DPR
Took a quick look at the problem provided in class of $z+2z^3$ . I think expanding it it is just basic algebra but I was wondering if there was an easier way maybe converting to polar with Euler's to avoid the cubic.
Here's what I tried quickly:
$$\begin{aligned}
&z+2z^3\\
&(a+bi)+2(a+bi)^3\\
&(a+bi)+2[(a^3-3ab^2)+i(3a^2b-b^3)]\\
&(a+bi)+2a^3-6ab^2+6a^2bi-2b^3i\\
&(2a^3-6ab^2+a)+i(6a^2b-2b^3+b)\end{aligned}$$
