Question

If $\sqrt{\left(a+ib\right)}=x+iy$, then the possible value of $\sqrt{\left(a-ib\right)}$ is

Answer 1

Finding the possible value of $\sqrt{\left(a-ib\right)}$:

Step 1:Square both the above equation,

Given relation $\sqrt{\left(a+ib\right)}=x+iy$

$\begin{array}{rcl}\left(a+ib\right)& =& {\left(x+iy\right)}^2\\ & =& x^2-y^2+i2xy\end{array}$

Step 2: Compare the real and imaginary parts, obtain

$\begin{array}{rcl}a& =& x^2-y^2\\ b& =& 2xy\end{array}$

From above

$\begin{array}{rcl}a-ib& =& x^2-y^2-i2xy\\ & =& {\left(x-iy\right)}^2\end{array}$

Take square roots on both the sides of the above equation,

$\begin{array}{rcl}\sqrt{\left(a-ib\right)}& =& \left(x-iy\right)\end{array}$

Hence, the value of $\sqrt{\left(a-ib\right)}$ is $\left(x-iy\right)$.