Question

Find the square root of the given complex number:
$z=-16+30i$

Answer 1

Hint: Here, we will assume that the square root of the given complex number to be x+iy. It means that the given complex number is the square of x+iy. Using this we can calculate the square root of -16 + 30i.

Complete step-by-step answer:

A complex number is a number that can be expressed in the form a+ib, where a and b are real numbers and I is the solution of the equation ${{x}^{2}}=-1$. Since, no real number satisfies this equation, $i$ is called an imaginary number.

 Let us assume that the square root of the given complex number that is $z=-16+30i$ is $x+iy$.

Since, $z=-16+30i$

We can write:

On comparing the real part and imaginary part on both sides of the above equation, we get:

$\begin{align}

  & {{x}^{2}}-{{y}^{2}}=-16........\left( 1 \right) \\

 & 2xy=30.............\left( 2 \right) \\

\end{align}$

Now, from equation (1), we can write:

$\begin{align}

  & \left( {{x}^{2}}-{{y}^{2}} \right)={{\left( -16 \right)}^{2}} \\

 & \Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-4{{x}^{2}}{{y}^{2}}=256 \\

\end{align}$

Using equation (2), that is putting the value of 2xy from equation (2) in the above equation, we get:

$\begin{align}

  & {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-{{\left( 30 \right)}^{2}}=256 \\

 & \Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-900=256 \\

 & \Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=1156 \\

 & \Rightarrow \left( {{x}^{2}}+{{y}^{2}} \right)=\sqrt{1156}=34.............\left( 3 \right) \\

\end{align}$

On adding equation (1) and (3), we get:

$\begin{align}

  & {{x}^{2}}-{{y}^{2}}+{{x}^{2}}+{{y}^{2}}=-16+34 \\

 & \Rightarrow 2{{x}^{2}}=18 \\

 & \Rightarrow {{x}^{2}}=9 \\

 & \Rightarrow x=\pm 3 \\

\end{align}$

When x=3, on putting x = 3 in equation (2), we get:

$\begin{align}

  & 2\times 3\times y=30 \\

 & \Rightarrow y=\dfrac{30}{6}=5 \\

\end{align}$

On putting x = -3, in equation (2), we get:

$\begin{align}

  & 2\times \left( -3 \right)\times y=30 \\

 & \Rightarrow y=\dfrac{-30}{6}=-5 \\

\end{align}$

So, the values of x are 3, -3 and that of y are 5 and -5.

Hence, the square root of the complex number $-16+30i$ are $3+5i$ and $-3-5i$.

Note: Students should note here that the value of ${{i}^{2}}$ is -1. Thus, while calculating the value of ${{\left( x+iy \right)}^{2}}$, it should be kept in mind that ${{i}^{2}}$ should be replaced by -1. The calculations must be done properly to avoid mistakes.