Question

Find the square root of the following complex number

$7 + 24i$
​

Answer 1

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

Let assume that

$\rm \: \sqrt{7 + 24i} = x + iy - - - - (1)$

On squaring both sides, we get

$\rm \: 7 + 24i = {(x + iy)}^{2}$

$\rm \: 7 + 24i = {x}^{2} + {i}^{2} {y}^{2} + 2ixy$

$\rm \: 7 + 24i = {x}^{2}-{y}^{2} + 2ixy \: \: \: \{ \: \because \: {i}^{2} = - 1 \}$

On comparing real and Imaginary parts, we get

$\rm \: {x}^{2} - {y}^{2} = 7 - - - (2)$

$\rm \: 2xy = 24 - - - (3)$

Now, we know that,

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

On substituting the values from equation (2) and (3), we get

$\rm \: {x}^{2} + {y}^{2} = \sqrt{ {(7)}^{2} + {(24)}^{2} }$

$\rm \: {x}^{2} + {y}^{2} = \sqrt{49 + 576}$

$\rm \: {x}^{2} + {y}^{2} = \sqrt{625}$

$\rm \: {x}^{2} + {y}^{2} = 25 - - - (4)$

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

$\rm \: {2x}^{2} = 32$

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

$\rm\implies \:x = \: \pm \: 4$

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

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

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

$\rm\implies \:y \: = \: \pm \: 3$

Hence, the values of x and y are as follow :-

$\rm \: As \: 2xy = 24$

$\rm\implies \:xy > 0$

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 4 & \sf 3 \\ \\ \sf - 4 & \sf - 3 \end{array}} \\ \end{gathered} </p><p>$

Hence,

$\rm\implies \:\rm \: \sqrt{7 + 24i} = \: \pm \: (4 + 3i)$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Complex \: number & \bf arg(z) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf x + iy & \sf  {tan}^{ - 1}\bigg |\dfrac{y}{x} \bigg|   \\ \\ \sf  - x + iy & \sf \pi - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf  - x - iy & \sf  - \pi + {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf x - iy & \sf  - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \end{array}} \\ \end{gathered}$