Given $z_1=12\big($cos$\displaystyle \frac{3\pi}{8}+i $sin$\displaystyle\frac{3\pi}{8}\big)$ and $z_2=4\big($ cos$\displaystyle\frac{\pi}{8}-i$ sin $\displaystyle\frac{\pi}{8}\big)$, find $z_1z_2$.
So, the general formula is that
if $z_1=r_1($ cos $\theta_1 +i$ sin $\theta_1)$ and $z_2=r_2($ cos $\theta_2+i $ sin $\theta_2)$,
then $z_1z_2=r_1r_2[$ cos $(\theta_1+\theta_2) +i$ sin $(\theta_1+\theta_2)]$
So, I do realise that I have to change the negative sign in $z_2=4\big($ cos$\displaystyle\frac{\pi}{8}-i$ sin $\displaystyle\frac{\pi}{8}\big)$ before I can use the formula, but how to do so?
$$\cos -\frac\pi 8= \cos\frac\pi 8$$ $$\sin -\frac\pi 8=-\sin\frac\pi 8$$ So $$4(\cos\frac\pi 8-i\sin\frac\pi 8)=4(\cos-\frac\pi 8+i\sin-\frac\pi 8)$$