Question

If $i{z}^3+z^2-z+i=0$, then show that $\left|z\right|=1$.

Answer 1

Dividing throughout by $i$ and knowing that $\frac{1}{i}=-i$
we get
$z^3-i{z}^2+iz+1=0$
or $z^2(z-i)+i(z-i)=0as1=-i^2$
or $(z-i)(z^2+i)=0$
$\therefore z=i$ or $z^2=-i$
$\therefore \left|z\right|=\left|i\right|=1$
$\left|z^2\right|={\left|z\right|}^2=|-i|=1$
$\therefore \left|z\right|=1$
Hence in either case $\left|z\right|=1$.