Find the square roots of the complex number $3 + 4i$.

Question

Accepted Answer

$\pm(2 + i)$. Let $(x + yi)^2 = 3 + 4i$. Then $x^2 - y^2 = 3$ and $2xy = 4$, so $xy = 2$. Trying $x = 2$, $y = 1$: $4 - 1 = 3$ and $2(2)(1) = 4$, which works. The roots are $\pm(2 + i)$. Check: $(2 + i)^2 = 4 + 4i + i^2 = 3 + 4i$. The pair $(1 + 2i)$ gives $1 - 4 + 4i = -3 + 4i$.

Answer

A) $\pm(1 + 2i)$

Answer

B) $\pm(2 - i)$

Answer

C) $\pm(\sqrt{3} + 2i)$

Answer

D) $\pm(2 + i)$

Find the square roots of the complex number $3 + 4i$.

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