Complex Numbers 2 Question 42
43. If $i z^{3}+z^{2}-z+i=0$, then show that $|z|=1$.
(1995, 5M)
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Solution:
- Given, $i z^{3}+z^{2}-z+i=0$
$$ \begin{aligned} & \Rightarrow \quad i z^{3}-i^{2} z^{2}-z+i=0 \quad\left[\because i^{2}=-1\right] \\ & \Rightarrow \quad i z^{2}(z-i)-1(z-i)=0 \\ & \Rightarrow \quad\left(i z^{2}-1\right)(z-i)=0 \\ & \Rightarrow \quad z-i=0 \text { or } i z^{2}-1=0 \\ & \Rightarrow \quad z=i \quad \text { or } \quad z^{2}=\frac{1}{i}=-i \\ & \text { If } \quad z=i \text {, then }|z|=|i|=1 \\ & \text { If } z^{2}=-i \text {, then }\left|z^{2}\right|=|-i|=1 \\ & \Rightarrow \quad|z|^{2}=1 \quad \Rightarrow \quad|z|=1 \end{aligned} $$
