by Guest » Thu Jan 30, 2025 10:00 am
for the first one, let [tex]z=a+ib[/tex]. Now,
[tex]|z+16|=4|z+1|\Longrightarrow|(a+16)+ib|=4|(a+1)+ib|\Longrightarrow{(a+16)}^2+b^2=16\left\{(a+1)^2+b^2\right\}[/tex]
[tex]\Longrightarrow a^2+32a+256+b^2=16a^2+32a+16+16b^2\Longrightarrow15\left(a^2+b^2\right)=240\Longrightarrow a^2+b^2=16[/tex]
[tex]\Longrightarrow\sqrt{a^2+b^2}=|z|=\boxed{4}[/tex]
for the second one, [tex]\frac{-1}{x-iy}=\frac{4+7i}{5-3i}\Longrightarrow x-iy=\frac{-3-5i}{4+7i}=\frac{-3-5i}{4+7i}\times\frac{4-7i}{4-7i}=\frac{-47+i}{16+49}[/tex]
[tex]\Longrightarrow\boxed{x=\frac{-47}{65},y=\frac{-1}{65}}[/tex]