the property that 0·x = 0 for every element x of the ring you work with, so division by 0 is undefined. What is the catch, then? The catch is that it has been proven that every *ring* (well, technically, every commutative ring, I think) can be suitably, rigorously, and naturally extended to a more general structure called a *wheel.* You can read more about this if you search online "wheel theory." Wheels use a suitable set of axioms which are generalizations of the axioms of a ring, and they append to the ring (R, +, ·, 0, 1) an involution / and the elements /0 and 0·/0, which can be abbreviated as simply 0/0. This forms the structure (W, 0, 1, +, ·, /) via a canonical construction from R. This constuction is called the *wheel of fractions* of R, which is analogous to idea of constructing a field called *the field of fractions* of R. In this regard, wheels are generalizations of commutative rings, but also generalizations of fields of fractions of a commutative ring. Funnily enough, while division by 0 is well defined in a wheel, the equation 0·x = 1 still has no solutions in such a wheel. In fact, I do not believe 0·x = 1 can be solvable in any type of structure outside the zero ring. But yes, division by 0 is totally allowed if you work in a wheel. So in a regard, this is not actually "an unsolved problem" any longer. The issue is that wheel theory is relatively new area of study of mathematics, and as such, not a well known one.

[@rorydaulton6858]

If you actually substitute some integer for n other than zero you will see that your answer for i^x=2 is wrong. Instead of adding 2n*pi*i to the logarithm of i in the denominator you should add that to the logarithm of 2 in the numerator. You end up with the answer x = 4n - 2i * ln(2) / pi. This works for all n, unlike your answer. I'm not exactly sure *why* your method is wrong.

[@pinklady7184]

Cool! I love to see you do more challenging logarithms in your next video. Brutally difficult logarithms, pretty please. 🙏🙏🙏🙏

[@sergeykaverin7460]

In the second prob you smth strange. There are 2 different things: 1) e in a power of cmplx z AND 2) exp() function If you mean exp() function, you can use Euler (or Taylor series exp(z)=1+z+...). The correct answer is: x = e^(-i*ln2) = e^(-i*ln2*(1+2*pi*i*k)) = e^(-2*pi*k - i*ln2) Only after that you can use Euler: x = e^(-2*pi*k) * (cos(ln2) - sin(ln2))

[@medotedo8410]

I do not get it. When I try to plug in the value of x I do not get the answer as 2 why is that ?

[@engomarmaysara5475]

Hi 😊😊 I love your explain and I felt comfortable feeling😍😍 And I have a small qu :_ How I can use the cadan way to solve the equation in the third degree 😊😊 Can you make the veido .. and thanks😊😊

[@particleonazock2246]

bprp = BlackPenRedPen = Steve Chow = Laoshi Chow = Absolute Legend

[@astherphoenix9648]

so the anwser for question 2 is in the form e^(i x argument). already, can you not add multiples of 2pi indefinitely to produce an infinite number of anwsers.

[@gamerdio2503]

Unlike the first question, adding multiples of 2π to the second equation doesn't produce any different numbers. The numerical value is the exact same

[@nanamacapagal8342]

What about the i-th root of I?

[@azzanporter4377]

You can also do the natural log for the first one I think it was much easier and better

[@shivansh668]

I appreciate your efforts 😊

[@jperez7893]

how do you produce a table of logarithms from scratch?

[@szehwong5304]

how about i^x=x^i ?

[@akshinbarathi8914]

is it possible to solve sing takilg log?

[@AndDiracisHisProphet]

aren't the other solutions just exp(-ln(2)+2n*pi)?

[@aliasgharheidaritabar9128]

Thank you for your amazing videos. Please make more videos.

[@josephmulligan3247]

You know that your explanations are amazing if a 15 yr old can understand this! Thanks for making math fun!

[@blackpenredpen]

Happy to help!

[@Sesquipedalia]

@@blackpenredpen same im 15 and u got me loving math.i love especially the sin(z) = 2

[@math_the_why_behind]

Excited to watch this video!

[@rijjomagno8328]

Sir answer this question what is 20% of 20% of 20?

[@NikhilS-1]

Can you integerate 1.dx/x^5+1?? 😕😕😕

[@alexpotts6520]

I think the second once has infinite answers too, because that ln2 is really a ln(2*e^2*i*pi*n).

[@freddyfozzyfilms2688]

if you look at 2 in polar form, wouldnt you get infinite solutions for ln(2)?

[@gamerdio2503]

Yes, but once you do e^-i*that value, you end up with the same numerical value at the end.

[@d.farfan98]

I propose to you to solve the integral sec(x)

[@mediaevalmonster]

Why the heck he has a pokemon granade in his hand?

[@serchsource8744]

Microphone...he thought it might be cool!

[@shaunmodipane1]

Can't you just add zero ( in this case 2πn) in the circular function (cos(In(2) +2πn) -i•sin(In(2)+2πn))

[@happypiano4810]

Umm... X^i=2 Square both sides X^i^2 = 2^2 Simplify X^-1 = 4 1/x = 4 4x = 1 x = 1/4 Check: (1/4)^i=2 Square both sides (1/4)^-1=4 4=4