If $i\mathcal{z}^{3} + \mathcal{z}^{2}\mathcal{- z +}i = 0$, then $\mathcal{|z|}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(a)
$\mathcal{z}^{3} + \frac{1}{i}\mathcal{z}^{2} - \frac{\mathcal{z}}{i} + 1 = 0$
$\Rightarrow \ \ \mathcal{z}^{3} - i\mathcal{z}^{2} + i\mathcal{z} + 1 = 0$
$\Rightarrow \ \ \mathcal{z}^{2}\left( \mathcal{z -}i \right) + i\left( \mathcal{z -}i \right) = 0$
$\Rightarrow \ \ \left( \mathcal{z -}i \right)\left( \mathcal{z}^{2} + i \right)\ \  \Rightarrow \ \left| \mathcal{z} \right| = 1$

Q. If $i\mathcal{z}^{3} + \mathcal{z}^{2}\mathcal{- z +}i = 0$, then $\mathcal{|z|}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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