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 \left[ \sqrt{\frac{1}{2}(\sqrt{{{a}^{2}}+{{b}^{2}}+a})} \right. $
$ \left. -i\sqrt{\frac{1}{2}(\sqrt{{{a}^{2}}+{{b}^{2}}-a})} \right] $
$ \therefore $ $ \sqrt{5-12i}=\pm \left[ \sqrt{\frac{1}{2}(\sqrt{25+144+5})} \right. $
$ \left. -i\sqrt{\frac{1}{2}(\sqrt{25+144}}-5) \right] $
$ =\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) $