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

Last updated date: 18th Jul 2024
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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.

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 $\sqrt{-7+24i}=x+iy$.