How do I find all solutions of (2z + 6)^5= 1 + i
Do I have to find the polar form and determine what angle it has and then use
z=r(cosx + isinx)? Or is that not helpful in this situation?
Thanks.
Oh, yes, yes: the polar form is very helpful in this case. So you have
(I am using here as the "polar form" of z, instead of , because it is easier to type.)
Next you can take the fifth root by dividing the exponents on both sides by 5, and you get
Finally subtract 6 and divide by 2: done. - Except that you still need to figure out how many different solutions you have now got...