Question

How do you find the square root of a complex number?

Answer 1

To find the square root of a complex number:

Step1. Assume the complex number :

Now, let us derive the formula to find the square root of a complex number $a+ib$. Assume the square root of complex number $a+ib$ to be $x+iy$, that is, $\sqrt{a+ib}=\pm (x+iy)$.

Now squaring both sides of the equation, we have

${\left[\sqrt{a+ib}\right]}^2={(x+iy)}^2$

$\Rightarrow a+ib=x^2+{\left(iy\right)}^2+i2xy$

$\Rightarrow a+ib=x^2-y^2+i2xy$ [Because ${\mathbf{i}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{1}$]

Step2. To find the real and imaginary parts:

Comparing real and imaginary parts on both sides, we have

$a=x^2-y^2.........\left(i\right)$

and $b=2xy$

We know that ${(x^2+y^2)}^2={(x^2-y^2)}^{}+4x^2{y}^2$

$\Rightarrow {(x^2+y^2)}^2=a^2+b^2$

$\Rightarrow x^2+y^2=\sqrt{a^2+b^2}.........\left(ii\right)$ [Because $x^2+y^2$ is always positive as sum of squares of non-zero real numbers is always greater than zero]

Now, Solving the equation (i) and (ii), we get

$x=\pm \sqrt{\frac{(a^2+b^2)+a}{2}}$ and $y=\pm \sqrt{\frac{(a^2+b^2)-a}{2}}$

Since$2xy=b$, therefore we have

1. $x$ and $y$ have the same sign if $b>0$
2. $x$ and $y$ have opposite signs if $b<0$

Therefore the square root of a complex number $a+ib(b\ne 0)$ is given by

$\sqrt{a+ib}=\pm \sqrt{\frac{(a^2+b^2)+a}{2}}+\left(\frac{ib}{\left|b\right|}\right)\sqrt{\frac{(a^2+b^2)-a}{2}}$

Hence, the formula to determine the square root of a complex number is:

$\sqrt{\mathrm{a}+\mathrm{ib}}=\pm (\sqrt{\frac{\left|\mathrm{z}\right|+\mathrm{a}}{2}}+\mathrm{i}\frac{\mathrm{b}}{\left|\mathrm{b}\right|}\sqrt{\frac{\left|\mathrm{z}\right|-\mathrm{a}}{2}})$

Where $z=a+ib$and $b$ not equal to zero.