A) \[\pm (3+2i)\]
B) \[\pm (3-2i)\]
C) \[\pm (2+3i)\]
D) \[\pm (2-3i)\]
Correct Answer: B
Solution :
We know that, \[\sqrt{a-ib}=\pm \left[ \sqrt{\frac{1}{2}(\sqrt{{{a}^{2}}+{{b}^{2}}+a})} \right.\] \[\left. -i\sqrt{\frac{1}{2}(\sqrt{{{a}^{2}}+{{b}^{2}}-a})} \right]\] \[\therefore \]\[\sqrt{5-12i}=\pm \left[ \sqrt{\frac{1}{2}(\sqrt{25+144+5})} \right.\] \[\left. -i\sqrt{\frac{1}{2}(\sqrt{25+144}}-5) \right]\] \[=\pm \left[ \sqrt{\frac{1}{2}(13+5)}-i\sqrt{\frac{1}{2}(13-5)} \right]\] \[=\pm [\sqrt{9}-i\sqrt{4}]\] \[=\pm (3-2i)\]
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