Complex Numbers for you

If $iz^{3}+z^{2}-z+i=0$, then $|z|$ equals

(a) 4

(b) 3

(c) 2

(d) 1.

Solution.

We can write the given equation as

$z^{3}+\frac{1}{i}z^{2}-\frac{1}{i}z+1=0$, or

$z^{3}-iz^{2}+iz-i^{2}=0$

$\Longrightarrow z^{2}(z-i)+i(z-i)=0$

$\Longrightarrow (z^{2}+i)(z-i)=0 \Longrightarrow z^{2}=-i, z=i$

$\Longrightarrow |z|^{2}=|-i|$ and $|z|=|i|$

$\Longrightarrow |z|^{2}=1$ and $|z|=1$

$\Longrightarrow |z|=1$