Question

$find the value of \sqrt{ - 8 - 6i}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\: \sqrt{ - 8 - 6i}$

Let assume that

$\rm :\longmapsto\: \sqrt{ - 8 - 6i} = x + iy - - - (1)$

On squaring both sides, we get

$\rm :\longmapsto\: - 8 - 6i = {(x + iy)}^{2}$

$\rm :\longmapsto\: - 8 - 6i = {x}^{2} + {i}^{2} {y}^{2} + 2xyi$

$\rm :\longmapsto\: - 8 - 6i = {x}^{2} - {y}^{2} + 2xyi$

On comparing, we get

$\rm :\longmapsto\: {x}^{2} - {y}^{2} = - 8 - - - (2)$

and

$\rm :\longmapsto\:2xy = - 6 - - (3)$

Now,

We know that,

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = \sqrt{ {( {x}^{2} - {y}^{2})}^{2} + {(2xy)}^{2} }$

$\rm \: = \: \: \sqrt{ {( - 8)}^{2} + {( - 6)}^{2} }$

$\rm \: = \: \: \sqrt{64 + 36}$

$\rm \: = \: \: \sqrt{100}$

$\rm \: = \: \: 10$

$\bf\implies \: {x}^{2} + {y}^{2} = 10 - - - (4)$

On adding equation (2) and equation (4), we get

$\rm :\longmapsto\: {2x}^{2} = 2$

$\rm :\longmapsto\: {x}^{2} = 1$

$\bf\implies \:x = \: \pm \: 1 - - - (5)$

On Subtracting equation (2) from equation (4), we get

$\rm :\longmapsto\: {2y}^{2} = 18$

$\rm :\longmapsto\: {y}^{2} = 9$

$\bf\implies \:y= \: \pm \: 3 - - - (6)$

Since,

From equation (3), we have

$\rm :\longmapsto\:2xy = - 6$

$\rm :\implies\:xy < 0$

$\rm :\implies\:x > 0 \: and \: y < 0 \: \: or \: \: x < 0 \: and \: y > 0$

Hence,

The values of x and y are as follow :

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 1 & \sf - 3 \\ \\ \sf - 1 & \sf 3 \end{array}} \\ \end{gathered}$

Therefore,

$\rm :\longmapsto\: \sqrt{ - 8 - 6i} = \: \pm \: (1 - 3i)$

Additional Information :-

If z is a complex number such that z = x + iy then

$\rm :\longmapsto\:Re(z) = x$

$\rm :\longmapsto\:Im(z) = y$

$\rm :\longmapsto\: |z| = \sqrt{ {x}^{2} + {y}^{2} }$

$\rm :\longmapsto\: \overline{z} \: = \: x - iy$

$\rm :\longmapsto\: z\overline{z} \: = \: { |z| }^{2}$