Question

For z1 = 2cis120degrees and z2 = 4cis30degrees, find z1 · z2 in rectangular form.

Answer 1

I'm assuming \(\newcommand{cis}{\,\text{cis}\,}\cis\theta=e^{i\theta}\).
\[z_1=2\cis120^\circ=2(\cos120^\circ+i\sin120^\circ)\\
z_2=4\cis30^\circ=4(\cos30^\circ+i\sin30^\circ)\]
The convenient thing about polar forms of complex number is that multiplying them amounts to multiplying their moduli and adding their arguments, i.e. for two complex numbers \(\large a=re^{i\alpha}\) and \(\large b=se^{i\beta}\), the product is \(\large ab=rse^{i(\alpha+\beta)}\).

So,
\[z_1z_2=2\times4\cis(120^\circ+30^\circ)=8\cis150^\circ\]
From here you can convert to rectangular form.