This Just Can't Be Real 

Hello!

i^i^... i times

really simple one, 1/i = -i

Zeta(i) lol (idk how dis works but yeah)

i-th root of i

Yeah, it’s pretty hard to…imagine. 😎

@don'tfeel I’m curious, and as this question shows I am not very knowledgeable: how is the fact that i^i is equal to a real number related to hyperbolic geometry?

@don'tfeel Interesting; thanks for letting me know!

@@PunmasterSTP Joke aside, its actually not hard to imagine at all. 4^(1/3) both of these numbers are rational the output is irrational but it's fine because both rational and irrational numbers are Real similarly, both imaginary and real numbers are Complex!

@@wraithlordkoto Oh I quite agree; I just try to sneak in puns where I can 😎 I also think it’s interesting how one irrational number can be raised to the power of another irrational number, and end up being an integer!

Kinda easy ngl

@@mcsyllesen5183 do it then.

@@shaunthedcoaddict1656 look, every complex number z = x + iy can be represented as it's magnitude |z|(which is a real value), times it's "rotation" which is expressed as e^{i*theta}, where theta is the angle the line from 0,0 to the point z in the argand plane, or tan theta = y/x i = 1 * e^{i* pi/2} so when you do i^i it is 1^i * e^{i* pi/2 * i} = e^{-1 * pi/2} which is a real quantity

@@Linzz_1213 not necessarily. it is VERY EASY. to the point that you CAN understand it. maybe you don't have the intuition when it comes to complex numbwrs, which is okay, you could always learn. It seems like you're locking the concept behind some arbitrary wall you call "difficult math". No hate to anyone.

same

There’s even a proof that does not rely on Euler’s formula. Just take conjugate of i^i and manipulate to get i^i again. You could also first take the log before taking conjugate. Either way, only real numbers are unaffected by the complex conjugate.

That's 2 lines

Multiplying the exponents may or may not apply to complex numbers

I would ask your complex math teacher if this answer works for him.. If it does I would be concerned, since i is a ninety degree phase shift and a 90 degree phase shift of a ninety degree phase shift is meaningless in the complex plane. This should be solved in hypercomplex space and the question needs to have more defined boundary conditions or flakey math can be used to arrive at meaningless however clever answers.

@johnnychinstrap what a bitchy, unnecessary answer 🙄

Math without numbers>math with numbers

E is a constant

@@lakshay-musicalscientist2144 i is also a number, just an imaginary one

I know that but i assumed that the guy meant a numerical coefficient of an expression

@@dpage446 yeah except in this case we had numbers that were so cherished among mathematicians that they specially decided to dress them up with fancy alphabetical outfits i, e, π

So then raising a not purely real complex no. to an imaginary power doesn't make sense in general, right?

@@anshumanagrawal346 It makes sense only in a limited context. To be frank, even just raising real numbers to real numbers is already problematic. For example, a^b is treated as exp[b·ln(a)], but this is inconsistent with letting (-1)^2 = 1. However, there are many things that work with the logarithm for real numbers that make it a useful concept that do not translate for complex numbers. For one, the logarithm is a very important function in asymptotic analysis, and also very important in analytic number theory. Also, the natural logarithm can serve as an antiderivative of the reciprocal function. But for the complex numbers, the logarithm is not an antiderivative of the reciprocal function: the reciprocal function has no antiderivatives. And while the logarithm does not have the real numbers as the domain, only a proper subset instead, it at least is well-defined on that domain. In this sense, extending the logarithm to the complex numbers, rather fixing problems, only creates more. ln(0) is still undefined, but now ln is not even well-defined otherwise. Complex analysis has more power and theoretical elegance than real analysis, but it is not as though no sacrifices are being made when we work with complex numbers. I consider the concept of the logarithm to be one such sacrifice. In fact, the idea of restricting functions to some subset of the complex numbers for injectivity is less natural than it is with real numbers. Branch cuts are an extremely important idea in other areas of mathematics, but as far as functional analysis and complex arithmetic are concerned, it is very artificial. I am not saying the principal logarithm should never be used. I am saying the idea is nowhere near as natural an extension as many people think it is. And so, talking about it as if it was can be misleading at best.

Found the fun-ruiner

Logarithms are multifunctions.

you dont need the log tho. u can jus substitute using eulers identity. e^iπ=-1, e^iπ/2=i then i^i=(e^iπ/2)^i=e^(i^2)π/2=e^-π/2. no logs

nerd

I agree, if you only want to know what i^i is. I do think that Bri wanted to also show what ln(x+iy) would be, which makes this approach give the viewer a bit more knowledge than just the answer to the question in the thumbnail.

Since i is defined as the square root of a real number, it's interesting, but also obvious at least in retrospect.

🤓

Me too My mind is blown Being a 9th grader, Loving Math... not studied this but somehow was able to understand this thing...Its just Amazing I have made my own math book of math tricks... hopefully it will be published very soon and will be distributed to all math students 😀

@@IS-py3dk OH MY! Can you please publish it?

@Pine Delgado and you have another bookcase which contains the books which describe the places of your books. Right?

@@demircan_demirbag I'm not sure what you mean. I have four bookcases full of general/popular math books and math textbooks and one bookcase for math history.

That's what i was thinking when i saw the thumbnail

U are missing a minus on exp(pi/2)

Yes, this is correct and a much easier way to solve the problem (assuming you put in the aforementioned minus sign)

This is wrong!!

That polar form is jus tone of infinite answers, the real substitution is i=exp(i*(pi/2+2*pi*k)), where k is an integer.

_Complexer Numbers_

@@cameronbigley7483 i don't think in any of these functions y and x are separable in any way because each others are arguments of each other

f'(x) will be the inverse of g'(y)

@@Blaqjaqshellaq no, it is not. A lot of people are thinking that g(x) will be an inverse or an involution but it is none. The easiest example of what I'm saying is say f(x)= 5y^2+7y+1 and then let g(y)=3x^3+6. These functions are two independent algebraic functions, not inverses or involutions yet related to each other

@@Glitch-cp8wz If those two functions really depend on each other, then the resultant graph is just two points: (-1.32, -0.86) and (-1.30, -0.52). These are the points where the individual graphs intersect. The graph of the function is just a plot of two points and is thus not differentiable. You are effectively trying to differentiate a discrete list

@@AuroraNora3 oh okay. Also how did you get these points? I'm curious about that (I'm just a high school student doing some calculus, so i don't know anything much about it yet)

He did not apply logarithm to i but instead use Euler's identity to decompose i into polar form having arguments of r=1 and theta=pi/2. Both 1 and pi/2 are positive numbers so you can use the logarithmic function on both arguments.

Well, using Eulers Identify (e^ipi = -1) you can take the natural log of -1, as ln(-1) would be i*pi, you can use this to take the natural log of powers and multiples of i and negative numbers, and the natural log of -i (which also happens to be 1/i and i^3) as (3i*pi)/2. The natural log of i can be derived using properties of exponents and logs, as log (a^r) = r log (a), so ln (-1)^1/2 (which is square root of negative one), being 1/2 *ln (-1) which was stated to be i*pi earlier, giving up 1/2 * i*pi, giving up (i*pi)/2

@@thegddevil5664 nuff said.

That's just how ln() works. he even put the line in yellow to tell you :w

bro.. that's not the π symbol..

π^π^π^π unknown integer or no, we ever don`t know e^e^e^e^e is integer or not

as if it wasn't already stated in the video...

yes but no

He did make a video on it.

He substitutes ln(i) as i(pi)/2 in the expression e^i * ln(i)

the inverse of ln is e, therefore raising e to the ln(i) cancels the ln and the e

irrational complex number