For any non zero complex z, if you apply 4 times the rise to power i, you fall back on the z you started with. Not difficult to see with polar coordinates : I mean z=r . e^(i theta). That means it is a bijection C* -> C* of order 4. Noice. In fact we could see this function as a group isomorphism of (C*, x) to itself. OH I just realised : it is a square root of the inversion. Well after all it's not that surprising since i*i=-1. Is this function holomorphic ? Hmmmm Oh shit there is a problem. This function depends on the theta you choose in z=r e^(i theta). In other words it has the same problems of definition as the complex logarithm. I guess we could save what i said before by saying we define this function on C - (R^-). And it would have its values in the set..... OH how nice is that !! The image set is the union of an opened disc and the complex plane without the closed disc. So pretty. It transformed the (half of) straight line in a circle. And it must probably be holomorphic wherever we define it. I don't know what there is left to say ^^'
I wish I could understand this but I enjoy watching these videos just for hearing the beauty of mathematical language nonetheless. Thanks for making my day :)
Oh I was thinking we could generalize the formula to i^(i^n), where n varying on a real numberline would enclose a different shape. The finishing formula is (i^(npi/2))*e^(-pi/2 *sin(npi/2)), and idk what sort of a shape that would generate, but it's like a finer, "less sharp" version of this
It's an arrowhead for sure. A spear head has those two smaller pointy corners facing into it so it makes a kind of diamond that got lengthened on one side
Great video, I was inspired to look at what happens when you raise a general complex number to the power of i. Let z = r e^it be the polar form of z, z^i = r^i e^(-theta) To determine r^i, let's try solve r^i = c, this gives us i = log_r(c) = ln(c)/ln(r), where "log_r" is the logarithm base r, "ln" is the natural logarithm, and "log_r(c) = ln(c)/ln(r)" is the change of base formula. So ln(c) = i ln(r), so r^i = c = e^(i ln(r)) So putting it all together z^i = e^(-theta) e^(i * ln(r)) = e^-t (cos(ln r) + i sin(ln r)) So z^i results in a complex number with magnitude e^(-theta) and angle ln(r) In other words this operation swaps the role of the magnitude and angle (with some exp and log thrown in the mix)!
Look at f(t)=i^{e^{iπt/2}} as t goes from 0 to 4. The cases shown in the video are just the integer values for t. The same computation at the beginning of the video will give the formula f(t)=e^{e^{iπt/2}iπ/2}. It'd be cool to plot this but I'm too lazy. Anyone wanna make one of those fancy Desmos pages you can link to as a reply?
The area of the arrowhead is 2sinh(pi/2), quite interesting! The perimeter is less elegant, it is: 2*(sqrt(e^pi + 1) + sqrt(e^-pi + 1)) which is about 11.87.
The actual shape of i^i^n can be described as follows: let a(t) = e^(pi/2 • cos(pi/2 • x)) let b(t) = cos(pi/2 • sin(pi/2 • x)) let c(t) = sin(pi/2 • sin(pi/2 • x)) Then i^i^(t-1) = a(t)b(t) + i•a(t)c(t). In the video we draw the shape with t=2,3,4,5. To get a closed shape we can draw this from 2 to 6 instead (as a continuous interval of values for t). What we get resembles r=7/4•cos(t)+7/3 shifted to the right by 1. So it’s more apple shaped, rather then an arrow. Complete derivation: I will start with i^i^(t-1) instead. Let’s use i=e^(i•pi/2). i^i^(t-1) = (e^(i•pi/2))^i^(t-1) = e^(pi/2 • i^t) = e^(pi/2 • (e^(i•pi/2))^t) = e^(pi/2 • e^(i•t•pi/2)) = e^(pi/2 • (cos(t•pi/2)+i•sin(t•pi/2))) = e^(pi/2 • cos(t•pi/2)) • e^(i•pi/2 • sin(t•pi/2)) = a(t)•e^(i•pi/2 • sin(t•pi/2)) = a(t)•(cos(pi/2 • sin(t•pi/2))+i•sin(pi/2 • sin(t•pi/2))) = a(t)•(b(t)+i•c(t)) = a(t)b(t)+i•a(t)c(t)
the height is 0 because it's i +(-i) = 0 and the area is also 0 because it's bh/2, it looks like something going in a circular motion because of the coordinates of the points i, e^pi/2, -i, e^-pi/2
Would be fun to see if raising it by the increasing fractional values of i through each step would yield the same shape. I gotta run errands, maybe when I get home.
i^i^i is ambiguous, since exponentiation is nonassociative; do you mean i^(i^i) or (i^i)^i? In your video, you use the second interpretation. Since (a^b)^c = a^(b*c), this results in i^i^i^•••^i = i^(i^n), where the term on the left has n+1 i's. Expressed this way, it's clear why your expression repeats with period 4. I have found, however, that a^b^c = a^(b^c) occurs more commonly, especially when defining tetration. I don't think this will repeat, but I'm not sure.
Forgetting about the 2πni we have i=e^(iπ/2) i^i=e^(-π/2) then (i^i)^i=e^(-πi/2) and generally with n i‘s where n is natural we get e^(-i^(n-2) π/2) because since e is nonnegative (e^a)^b=e^(ab). And we get this shape because i^n is periodically i,-1,-i,1. The circumference is easily getable since you can calculate the distance between 2 complex numbers z_1=a_1+b_1i and z_2=a_2+b_2i where a_1,a_2,b_1 and b_2 are real. As sqrt((a_1-a_1)^2+(b_1,-b_2)^2) You can also get the area since putting stuff into polar cords gives you the angles pretty easily to.
just a suggestion: can you record your audio in mono next time? The audio switching from ear to ear gets a little bit distracting, and I think mono audio would be a better experience.
@@DylanYoung in my discipline j is used as the imaginary unit - to not confuse it with other values (you could still confuse it with the current density)
