Chapter 7.2, Problem 37WE

To determine

Tosolve:The given radical equation.

Answer to Problem 37WE

The solution of the given radical equation in the simplest radical form is

  $\begin{array}{l}z=\frac{3}{2}+\left(1-i\frac{1}{2}\right)i\\ z=\frac{3}{2}+\left(1+i\frac{1}{2}\right)i\\ z=2+i\\ z=1+i\end{array}$ .

Explanation of Solution

Given:

The given radical equation is

  $z^2-\left(3+2i\right)z+\left(1+3i\right)=0$

Calculation:

The solution of the given radical equation is obtained in the following manner.

Substitute $z=\left(x+iy\right)$

The equation becomes:

  ${\left(x+iy\right)}^2-\left(3+2i\right)\left(x+iy\right)+\left(1+3i\right)=0$

Expand the above expression:

  $\begin{array}{l}{\left(x+iy\right)}^2-\left(3+2i\right)\left(x+iy\right)+\left(1+3i\right)=0\\ \left(x^2-y^2-3x+2y+1\right)+i\left(3-2x-3y+2xy\right)=0+i0\end{array}$

Equate the real and imaginary part of the complex number. The complex numbers are equal only when their real and imaginary parts are equal.

Develop the system of equations and then solve these.

  $\left[\begin{array}{l}{x}^2-y^2-3x+2y+1=0\\ 3-2x-3y+2xy=0\end{array}\right]$

The solution of these system of equation is

  $\left[\begin{array}{l}x=\frac{3}{2}y=1-i\frac{1}{2}\\ x=\frac{3}{2}y=1-i\frac{1}{2}\\ x=2y=1\\ x=1y=1\end{array}\right]$

Again substitute $z=\left(x+iy\right)$

The solution is

  $\begin{array}{l}z=\frac{3}{2}+\left(1-i\frac{1}{2}\right)i\\ z=\frac{3}{2}+\left(1+i\frac{1}{2}\right)i\\ z=2+i\\ z=1+i\end{array}$

Therefore, the solution of the given radical equation is

  $\begin{array}{l}z=\frac{3}{2}+\left(1-i\frac{1}{2}\right)i\\ z=\frac{3}{2}+\left(1+i\frac{1}{2}\right)i\\ z=2+i\\ z=1+i\end{array}$ .