An Interesting Shape in the Complex Plane (from i^i).

I decided to plot a cyclical pattern on the complex plane, and it looked cool. So, well, here's a video. I bet there are some cool things to discover about this shape. I leave that to you!

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[Felixkeeg]

Man, the editing is god-tier 90's technology

[kalkal]

When Starlord is your maths professor.

[alkankondo]

LOL at 3:17. Glad you DIDN'T edit that out!

[korayacar]

Its area is exp(pi/2)-exp(-pi/2), which turns out to be 2i*sin(-i*pi/2).

[CCequalPi]

"If something feels good you should do it again" ... *Me on crack agreeing

[cptcaptain]

Ancient greek astronomer: ... thats a Lobster!

[oliviasoutlook]

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 :)

[-sideddice]

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 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 ^^'

[joshuaisemperor]

The quality of the videos over the time get's yet better and better.

[ConnorMooneyhan]

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

[drandrewsanchez]

Man I just fricken love your videos man

[fountainovaphilosopher]

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

[djsmeguk]

Whenever I see iteration in the complex plane producing cardioid type shapes, my Mandelbrot senses start tingling.. is this perhaps a transformation?

[thedoublehelix]

The perimeter is 2sqrt(1+e^pi) + 2sqrt(1+e^(-pi)) = 2sqrt(y+2sqrt(y)) where y=e^pi +1/e^pi + 2 which is sort of interesting.

[shoopinc]

"If something feels good, you should always do it again." - Epic Math Time

[FrostDirt]

That's neither a spearhead nor an arrowhead. That's a spaceship!!

[Celastrous]

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.

[fullfungo]

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)

[xyBubu]

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

[tomkerruish]

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.