Square Root Of A Complex Number

How to find Square root of a complex number

A number of the form $z=x+iy$ where $x,y \in \mathbb{R}$

and $i=\sqrt{-1}$ is called a complex number where $x$

is called as real part and $y$ is called imaginary part of complex number. To find the square root of a complex number, we will assume that $\sqrt{x+iy}=a+ib$

. Then square both the sides and compare real and imaginary part to find the value of $a$ and $b$

, which will give us the square root.

Formula for finding square root of a complex number:

The square root of $z=x+iy$ is

$\sqrt{x+iy}= \pm \left [ \sqrt{\frac{\left | z \right |+x}{2}}+i\sqrt{\frac{\left | z \right |-x}{2}} \right ]$

for $y>0$ and

$= \pm \left [ \sqrt{\frac{\left | z \right |+x}{2}}-i\sqrt{\frac{\left | z \right |-x}{2}} \right ]$

for $y< 0$

Proof. Let square root of $x+iy \, \, is \, \, a+ib$

. That is $\sqrt{x+iy}=a+ib$ where $a$

and $b$ in $\mathbb{R}$

. Now square both the sides we get

$x+iy= (a+ib)^{2}=a^{2}-b^{2}+2iab$

Equating real and imaginary parts we get $a^{2}-b^{2}=x.....(i)\, \,\, and\,\,\, 2ab=y......(ii)$

Now $(a^{2}+b^{2})^{2}=(a^{2}-b^{2})^{2}+4a^{2}b^{2}$ $\Rightarrow a^{2}+b^{2}=\sqrt{x^{2}+y^{2}} ............(iii)$

Solving (i) and (iii) we get $2a^{2}=(\sqrt{x^{2}+y^{2}})+x=\left | z \right |+x$

$a^{2}=\frac{\left | z \right |+x}{2}\Rightarrow a=\pm \sqrt{\frac{\left | z \right |+x}{2}}$

Similarly $b=\pm \sqrt{\frac{\left | z \right |-x}{2}}$ . Since 2ab=y , it is clear that both $a$

and $b$ have the same sign when $y$

is positive and $a$ and $b$

have different sign when $y$ is negative. Therefore

$\sqrt{x+iy}= \pm \left [ \sqrt{\frac{\left | z \right |+x}{2}}+i\sqrt{\frac{\left | z \right |-x}{2}} \right ]$

for $y>0$ and

$= \pm \left [ \sqrt{\frac{\left | z \right |+x}{2}}-i\sqrt{\frac{\left | z \right |-x}{2}} \right ]$

for $y< 0$

Example : Find the square root of (-5-12i) .

Sol. Here $\left | (-5-12i)\right |= \sqrt{(-5)^{2}+(-12)^{2}}=\sqrt{25+144}=13$

Applying the formula for square root we get

$\sqrt{(-5-12i)}=\pm \left [ \sqrt{\frac{\left | z \right |+x}{2}} - i \sqrt{\frac{\left | z \right |-x}{2}}\right ]$ ( $\because b$

is negative)

$\, \, \, \, \, \, \, \, \, \, =\pm \left ( \sqrt{\frac{13-5}{2}}-i\sqrt{\frac{13+5}{2}} \right )$

$=\pm (2-3i)$

Trick to find the square root of a complex number:

To find $\sqrt{x+iy}$ , follow the following steps:

First find the number $\frac{\left | y \right |}{2}$ .
Now factorise the given number in such a way that difference of square of these factors is equal to the real number $(x)$ .