Hi.

I wonder if anyone know how to solve this math problem?

z^9 = 1 + i * square root of 3 , answer on the form a + bi

Should I solve it like this:

z^9= 2^9(cos (pi/3 *9) + isin (pi/3 *9)) = -512

or like this:

r^9(cos 9u + isin9u) = 2(cos 60 + isin 60)

n=0 zo=2^(1/9)(cos 60/9 + isin 60/9) = 2^(1/9) *(cos( 60/9) + isin (60/9))

n=1

.

.

.

n=9

Which is the right way to solve this problem?

Tank you.