Answer

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Hint: For solving this problem, first assume the square root of the desired number in the form of two variables in the rectangular system. Now for obtaining a relationship between these two variables, we use squaring of both sides technique and then equating the real part with real number and imaginary part with imaginary number. By using this methodology, we can easily evaluate the two variables.

According to the problem statement, we are required to find the square root of complex number $-8-6i$.

Let the square root of the complex number be $a+ib$ such that $\sqrt{-8-6i}=a+ib\ldots (1)$

Now for obtaining the relationship between a and b, we square both the sides of equation (1)

${{\left( \sqrt{-8-6i} \right)}^{2}}={{\left( a+ib \right)}^{2}}$

Now, by using the expression ${{(x+y)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy$, the above expression can be simplified as:

$\begin{align}

& -8-6i={{a}^{2}}+{{\left( ib \right)}^{2}}+2\cdot a\cdot \left( ib \right) \\

& -8-6i={{a}^{2}}-{{b}^{2}}+2\cdot a\cdot \left( ib \right) \\

\end{align}$

Now, on equating the terms in such a manner that real part is equated with real number and imaginary part is equated with imaginary number, we get

$\begin{align}

& {{a}^{2}}-{{b}^{2}}=-8\ldots (2) \\

& 2ab=-6 \\

& ab=\dfrac{-6}{2}=-3\ldots (3) \\

\end{align}$

By using equation (3) $\left( b=\dfrac{-3}{a} \right)$, equation (2) can be reduced as

$\begin{align}

& {{a}^{2}}-{{\left( \dfrac{-3}{a} \right)}^{2}}=-8 \\

& {{a}^{2}}-\dfrac{9}{{{a}^{2}}}=-8 \\

& \dfrac{{{a}^{4}}-9}{{{a}^{2}}}=-8 \\

& {{a}^{4}}+8{{a}^{2}}-9=0\ldots (4) \\

\end{align}$

Now, by using splitting the middle term method, we factorise the above equation (4) as

$\begin{align}

& {{a}^{4}}-{{a}^{2}}+9{{a}^{2}}-9=0 \\

& {{a}^{2}}\left( {{a}^{2}}-1 \right)+9\left( {{a}^{2}}-1 \right)=0 \\

& \left( {{a}^{2}}+9 \right)\left( {{a}^{2}}-1 \right)=0 \\

& {{a}^{2}}=-9,{{a}^{2}}=1 \\

& a=\sqrt{-9},a=\sqrt{1} \\

& a=\pm 3i,a=\pm 1 \\

\end{align}$

Therefore, the value of a is $a=\pm 3i,a=\pm 1$.

Putting the values of a in equation (3), we get

$\begin{align}

& b=\dfrac{-3}{\pm 3i},b=\dfrac{-3}{\pm 1} \\

& b=\mp \dfrac{1}{i},b=\mp 3 \\

& b=\mp \dfrac{i}{{{i}^{2}}},b=\mp 3 \\

& b=\pm i,b=\mp 3 \\

\end{align}$

The value of b is $b=\pm i,b=\mp 3$

$\begin{align}

& Z=a+ib \\

& Z=\pm \left( 3i+i\times i \right) \\

& Z=\pm \left( 3i+{{i}^{2}} \right) \\

& Z=\pm \left( -1+3i \right) \\

& \text{Also, Z=}\pm \left( 1-3i \right) \\

\end{align}$

Hence, the square root of complex numbers $-8-6i$ are $\pm \left( -1+3i \right)\text{ and }\pm \left( 1-3i \right)$.

Note: This problem can also be solved alternatively by first converting the complex number into Euler's form and then taking the square root of the number. The Euler form of $-8-6i$ is $r{{e}^{i\theta }}=\pm 10{{e}^{i0.20555\pi }}$. In this way, a solution is obtained with less calculations.

__Complete step-by-step solution -__According to the problem statement, we are required to find the square root of complex number $-8-6i$.

Let the square root of the complex number be $a+ib$ such that $\sqrt{-8-6i}=a+ib\ldots (1)$

Now for obtaining the relationship between a and b, we square both the sides of equation (1)

${{\left( \sqrt{-8-6i} \right)}^{2}}={{\left( a+ib \right)}^{2}}$

Now, by using the expression ${{(x+y)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy$, the above expression can be simplified as:

$\begin{align}

& -8-6i={{a}^{2}}+{{\left( ib \right)}^{2}}+2\cdot a\cdot \left( ib \right) \\

& -8-6i={{a}^{2}}-{{b}^{2}}+2\cdot a\cdot \left( ib \right) \\

\end{align}$

Now, on equating the terms in such a manner that real part is equated with real number and imaginary part is equated with imaginary number, we get

$\begin{align}

& {{a}^{2}}-{{b}^{2}}=-8\ldots (2) \\

& 2ab=-6 \\

& ab=\dfrac{-6}{2}=-3\ldots (3) \\

\end{align}$

By using equation (3) $\left( b=\dfrac{-3}{a} \right)$, equation (2) can be reduced as

$\begin{align}

& {{a}^{2}}-{{\left( \dfrac{-3}{a} \right)}^{2}}=-8 \\

& {{a}^{2}}-\dfrac{9}{{{a}^{2}}}=-8 \\

& \dfrac{{{a}^{4}}-9}{{{a}^{2}}}=-8 \\

& {{a}^{4}}+8{{a}^{2}}-9=0\ldots (4) \\

\end{align}$

Now, by using splitting the middle term method, we factorise the above equation (4) as

$\begin{align}

& {{a}^{4}}-{{a}^{2}}+9{{a}^{2}}-9=0 \\

& {{a}^{2}}\left( {{a}^{2}}-1 \right)+9\left( {{a}^{2}}-1 \right)=0 \\

& \left( {{a}^{2}}+9 \right)\left( {{a}^{2}}-1 \right)=0 \\

& {{a}^{2}}=-9,{{a}^{2}}=1 \\

& a=\sqrt{-9},a=\sqrt{1} \\

& a=\pm 3i,a=\pm 1 \\

\end{align}$

Therefore, the value of a is $a=\pm 3i,a=\pm 1$.

Putting the values of a in equation (3), we get

$\begin{align}

& b=\dfrac{-3}{\pm 3i},b=\dfrac{-3}{\pm 1} \\

& b=\mp \dfrac{1}{i},b=\mp 3 \\

& b=\mp \dfrac{i}{{{i}^{2}}},b=\mp 3 \\

& b=\pm i,b=\mp 3 \\

\end{align}$

The value of b is $b=\pm i,b=\mp 3$

$\begin{align}

& Z=a+ib \\

& Z=\pm \left( 3i+i\times i \right) \\

& Z=\pm \left( 3i+{{i}^{2}} \right) \\

& Z=\pm \left( -1+3i \right) \\

& \text{Also, Z=}\pm \left( 1-3i \right) \\

\end{align}$

Hence, the square root of complex numbers $-8-6i$ are $\pm \left( -1+3i \right)\text{ and }\pm \left( 1-3i \right)$.

Note: This problem can also be solved alternatively by first converting the complex number into Euler's form and then taking the square root of the number. The Euler form of $-8-6i$ is $r{{e}^{i\theta }}=\pm 10{{e}^{i0.20555\pi }}$. In this way, a solution is obtained with less calculations.

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