Chapter 4.3, Problem 1AYU

1. Find the sum and the product of the complex numbers $3-2i$ and $-3+5i$. (pp. A58-A61)

To determine

To find: The sum and product of the complex numbers $\left(3-2i\right)$ and $\left(-3+5i\right)$.

Answer to Problem 1AYU

$\left(3-2i\right)+\left(-3+5i\right)=0+3i$

$\left(3-2i\right).\left(-3+5i\right)=1+21i$

Explanation of Solution

Given:

The complex numbers.

$\left(3-2i\right)$ and $\left(-3+5i\right)$

Sum of the complex numbers.

$\left(a+bi\right)+\left(c+di\right)=\left(a+c\right)+\left(b+d\right)i$

$\left(3-2i\right)+\left(-3+5i\right)=\left[3+\left(-3\right)\right]+\left[-2+5\right]i$

$=0+3i$

Product of the complex numbers.

$\left(a+bi\right).\left(c+di\right)=a.\left(c+di\right)+bi\left(c+di\right)$

Using distributive law.

$\left(3-2i\right).\left(-3+5i\right)=3.\left(-3+5i\right)+\left(-2i\right).\left(-3+5i\right)$

$=-9+15i+6i-10i^2$

$=-9+21i-10\left(-1\right)$

$=1+21i$