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
$\displaystyle (2z+6)^5=1+i=\sqrt{2}\cdot e^{\displaystyle i\tfrac{\pi}{4}+i2n\pi}, n\in\mathbb{Z}$
(I am using $\displaystyle z=|z|\cdot e^{i\varphi}$ here as the "polar form" of z, instead of $\displaystyle z=|z|\cdot\big(\cos\varphi+i\sin\varphi\big)$, 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
$\displaystyle 2z+6=\sqrt[10]{2}\cdot e^{\displaystyle i\tfrac{(8n+1)\pi}{20}}, n\in \mathbb{Z}$
Finally subtract 6 and divide by 2: done. - Except that you still need to figure out how many different solutions you have now got...