Find all solutions to Log(z^3-9)=(pi)*i ,z = a +bi

[squaremeplz]

Homework Statement

Sorry to repost this but no one replied to my other thread.

find all solutions to Log(z^3-9)=(pi)*i ,z = a +bi

The Attempt at a Solution

z^3-9 = e^(pi*i)

e^(pi*i)=-1

z^3 = -1 + 9

then the problem becomes finding the roots of

z^3 = 8

which are -1 - sqrt(3)i, -1 + sqrt(3)i, and 2

did I get this right? Thanks.

 

[HallsofIvy]

Yes, that is correct.