Question

Solve the attached sum...
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Answer 1

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Answer refer to the attachment.

Answer 2

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Here is Your Answer...!!

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Step by step solution:

$Given \ z_{1}=2-i \ and \ z_{2}=-2+i\\\\First \ let \ us \ find \ z_{1}z_{2}\\\\z_{1}z_{2}=(2-i)(-2+i)\\\\z_{1}z_{2}=-4+2i+2i-1\\\\z_{1}z_{2}=-3+4i\\\\Now \ we \ have \ to \ find \ \frac{z_{1}z_{2}}{z_{1}^-}$

$\frac{z_{1}z_{2}}{z_{1}^-}=\frac{(-3+4i)(2-i)}{(2+i)(2-i)}\\\\ \frac{z_{1}z_{2}}{z_{1}^-}=\frac{-6+3i+8i-4i^2}{4-(-1)}\\\\\frac{z_{1}z_{2}}{z_{1}^-}=\frac{-6+11i-4(-1)}{4+1}\\\\\frac{z_{1}z_{2}}{z_{1}^-}=\frac{-2+11i}{5}=\frac{-2}{5} +\frac{11i}{5}\\\\Re(\frac{z_{1}z_{2}}{z_{1}^-})=\frac{-2}{5}$

Hope it is clear to you.