Помогите решить алгебру

Решите задачу:

$z=-6-6i\\Re\ z=-6\ ;Im\ z=-6\\\phi=arctg\frac{Im\ z}{Re\ z}-\pi=arctg\frac{-6}{-6}-\pi=arctg1-\pi=-\frac{3\pi}{4}\\r=\sqrt{Re\ z^2+Im\ z^2}=\sqrt{(-6)^2+(-6)^2}=\sqrt{72}=6\sqrt2\\z=6\sqrt2(cos(-\frac{3\pi}{4})+i\sin(-\frac{3\pi}{4}))$

$z=0+(\sqrt3+1)i\\Re\ z=0\ ;Im\ z=(\sqrt3+1)\\\phi=\frac{\pi}{2}\\r=\sqrt{Im\ z^2}=\sqrt{(\sqrt3+1)^2}=|\sqrt3+1|\\z=|\sqrt3+1|(cos\frac{\pi}{2}+i\sin\frac{\pi}{2})$

$z_1z_2=6\sqrt2(cos(-\frac{3\pi}{4})+i\sin(-\frac{3\pi}{4}))*|\sqrt3+1|(cos\frac{\pi}{2}+i\sin\frac{\pi}{2})=\\=6\sqrt2*|\sqrt3+1|(-\frac{\sqrt 2}{2}-i\frac{\sqrt 2}{2})i=-6|\sqrt3+1|i+6|\sqrt3+1|=\\=-6|\sqrt3+1|(1-i)\\\\\frac{z_1}{z_2}=\frac{6\sqrt2(cos(-\frac{3\pi}{4})+i\sin(-\frac{3\pi}{4}))}{|\sqrt3+1|(cos\frac{\pi}{2}+i\sin\frac{\pi}{2})}=\frac{6\sqrt2(-\frac{\sqrt 2}{2}-i\frac{\sqrt 2}{2})*i}{|\sqrt3+1|i*i}=-\frac{6\sqrt2(-\frac{\sqrt 2}{2}i+\frac{\sqrt 2}{2})}{|\sqrt3+1|}=\\=-\frac{-6i+6}{|\sqrt3+1|}$