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Complex Numbers

If z=2-3i, ...

Question

If $z = 2 - 3 i$ , then show that, $z^{2} - 4 z + 13 = 0$ .

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Solution

It is given that,

$z = 2 - 3 i$

$⟹ z - 2 = - 3 i$

On squaring both side, we get
$(z - 2)^{2} = (- 3 i)^{2}$

$⟹ z^{2} + 4 - 4 z = - 9$

$⟹ z^{2} - 4 z + 13 = 0$

Hence proved.

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