Question

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

• A. $\left|z\right|<1$
• B. $\left|z\right|>1$
• C. $\left|z\right|=1$
• D. $\left|z\right|=0$

Answer 1

Answer: (C)

$\left|z\right|=1$

Given $i{z}^3+z^2-z+i=0\Rightarrow i{z}^2(z-i)-(z-i)=0$

$\Rightarrow (z-i)(i{z}^2-1)=0\Rightarrow z=i$ or $z^2=\frac{1}{i}=-i$

Now $z=i\Rightarrow \left|z\right|=\left|i\right|=1$

and $z^2=-i\Rightarrow \left|z^2\right|=|-i|\Rightarrow \left|z^2\right|=1\Rightarrow \left|z\right|=1$

Thus, in both cases $\left|z\right|=1$