Chapter 4.6, Problem 1AYU

To determine

To calculate:

The sum and product value of the following complex numbers:

  $3-2i$ and $-3+5i$

Answer to Problem 1AYU

The given complex numbers has:

Addition: $3i$

Multiplication: $1+21i$

Explanation of Solution

Given information:

  $\begin{array}{l}3-2i\\ -3+5i\end{array}$

Formula Used:

The addition of the complex numbers:

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

The multiplication of complex numbers:

  $\left(a+bi\right)\times \left(c+di\right)=\left(ac-bd\right)+\left(ad+bc\right)i$

Calculation:

The given two complex numbers get compared as:

  $\begin{array}{l}a+bi=3-2i\\ a=3,b=-2\end{array}$

And

  $\begin{array}{l}c+di=-3+5i\\ c=-3,d=5\end{array}$

Now the addition of the complex numbers:

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

Put all values:

  $\begin{array}{l}\left(a+bi\right)+\left(c+di\right)=\left(a+c\right)+\left(b+d\right)i\\ \left(3-2i\right)+\left(-3+5i\right)=\left(3-3\right)+\left(-2+5\right)i\\ \left(3-2i\right)+\left(-3+5i\right)=0+3i\\ \left(3-2i\right)+\left(-3+5i\right)=3i\end{array}$

The multiplication of complex numbers:

  $\left(a+bi\right)\times \left(c+di\right)=\left(ac-bd\right)+\left(ad+bc\right)i$

Put all values:

  $\begin{array}{l}\left(a+bi\right)\times \left(c+di\right)=\left(ac-bd\right)+\left(ad+bc\right)i\\ \left(3-2i\right)\times \left(-3+5i\right)=\left[\left(3\times -3\right)-\left(-2\times 5\right)\right]+\left[\left(3\times 5\right)+\left(-2\times -3\right)\right]i\\ \left(3-2i\right)\times \left(-3+5i\right)=1+21i\end{array}$

Hence, the addition and multiplication of complex numbers is $3i$ and $1+21i$ respectively.