# Question 6, Exercise 1.1

Solutions of Question 6 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

## Question 6(i)

Perform the indicated division $\dfrac{4+i}{3+5i}$ and write the answer in the form $a+ib$.

### Solution

\begin{align}\dfrac{4+i}{3+5i}&=\dfrac{4+i}{3+5i}\times \dfrac{3-5i}{3-5i}\\ &=\dfrac{\left( 12+5 \right)+\left( 3-20 \right)i}{9-25{{i}^{2}}}\\ &=\dfrac{17-17i}{9+25}\\ &=\dfrac{17}{34}-\dfrac{17}{34}i\\ &=\dfrac{1}{2}-\dfrac{1}{2}i\end{align}

## Question 6(ii)

Perform the indicated division $\dfrac{1}{-8+i}$ and write the answer in the form $a+ib$.

### Solution

\begin{align}\dfrac{1}{-8+i}&=\dfrac{1}{-8+i}\times \dfrac{-8-i}{-8-i}\\ &=\dfrac{-8-i}{64-{{i}^{2}}}\\ &=\dfrac{-8-i}{64+1}\\ &=\dfrac{-8}{65}-\dfrac{1}{65}i\end{align}

## Question 6(iii)

Perform the indicated division $\dfrac{1}{7-3i}$ and write the answer in the form $a+ib$.

### Solution

\begin{align}\dfrac{1}{7-3i}&=\dfrac{1}{7-3i}\times \dfrac{7+3i}{7+3i}\\ &=\dfrac{7+3i}{49-9{{i}^{2}}}\\ &=\dfrac{7+3i}{49+9}\\ &=\dfrac{7+3i}{58}\\ &=\dfrac{7}{58}+\dfrac{3}{58}i\end{align}

## Question 6(iv)

Perform the indicated division $\dfrac{6+i}{i}$ and write the answer in the form $a+ib$.

### Solution

\begin{align}\dfrac{6+i}{i}&=\dfrac{6+i}{i}\times \dfrac{-i}{-i}\\ &=\dfrac{-6i-{{i}^{2}}}{-{{i}^{2}}}\\ &=\dfrac{-6i+1}{1}\\ &=1-6i\end{align}