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Question

If $\left(-7-24i{\right)}^{\frac{1}{2}}=x-iy$, then ${x}^{2}+{y}^{2}$ is equal to

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Solution

Step 1. Remove square root part :Given that $\left(-7-24i{\right)}^{\frac{1}{2}}=x-iy$Squaring on both sides $\left(-7-24i\right)={\left(x-iy\right)}^{2}$$⇒$$\left(-7-24i\right)={x}^{2}-2xyi-{y}^{2}$Step 2. Equating real and imaginary part ${x}^{2}-{y}^{2}=-7$ and $2xy=24$ $\begin{array}{rcl}& ⇒& xy=\frac{24}{2}\\ & =& 12\end{array}$As we know,$\begin{array}{rcl}{\left({x}^{2}+{y}^{2}\right)}^{2}& =& {\left({x}^{2}-{y}^{2}\right)}^{2}+4{x}^{2}{y}^{2}\\ & =& {\left(-7\right)}^{2}+{\left(2xy\right)}^{2}\\ & =& 49+{\left(24\right)}^{2}\\ & =& 625\end{array}$Taking Square root on both sides$\left({x}^{2}+{y}^{2}\right)=25$Hence, the value of ${\mathbit{x}}^{\mathbf{2}}\mathbf{+}{\mathbit{y}}^{\mathbf{2}}$ is $\mathbf{25}$.

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