Chapter 9.8, Problem 23WE

To determine

To calculate: To find the square roots of complex number

Answer to Problem 23WE

Square roots of complex number are $\left(-3+2i\right)$ and $\left(3-2i\right)$

Explanation of Solution

Given information: Complex number is $5+12i$

Calculation:

Given the complex number as follows:

  $5+12i$

Let the square root of complex number be $x+yi$

Thus,

  $\begin{array}{l}{\left(x+yi\right)}^2=5+12i\\ {\left(x\right)}^2+{\left(yi\right)}^2-2xyi=5+12i\\ \left(x^2-y^2\right)-2xyi=5+12i\end{array}$

Equating the real and imaginary components:

  $\begin{array}{l}{x}^2-y^2=5---(1)\\ -2xy=12----(2)\end{array}$

From equation (2),

  $\begin{array}{l}-2xy=12\\ x=-\frac{6}{y}\end{array}$

Substituting the value in (1),

  $\begin{array}{l}{\left(-\frac{6}{y}\right)}^2-y^2=5\\ \frac{36}{y^2}-y^2=5\\ 36-y^4=5y^2\\ y^4+5y^2-36=0\\ y^4+9y^2-4y^2-36=0\\ \left(y^2+9\right)\left(y^2-4\right)=0\\ y^2=-9,4\end{array}$

  $y^2\ne -9$ as its imaginary

Thus, $y^2=4$

  $y=\pm 2$

When $y=2$

  $\begin{array}{l}x=-\frac{6}{2}\\ x=-3\end{array}$

When $y=-2$

  $\begin{array}{l}x=-\frac{6}{\left(-2\right)}\\ x=3\end{array}$

Conclusion:

Hence, square roots of complex number are $\left(-3+2i\right)$ and $\left(3-2i\right)$