Find the square root of complex number $z = 3 + 4 i$

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Solution

Given, $z = 3 + 4 i$

$(x + i y)^{2} = 3 + 4 i$

$x^{2} - y^{2} + i (2 x y) = 3 + 4 i$

$\Rightarrow x^{2} - y^{2} = 3$ and $2 x y = 4$

$\Rightarrow x^{2} - y^{2} = 3$ and $x y = 2$

$∴ x = 2$ and $y = 1$

$\Rightarrow \sqrt{3 + 4 i} = 2 + i$

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