Question

Помогите решить алгебру

Answer 1

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

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

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