July 16, 2024

# How to find Square root of a complex number

How to find Square root of a complex number

A number of the form $z=x+iy$

where $x,y&space;\in&space;\mathbb{R}$ 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

$\sqrt{x+iy}=&space;\pm&space;\left&space;[&space;\sqrt{\frac{\left&space;|&space;z&space;\right&space;|+x}{2}}+i\sqrt{\frac{\left&space;|&space;z&space;\right&space;|-x}{2}}&space;\right&space;]$       for  $y>0$    and

$=&space;\pm&space;\left&space;[&space;\sqrt{\frac{\left&space;|&space;z&space;\right&space;|+x}{2}}-i\sqrt{\frac{\left&space;|&space;z&space;\right&space;|-x}{2}}&space;\right&space;]$          for       $y<&space;0$

Proof. Let square root of $x+iy&space;\,&space;\,&space;is&space;\,&space;\,&space;a+ib$. That is  $\sqrt{x+iy}=a+ib$  where $a$ and $b$ in $\mathbb{R}$. Now square both the sides we get

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

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

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

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

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

Similarly $b=\pm&space;\sqrt{\frac{\left&space;|&space;z&space;\right&space;|-x}{2}}$. 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

$\sqrt{x+iy}=&space;\pm&space;\left&space;[&space;\sqrt{\frac{\left&space;|&space;z&space;\right&space;|+x}{2}}+i\sqrt{\frac{\left&space;|&space;z&space;\right&space;|-x}{2}}&space;\right&space;]$       for  $y>0$    and

$=&space;\pm&space;\left&space;[&space;\sqrt{\frac{\left&space;|&space;z&space;\right&space;|+x}{2}}-i\sqrt{\frac{\left&space;|&space;z&space;\right&space;|-x}{2}}&space;\right&space;]$          for       $y<&space;0$

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

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

Applying the formula for square root we get

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

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

$=\pm&space;(2-3i)$

Trick to find the square root of a complex number:

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

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

Ex.  Find the square root of 7+4i.

Sol. Find the number $\frac{\left&space;|&space;y&space;\right&space;|}{2}$  which is Equal to 12. Now factor 12 in such a way that difference of square of these factors is equal to the real number  $x=7$.

12= 4×3   and $4^{2}-3^{2}=16-9=7$. Therefore $\sqrt{7+4i}=\pm&space;(4+3i)$.

Ex.   Find the square root of $(-6+8i)$.

Sol. Here $\frac{\left&space;|&space;y&space;\right&space;|}{2}=\frac{8}{2}=4$   and $4=\sqrt{2}\times&space;\sqrt{8}$ such that $(\sqrt{2})^{2}-(\sqrt{8})^{2}=2-8=-6$

Thus    $\sqrt{(-6+8i)}=\pm&space;(\sqrt{2}+\sqrt{8}i)=\pm&space;(\sqrt{2}+2\sqrt{2}i)$

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