@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?
@@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!
@@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.
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.
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.
Technically, we regard the complex logarithm as ill-defined. It just makes no sense as a concept that extends the real logarithm, for many issues. Coterminal angles are not the only issue. Contour integration makes this much more damning. That being said, the function Log(z) = ln(|z|) + atan2[Im(z), Re(z)]·i can be a useful one to define in some limited contexts, and since it is mildly analogous to the logarithm, the notation has stuck on, despite it being somewhat misleading. However, I think it would be healthier to avoid thinking of the above function as an actual logarithm. Superficially, it has some of the same properties, but it causes more conceptual problems than it solves. Anyway, if you interpret the notation ζ^ψ specifically as referring to exp[ψ·Log(ζ)], then it makes perfect sense to think of i^i as being equal to exp(-π/2). However, this interpretation of the notation is inconsistent with how exponents are typically treated. For example, it makes powers with base 0 undefined, and it also changes the odd fractional powers. Keeping in mind this inconsistency is something that many people need to be reminded of.
@@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.
@@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
I so FRICKEN LOVE THIS!!!! CRAZY AWESOME!!! I'm being serious here as I LOVE math and I have a huge math library full of math books in my home with four bookcases full of these books not counting an additional bookcase for my books on math history!!😁
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 😀
@@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.
Wouldn't it be easier (and maybe less insightful) to substitute i by its polar form, exp(-i*pi/2)? Then you'll quickly find i^i = exp(i*-i*pi/2)=exp(pi/2) Edit: I missed a minus sign! Thanks for noticing!
You did it correctly. I have seen other videos this subject and they skipped Ln(i) portion, even though they got the same answer, but wrong reasoning or technique.👌
People need to realize that ln of x and ln of z are two different things. In the argand diagram complex functions are SURFACES. These are multi valued functions.
Why so complicated? i^i=e^(i*i*Pi/2) directly using the Euler formula with i*i=-1 and then you get the real value i^i=e^(-Pi/2), so I don't understand this way of solving!
Tbh i have an easier proof ,so it goes like this : i=√-1 Let's use eulers identity to take -1 to the complex world -1=e^(iπ) So i=√-1=e^(iπ/2) We are looking for i^i so lets's use other form of i we have just found (e^(iπ/2))^i In this expression we can multiply the exponent and we will have i × i which is -1 so it becomes e^(-π/2)which is also √(1/e^π) Which is a real number
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
I wonder what the significance of all possible rotations are for this number. Like why does it create exponential decay or growth of all things for when k is not 0 (usual convention).
Btw, it is not the only solution. You just had a principal solution, but you should note that it is multivalued function. Let me show it. i = e^(iπ/2) = e^(i5π/2) = e^(i9π/2) = ... So you will have: i^i = e^(iπ/2)i = e^(-π/2) i^i = e^(i5π/2)i = e^(-5π/2) i^i = e^(i9π/2)i = e^(-9π/2) ...
> we could travel around the circle another 2pi and we would get a different result > when people say i to the u they usually mean this Bro Comon It doesn't matter how many times you go aroubd the circle i to the i is e to the minus half pi. You're not going to get a different result.
Pretty obvious. Power of i means a 90deg counterclockwise rotation,,,, Rotation is thus from the imaginary direction pi/2 (90 deg) to the real one pi (180 deg)... True for every 360 deg additional rotation.
we can prove this without polar coordinates too. i^i=e^ln(i^i)=e^(ilni) ln(i)=ln((-1)^(1/2))=(1/2)(ln(-1)) ln(-1)=(i*pi) because of eulers identity =>(1/2)(ln(-1))=(i*pi)/2 =>i^i=e^i(i*pi/2)=e^-pi/2
Hey @brithemathguy, quick question, how do you find the derivative of y=f(x) while x=g(y). In this case, these functions are not inverses nor involutions. It looks like this function shoots up to infinity, can you help me with this?
@@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)
I have no idea what are you talking in video but I can solve it by e^πi =-1 Like this e^πi=i^2 e^π=i^(2/i) e^πi/2=i^1/I Since -1 is i^2 it caan be like this e^-π/2=i^i
Не факт, что так! Условие, что любое выражение равно числу "е" в степени логарифм натуральный от этого выражения - такое действует возможно на действительные числа только, а вот работает ли так же оно и на мнимые - не факт
Nice, but I find it a bit unsatisfying. You just reach a number in decimal notation which kind of obscures the meaning of the number. But I guess you got me thinking on a different direction so I'll give you that
I’m confused at 1:38. Sorry if this is a stupid question, im only in seventh grade. If r and e are multiplying, why is it ln(r) + ln(e^iθ), and not ln(r)*ln(e^iθ)?
i = exp( i *pi/2) = exp( i *5pi/2) = exp( i *7pi/2) = ...................... so basically i ^ i = exp(-pi/2) = exp(-5pi/2) = exp(-7pi/2) = ......... which means 0.20787 = 0.0003882 = 0.0000167758 mathematics are collapsing ........
