Question #313825

Find the square root of −5 − 12𝑖

1
Expert's answer
2022-03-21T00:38:17-0400

512i=(a+ib)2512i=a2b2+2abi{a2b2=52ab=12{b=6aa236a2=5{b=6aa4+5a236=0{b=6aa=±2{512i1=23i512i2=2+3i-5-12i=\left( a+ib \right) ^2\\-5-12i=a^2-b^2+2abi\\\left\{ \begin{array}{c} a^2-b^2=-5\\ 2ab=-12\\\end{array} \right. \Rightarrow \left\{ \begin{array}{c} b=-\frac{6}{a}\\ a^2-\frac{36}{a^2}=-5\\\end{array} \right. \Rightarrow \left\{ \begin{array}{c} b=-\frac{6}{a}\\ a^4+5a^2-36=0\\\end{array} \right. \Rightarrow \left\{ \begin{array}{c} b=-\frac{6}{a}\\ a=\pm 2\\\end{array} \right. \Rightarrow \\\Rightarrow \left\{ \begin{array}{c} \sqrt{-5-12i}_1=2-3i\\ \sqrt{-5-12i}_2=-2+3i\\\end{array} \right.


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