Square root of complex numbers - Practice Questions & MCQ

Let $z=x+i y$, is the complex number whose square root we have to find

Since the square root of a complex number must be a complex number, 

so let $\mathrm{z}^{1 / 2}=\mathrm{a}+\mathrm{ib}$
Now squaring both sides

$
z=x+i y=(a+i b)^2=a^2-b^2+2 i a b
$

Now comparing real and imaginary part and finding the value of a and b in terms of x and y 

$
\begin{aligned}
& a^2-b^2=x \\
& 2 a b=y \\
& a^2+b^2=\sqrt{\left(a^2-b^2\right)^2+4 a^2 b^2}=\sqrt{x^2+y^2}=|z|
\end{aligned}
$
Solving (i) and (iii) we get

$
2 \mathrm{a}^2=\mathrm{x}+|\mathrm{z}| \Rightarrow \mathrm{a}= \pm \sqrt{\frac{\mathrm{x}+|\mathrm{z}|}{2}}
$
Similarly we find $b= \pm \sqrt{\frac{|z|-x}{2}}$
So $\sqrt{\mathrm{z}}= \pm\left(\sqrt{\frac{|z|+\operatorname{Re}(\mathrm{z})}{2}}+\mathrm{i} \sqrt{\frac{|z|-\operatorname{Re}(\mathrm{z})}{2}}\right)$

if $\operatorname{Im}(z)>0$ otherwise there will be a -ve sign between the real and imaginary parts of the square root of $z$.

Note:
1. Students do not need to remember this formula. But, they are required to know the procedure to find the square root of a complex number.
2. If $a+i b$ is one of the square roots of $z$, then the other square root must be $-(a+i b)$

Complex Equations
To find the solution of the complex equation we substitute $z=x+i y$ and find the value of $x$ and $y$ by comparing real and imaginary parts of the equation obtained. $\mathrm{z}=\mathrm{x}+\mathrm{iy}$ is the required solution.