Chapter 7.4, Problem 62SR

To determine

The solution of the two numbers.

Answer to Problem 62SR

  $6+2i$ and $6-2i$ .

Explanation of Solution

Given information:

Two complex conjugate numbers have a sum of $12$ and a product of $40$ . Find the two numbers.

Calculation:

From the given information, two complex conjugate numbers have a sum of $12$ and a product of $40$ .

The equation that represents the situation are:

  $\begin{array}{l}(a+bi)+(a-bi)=\mathrm{12.........}(1)\\ (a+bi)+(a-bi)=\mathrm{40.........}(2)\end{array}$

Solve equation $(1)$ :

  $\begin{array}{l}(a+bi)+(a-bi)=12\\ 2a=12\\ a=6\end{array}$

Now, solve equation $(2)$ :

  $\begin{array}{l}(a+bi)+(a-bi)=40\\ a^2+b^2=40\text{ {where, }a\text{=6}}\\ {\text{6}}^2+b^2=40\\ b^2=40-36\\ b^2=4\\ b=\pm 2\end{array}$

Thus, the two numbers are $6+2i$ and $6-2i$ .