If $z = a + i b$ is a complex number such that $z = {(a + i b)}^{2}$ then
show that $x^{2} + y^{2} = {(a^{2} + b^{2})}^{2}$

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Solution

$z = (a + i b)$ (Given)
$z = (a + i b)^{2}$ (Given)
$⟹ a + i b = a^{2} + b^{2} + 2 a b i$
$a = a^{2} - b^{2}$ (Real)
$b = 2 a b$ (Imaginary)
$a^{2} + b^{2} = (a^{2} - b^{2})^{2} + (2 a b)^{2} = a^{4} + b^{4} - 2 a^{2} b^{2} + 4 a^{2} b^{2} = a^{4} + b^{4} + 2 a^{2} b^{2} = (a^{2} + b^{2})^{2}$
Hence Proved.

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