Thread: Comlex numbers

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.

Originally Posted by fgg

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.

Does giving one root solve a problem with 9?

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So



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This is the first of the nine square roots. The rest are evenly spaced around a circle, so have the same magnitude and are all separated by an angle of .

So the roots are

Evaluate the rest of the roots, then convert them back to Cartesians.

Thank you very much for your help!