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

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Solution

We have
$i z^{3} + z^{2} - z + i = 0$
$\Rightarrow z^{3} - i z^{2} + i z + 1 = 0$ [On dividing both sides by i]
$\Rightarrow z^{2} (z - i) + i (z - i) = 0$
$\Rightarrow (z - i) (z^{2} + i) = 0$
$\Rightarrow z = i$ or $z^{2} = - i$ .
And $z^{2} = - i \Rightarrow | z^{2} | = | - i | = 1 \Rightarrow | z |^{2} = 1 \Rightarrow | z | = 1$ .
Hence, in either case, we have |z|=1.

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