Question

The square root of $5-12i$ is

• A. $\pm (3+2i)$
• B. $\pm (3-2i)$
• C. $\pm (2+3i)$
• D. $\pm (2-3i)$

Answer 1

Answer: (B)

We know that, $\sqrt{a-ib}=\pm [\sqrt{\frac{1}{2}(\sqrt{a^2+b^2+a})}$
$-i\sqrt{\frac{1}{2}(\sqrt{a^2+b^2-a})}]$
$\therefore$ $\sqrt{5-12i}=\pm [\sqrt{\frac{1}{2}(\sqrt{25+144+5})}$
$-i\sqrt{\frac{1}{2}(\sqrt{25+144}}-5)]$
$=\pm \left[\sqrt{\frac{1}{2}(13+5)}-i\sqrt{\frac{1}{2}(13-5)}\right]$
$=\pm [\sqrt{9}-i\sqrt{4}]$
$=\pm (3-2i)$