Convert $\mathrm{i}^\pi$ into trigonometric form How can I convert the complex number $\mathrm i^\pi$ to trigonometric form?
I usually do these steps:


*

*take $ Z = a + b\mathrm i $ form,

*find $ r = \sqrt{a^2 + b^2} $,

*$ \cos(\phi) = a / r, \sin(\phi) = b / r $,

*find $ \phi $ from the above 2 equations.


For $ \mathrm i^\pi $ I have $ a = 1, b = 1, r = 1 $, $ \cos(\phi) = 1, \sin(\phi) = 1 $.
There's no such $ \phi $.
The online Convert Complex Numbers to Polar Form gives the answer $ \phi = 77.2567 $ or just, $ \phi = \dfrac{180 \arg(\mathrm i^\pi)}{\pi}$
 A: That Gravity Guy answers it very nicely, but if you want to see a drawn out calculation, first get your value into Exp[x] form:
I^Pi == E^x

Log[%[[1]]] == Log[%[[2]]] // PowerExpand
(*(I π^2)/2 == x*)

Solve[%, x] // Flatten
(*{x -> (I π^2)/2}*)

Now we can just convert to trig
ExpToTrig[E^x] /. %
(*Cos[π^2/2] + I Sin[π^2/2]*)

Get r, ϕ from
{Re[%], Im[%]} // ToPolarCoordinates // N
(*{1., -1.34838}*)

If you like degrees better
{%[[1]], %[[2]]/°}
(*{1., -77.2567}*)

A: Some useful identities:
$i = e^{i\pi/2}$
$(a^b)^c = a^{(bc)}$
$e^{i\phi} = \cos{\phi} + i\sin{\phi}$
Your expression:
$i^\pi = (e^{i\pi/2})^\pi = e^{i\pi^2/2} = 1 (\cos{(\pi^2/2)} + i \sin{(\pi^2/2}))$
A: ComplexExpand [$i^\pi$]

$ \cos(\frac{\pi^2}{2}) + i\sin(\frac{\pi^2}{2})$

A: So, first of all, you made a small mistake : 
In your method, you have $a = 0$, and not $a=1$
that makes $b=1$, $r=1$, $\cos(\phi)=0$ and $\sin(\phi)=1$, that implies $\phi = \frac{\pi}{2} + 2k\pi$ for any $k \in \mathbb{Z}$.
The problem is all values of $k$ are correct (as $\cos(\alpha) = \cos(\alpha+2\pi)$, whatever the value of $\alpha$, and we have the same for $\sin$), and as such, $i^\pi$ doesn't make any mathematical sense, since $e^{i\frac{\pi^2}{2}} \neq e^{i\frac{5\pi^2}{2}}$, and mathematically speaking, you have no reason to chose one over the other. 
Math software usually ignore this by arbitrarily choosing the argument value that is between $-\pi$ and $\pi$ (or sometimes $0$ and $2\pi$, which even there can lead to different results if you took, for instance, "$(-i)^\pi$"  )
