Find the square root of −5 − 12𝑖
−5−12i=(a+ib)2−5−12i=a2−b2+2abi{a2−b2=−52ab=−12⇒{b=−6aa2−36a2=−5⇒{b=−6aa4+5a2−36=0⇒{b=−6aa=±2⇒⇒{−5−12i1=2−3i−5−12i2=−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.−5−12i=(a+ib)2−5−12i=a2−b2+2abi{a2−b2=−52ab=−12⇒{b=−a6a2−a236=−5⇒{b=−a6a4+5a2−36=0⇒{b=−a6a=±2⇒⇒{−5−12i1=2−3i−5−12i2=−2+3i
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