Question

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

Answer 1

Step 1. Remove square root part :

Given that $(-7-24i{)}^{\frac{1}{2}}=x-iy$

Squaring on both sides

$(-7-24i)={(x-iy)}^2$

$\Rightarrow$$(-7-24i)=x^2-2xyi-y^2$

Step 2. Equating real and imaginary part

$x^2-y^2=-7$ and

$2xy=24$

$\begin{array}{rcl}& \Rightarrow & xy=\frac{24}{2}\\ & =& 12\end{array}$

As we know,

$\begin{array}{rcl}{(x^2+y^2)}^2& =& {(x^2-y^2)}^2+4x^2{y}^2\\ & =& {(-7)}^2+{\left(2xy\right)}^2\\ & =& 49+{\left(24\right)}^2\\ & =& 625\end{array}$

Taking Square root on both sides

$(x^2+y^2)=25$

Hence, the value of ${\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}$ is $\mathbf{25}$.