====== 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} 
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