How to find all complex Z such that Z^5=-32

[grossgermany]

Homework Statement

How to find all complex number Z such that Z^5=-32

Homework Equations

Euler equation
e^it=cost+isint

The Attempt at a Solution

I guess a naive way to solve is that since Z^5=(-2)^5
Therefore Z=-2, but this obviously too good to be true.
I have no background in complex analysis.

 

[Dick]

Z=(-2) isn't too good to be true. You are right. It is a root. But there are four more. Express -32 in polar form. I.e. -32=32*e^(it). The use deMoivre. What are the possibilities for t? There are more than one. Give me one to start out with.

 

[grossgermany]

Thank you for your reply.
Sorry I do not understand that -32=32*e^(it)?
e^it=cost+isint
I mean t is a variable that takes on arbitrary values.

 

[Dick]

Thank you for your reply.
Sorry I do not understand that -32=32*e^(it)?
e^it=cost+isint
I mean t is a variable that takes on arbitrary values.

t isn't arbitrary if -32=32*e^(it). That means e^(it)=(-1). What's t?

 

[grossgermany]

t=180degree=-5pi,-3pi,-pi,pi,3pi,5pi,7pi,9pi,11pi?
Therefore -32=32*e^(it)=32*[cos(pi)+isin[pi]]

Sorry how do I use de Moivre's formula? Never taken complex analysis

 

[Dick]

t=180degree=-5pi,-3pi,-pi,pi,3pi,5pi,7pi,9pi,11pi?
Therefore -32=32*e^(it)=32*[cos(pi)+isin[pi]]

Sorry how do I use de Moivre's formula? Never taken complex analysis

You could look it up rather than me telling you. You multiply the arguments (the angles) by 1/5. How many roots do you get before they start repeating?