Question

obtain in polar form all values of z^pi
z = 2+i

Answer 1

\[z^{\pi}\]

Answer 2

@dumbcow any idea ?

Answer 3

i think so...
anyway you can rewrite z^pi in polar form as
\[\large n^{\pi}(\frac{2}{n} +i \frac{1}{n})^{\pi} = n^{\pi}(\cos \theta +i \sin \theta)^{\pi}\]
then solve for n and theta
cos = 2/n
sin = 1/n
--> n = sqrt5
--> theta = .4636 radians
so
\[ \large (2+i)^{\pi} = \sqrt{5}^{\pi}(\cos (.4636 \pi) +i \sin(.4636 \pi))\]

Answer 4

not sure if angle has exact form....its arctan(1/2)

Answer 5

okey i am checking

Answer 6

here is what wolfram says so it looks like i'm correct

http://www.wolframalpha.com/input/?i=%282%2Bi%29^pi

Answer 7

okey then :) thank u

Answer 8

no problem