Question

find the square root of 6+8i
​

Answer 1

Answer:

Since xy>0, so x and y are of the same sign. ∴ (x=2√2andy=√2)or(x=-2√2andy=-√2). Hence, √6+8i=(2√2+√2i)or(-2√2-√2i).

Answer 2

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

Given complex number is

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

Let we assume that

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

On squaring both sides, we get

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

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

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

So, on comparing Real and Imaginary parts, we get

$\bf\implies \:\boxed{ \tt{ \: {x}^{2} - {y}^{2} = 6 \: \: }} - - - (2)$

and

$\bf\implies \:\boxed{ \tt{ \: 2xy = 8\: \: }} - - - (3)$

We know,

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

So, on substituting the values, we get

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = \sqrt{ {6 }^{2} + {8}^{2} }$

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = \sqrt{36 + 64 }$

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = \sqrt{100 }$

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

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

$\rm :\longmapsto\:2 {x}^{2} = 16$

$\rm :\longmapsto\:{x}^{2} = 8$

$\bf\implies \:x \: = \: \pm \: 2 \sqrt{2} - - - - (5)$

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

$\rm :\longmapsto\: {2y}^{2} = 4$

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

$\bf\implies \:y \: = \: \pm \sqrt{2} - - - - (6)$

Hᴇɴᴄᴇ,

➢ Pair of points of the for x and y are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 2 \sqrt{2} & \sf \sqrt{2} \\ \\ \sf - 2 \sqrt{2} & \sf - \sqrt{2} \end{array}} \\ \end{gathered}$

Hence,

$\red{\bf\implies \:\boxed{ \tt{ \: \sqrt{6 + 8i} = \pm \: (2 \sqrt{2} + i \sqrt{2} )\: \: }}}$