Question

If $z=3-5i$, the show that $z^3-10z^2+58z-136=0$

Answer 1

$z=3-5i\Rightarrow |z{|}^2=3^2+5^2=34$

Multiplying the equation throughout with $\overline{z}$,

$|z{|}^2{z}^2-10|z{|}^{2}z+58|z{|}^2-136\overline{z}=0$

Dividing throughout with 34,

$z^2-10z-4\overline{z}+58=0$

So, $z^2-10z-4\overline{z}+58=0\iff z^3-10z^2+58z-136=0$

ie, proving $z^2-10z-4\overline{z}+58=0$ proves $z^3-10z^2+58z-136=0$

$\Rightarrow (3-5i)(3-5i-10)-4(3+5i)+58=-\left(\right(21+25)+(15-35\left)i\right)+(46-20i)=0$

Q.E.D