# So it's well-known that "i^i" is surprisingly a real number

RumChicken

So it's well-known that "i^i" is surprisingly a real number.

This got me wondering, what is "i to the ith root" (I'm not sure if I worded that correctly, pic related). Can anyone help me with this?

Attached: i-to-the-i.png (10 KB, 204x196)

DeathDog

insert 1/i into the following and solve

$\displaystyle e^{ix} = cos(x) + i \cdot sin(x) \\ e^{i \frac{ \pi}{2}} = cos \left ( \frac{ \pi}{2} \right ) + i \cdot sin \left ( \frac{ \pi}{2} \right ) = 0 + i \cdot 1 = i \\ i^i= ({e^{i \frac{ \pi}{2}}})^i = e^{i^2 \frac{ \pi}{2}} = e^{- \frac{ \pi}{2}} = \frac{1}{e^{ \frac{\pi}{2}}} = \frac{1}{(e^ \pi)^ { \frac{1}{2}}} = \frac{1}{ \sqrt{e^ \pi }} \approx 0.20788 \\ i^i = e^{ln(i^i)} = e^{i \cdot ln(i)} = e^{i \cdot ln(e^{i \frac{ \pi}{2}})} = e^{i^2 \cdot \frac{ \pi}{2} \cdot ln(e)} = e^{- \frac{ \pi}{2}} \approx 0.20788$

AwesomeTucker

[eqn]\sqrt[i]{i} = i^{\frac{1}{i}} = i^{-i} = \frac{1}{i^i}[/eqn]

You have a brain, use it.

PurpleCharger

What the fuck? this is nonsense. Why not just do the thing below?

Oh.

Yeah, that makes sense.

Thanks!

Illusionz

what's the fun in that

BlogWobbles

LOL OP BTFO
YOU ARE TARDED OP
GTFO OF THIS BOARD
YOU BRAINLET

Boy_vs_Girl

one derives it from something known one derives it from the identity
don't really know what your issue is here

[eqn] \sqrt [i] { i } = i^{ \frac { 1 } { i } } = \ln \left( i^{ \frac { 1 } { i } }\right) = \frac { 1 } { i } \frac { i \pi } { 2 } = \frac { \pi } { 2 } [/eqn]

Booteefool

it's ugly
uh excuse me the complex log is not a function. please specify if you're using the principal branch :)

Techpill

TreeEater

LuckyDusty

Wait am I missing something? Why does 1/i = -i

MPmaster

1/i * i/i = i/-1 = -i

Well I'll be darned, ya learn something new every day

Skullbone

KILL YOURSELF NIGGER

LuckyDusty

crazy

TechHater

it's ugly

Attached: 1521237121653.png (13 KB, 400x400)

eGremlin

crazy
why the ableism?

BlogWobbles

\begin{align*} i^2&=-1 \\ -i^2&=1 \\ -i \cdot i&=1 \\ -i&= \dfrac{1}{i} \end{align*}

King_Martha

Well, the nth root of some number k can be expressed as k^(1/n). So the ith root of i is the same as i^(1/i). 1/i becomes -i if you multiply it by i/i, so therefore, i^1/i = i^-i. We're almost there now. Per Euler's Identity, i=e^(pi/2*i). -i = e^(3*pi/2), so i^-i = e^((pi/2*i)*3*pi/2*i) = e^(-3*pi^2/4), which is about .0006099.

WebTool

try again

CouchChiller

the correct answer is $\sqrt{e^ \pi} \approx 4.81$

try again

If you want to bluff about expressing simple stuff in complex words then sociology is probably a better fit major for you. We as mathematicians must strive for simplicity.

TreeEater

there is nothing simplistic about mathematics.

Emberburn

This is sort of backwards thinking. For a number a, the number 1/a is defined as a number b such that a*b=1. now i*(-i)=-i^2=1, you do the same process but in reverse for some reason.

RavySnake

i = e^(i pi/2)
i^(-i) = e^(-i^2 pi/2) = e^(pi / 2)

you were on the right track

DeathDog

in reverse
It's an equation, not a sentence in English.
You're manipulating both sides of a scale,
not reciting a poem of Shakespeare.

AwesomeTucker

No, you're verifying a statement "-i is the multiplicative inverse of i" and rather than solving an equation you can just show that it satisfies the criteria for being that. Your approach is exploratory however and shows how one would come up with the identity in the first place which could be useful for OP.

PurpleCharger

fuck off to /lit/

Fuzzy_Logic

I'm just helping you get better at math, throwing everything in an equation blender is not the best solution every time.

Illusionz

rude

oof