Question

\[i^i\]where i is the imaginary unit

Answer 1

What to do with this?

Answer 2

simplify, ie show it it real

Answer 3

i know the answer i just have forgotten how to get there

Answer 4

\[\approx 0.21\]

Answer 5

real because
\[i^i=e^{i\log(i)}\]

Answer 6

it has a simpler form

Answer 7

for the principle branch of the log, \(log(z)=\ln(|z|+i \arg(z)\) and so \(\log(i)=\frac{\pi}{2}i\) making
\[i^i=e^{i\log(i)}=e^{i\times\frac{\pi}{2}i}=e^{-\frac{\pi}{2}}\] a real number

Answer 8

\[i^{i}=((Cos(2n+1)pi/2)+(Sin(2n+1)pi/2))^{i}\]
\[i^{i}=e^{(i(2n+1)pie/2)i}\]
\[i^{i}=e^{-(4n+1)pie/2}\]
this is real

Answer 9

im not sure i follow your logic with the principle branch argument \[\log(i)=\fracπ2i\]

But you have solved it correctly @satellite73

Answer 10

in general you usually see \[\log(z)=\ln|z|+i\arg(z)\] but this makes the log a multi-valued function because \(\arg(z)\) is not unique
you can make it single values be saying for example that you restrict \(\arg(z)\) to the interval \([0,2\pi)\) and htne you get single valued function. this is sometimes denoted as \(Log(z)\)

Answer 11

sorry for n=0 my solution reduces to e^(-pie/2) that is approximately equal to 0.21

Answer 12

i wish i could you a second prize @rs32623

Answer 13

but at the outset you need to know that by definition, for complex numbers and for real numbers, that \(b^z=e^{z\log(b)}\)

Answer 14

i will award

Answer 15

courtesy of mr. van vliet

Answer 16

i thinkwe can make it more short
\[i^{i}=e^{(i*pie/2)*i}=e^{-pie/2}\approx0.21\]

Answer 17

Do you actually like Trout Mask Replica,
@satellite73
you must be the first (other) person i know.

Answer 18

@UnkleRhaukus this question is simply awesome beacuse a lot of peoples complain why do we study complex numbers & imaginary things; i think this is answer to them
\[i^{i} \] is real & that shows the beauty of mathematics!!

Answer 19

i makes a lot a of sense in quantum mechanics i experienced some really complex revelations