====== Question 7, Exercise 1.1 ======
Solutions of Question 7 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 7(i)=====
If ${{z}_{1}}=1+2i$ and ${{z}_{2}}=2+3i$, evaluate $|{{z}_{1}}+{{z}_{2}}|$.
====Solution====
We know that $z_1=1+2i$ and $z_2=2+3i$, then
\begin{align}
{{z}_{1}}+{{z}_{2}}&=1+2i+2+3i\\
&=1+2+2i+3i\\
&=3+5i
\end{align}
Now
\begin{align}
|z_1+z_2|&=\sqrt{3^2+5^2}\\
&=\sqrt{9+25}\\ 
&=\sqrt{34}\end{align} 


=====Question 7(ii)=====
If ${{z}_{1}}=1+2i$ and ${{z}_{2}}=2+3i$, evaluate $|{{z}_{1}}{{z}_{2}}|$.
====Solution====
We know that $z_1=1+2i$ and $z_2=2+3i$, then
\begin{align}
{{z}_{1}}{{z}_{2}}&=\left( 1+2i \right)\times \left( 2+3i \right)\\
&=\left( 2-6 \right)+\left( 3+4 \right)i\\
&=-4+7i.
\end{align}
Now
\begin{align}
|z_1 z_2|&=\sqrt{(-4)^2+7^2}\\
&=\sqrt{16+49}\\
&=\sqrt{65}
\end{align} 

=====Question 7(iii)=====
If ${{z}_{1}}=1+2i$ and ${{z}_{2}}=2+3i$, evaluate $\left|\dfrac{z_1}{z_2}\right|$.
====Solution====
We know that $z_1=1+2i$ and $z_2=2+3i$, then
\begin{align}
\dfrac{z_1}{z_2}&=\dfrac{1+2i}{2+3i}\\
&=\dfrac{1+2i}{2+3i}\times \dfrac{2-3i}{2-3i}\\
&=\dfrac{\left( 2+6 \right)+\left( 4-3 \right)i}{4+9}\\
&=\dfrac{\left( 2+6 \right)+\left( 4-3 \right)i}{4+9}\\
&=\dfrac{8+i}{13} =\dfrac{8}{13}+\dfrac{1}{13}i.
\end{align}
Now
\begin{align}
\left|\dfrac{z_1}{z_2}\right|&=\sqrt{\left(\dfrac{8}{13}\right)^2+\left(\dfrac{1}{13}\right)^2}\\
&=\sqrt{\dfrac{65}{169}}\\
&=\dfrac{\sqrt{65}}{13}\end{align}

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