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Find the square root of $\left( -7+24i \right)$ .

Answer
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Hint: Take -7 + 24i = a + bi, where a = -7 and b = 24. Assume the square root of $\left( -7+24i \right)$ equal to $\left( x+iy \right)$. Take square and solve the equation obtained. Find the equation connecting x, y, a and b. Then find the roots.

Complete step-by-step answer:
Let consider a complex number a + ib.
Let the square root of (a + ib) be x + iy
That is $\sqrt{a+ib}=x+iy\text{, where }x,y\in R$
Now square on both sides,
$\begin{align}
  & {{\left( \sqrt{a+ib} \right)}^{2}}={{\left( x+iy \right)}^{2}} \\
 & \Rightarrow a+ib={{x}^{2}}+2iy+{{i}^{2}}{{y}^{2}} \\
\end{align}$
We know ${{i}^{2}}=-1$
$\begin{align}
  & \therefore a+ib={{x}^{2}}+2iy+\left( -1 \right){{y}^{2}} \\
 & a+ib={{x}^{2}}-{{y}^{2}}+2iyx...............\left( 1 \right) \\
\end{align}$
Let us take ${{x}^{2}}-{{y}^{2}}=a..............\left( 2 \right)\text{ }\because {{\left( a+b \right)}^{2}}={{\left( a-b \right)}^{2}}+4ab$
$2xy=b.......................\left( 3 \right)$
$\begin{align}
  & {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}={{\left( {{x}^{2}}={{y}^{2}} \right)}^{2}}+4{{x}^{2}}{{y}^{2}}={{a}^{2}}+{{b}^{2}} \\
 & \Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}={{a}^{2}}+{{b}^{2}} \\
\end{align}$
Take square root on both sides.
$\begin{align}
  & \sqrt{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}}=\sqrt{{{a}^{2}}+{{b}^{2}}} \\
 & {{x}^{2}}+{{y}^{2}}=\sqrt{{{a}^{2}}+{{b}^{2}}}................\left( 4 \right) \\
\end{align}$
Add equation (2) and equation (4).
\[\dfrac{\begin{align}
  & {{x}^{2}}+{{y}^{2}}=\sqrt{{{a}^{2}}+{{b}^{2}}} \\
 & {{x}^{2}}-{{y}^{2}}=a\text{ } \\
\end{align}}{2{{x}^{2}}=\sqrt{{{a}^{2}}+{{b}^{2}}}+a}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{matrix}
   \therefore {{x}^{2}}=\dfrac{\sqrt{{{a}^{2}}+{{b}^{2}}}+a}{2} \\
   \therefore x=\pm \sqrt{\dfrac{\sqrt{{{a}^{2}}+{{b}^{2}}}+a}{2}} \\
\end{matrix}\]
Now subtract equation (2) and equation (4).
\[\begin{align}
  & \dfrac{\begin{align}
  & {{x}^{2}}+{{y}^{2}}=\sqrt{{{a}^{2}}+{{b}^{2}}} \\
 & {}^{-}{{x}^{2\left( + \right)}}-{{y}^{2}}={}^{-}a \\
\end{align}}{2{{y}^{2}}=\sqrt{{{a}^{2}}+{{b}^{2}}}-a} \\
 & {{y}^{2}}=\dfrac{\sqrt{{{a}^{2}}+{{b}^{2}}}-a}{2}\ \ \ \ \therefore y=\pm \sqrt{\dfrac{\sqrt{{{a}^{2}}+{{b}^{2}}}-a}{2}} \\
\end{align}\]
Now we have got the value of x and y.
$x=\pm \sqrt{\dfrac{\sqrt{{{a}^{2}}+{{b}^{2}}}+a}{2}}\ \ and\ y=\pm \sqrt{\dfrac{\sqrt{{{a}^{2}}+{{b}^{2}}}-a}{2}}\ .............\left( 5 \right)$
We have been asked to find the square root of (-7 +24i)
$\sqrt{\left( -7+24i \right)}=x+iy$ i.e., take square root of (-7 + 24i) equal to x + iy
Where a + ib = -7 + 24i
$\therefore $ a = -7
b = 24
Now squaring on both sides;
$\begin{align}
  & {{\left( \sqrt{-7+24i} \right)}^{2}}={{\left( x+iy \right)}^{2}} \\
 & -7+24i={{x}^{2}}+2xyi+{{\left( iy \right)}^{2}} \\
 & \therefore {{i}^{2}}=-1 \\
 & \Rightarrow -7+24i={{x}^{2}}-{{y}^{2}}+2xyi..............\left( 6 \right) \\
\end{align}$
Now compare equation (1) and equation (6) which is similar.
$\begin{align}
  & \therefore {{x}^{2}}-{{y}^{2}}=a\Rightarrow {{x}^{2}}-{{y}^{2}}=-7 \\
 & 2xy=b\Rightarrow 2xy=24 \\
\end{align}$
Now substitute the value of a = -7 and b = 24 in equation (5).
$\begin{align}
  & x=\pm \sqrt{\dfrac{\sqrt{{{a}^{2}}+{{b}^{2}}}+a}{2}}=\pm \sqrt{\dfrac{\sqrt{{{\left( -7 \right)}^{2}}+{{24}^{2}}}+\left( -7 \right)}{2}}=\pm \sqrt{\dfrac{25-7}{2}}=\pm \sqrt{\dfrac{18}{2}}=\pm 3 \\
 & y=\pm \sqrt{\dfrac{\sqrt{{{a}^{2}}+{{b}^{2}}}-a}{2}}=\pm \sqrt{\dfrac{\sqrt{{{\left( -7 \right)}^{2}}+{{24}^{2}}}-\left( -7 \right)}{2}}=\pm \sqrt{\dfrac{25+7}{2}}=\pm \sqrt{\dfrac{32}{2}}=\pm 4 \\
 & \therefore x+iy=\pm 3\pm 4 \\
\end{align}$
$\therefore $The roots are $+\left( 3+i4 \right)\ \ and\ \ -\left( 3+i4 \right)$.

Note: The proof of $\sqrt{a+ib}=x+iy$ is similar to our question$\left( -7+24i \right)$. Compare the general solution to $\sqrt{-7+24i}=x+iy$.