====== Question 6, Exercise 1.2 ======
Solutions of Question 6 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. 
====Question 6====
Find the value of $\lambda$; if $\left|\dfrac{z_{1}}{z_{2}}+\lambda\right|=\sqrt{\lambda+2}$; where $z_{1}=3+i$ and $z_{2}=1+i$.

**Solution.** 

Given: \begin{align} &z_{1}=3+i\text{ and } z_{2}=1+i.\end{align}
Now 
\begin{align}
\dfrac{z_1}{z_2} &= \dfrac{3+i}{1+i}\\
&=\dfrac{(3+i)(1-i)}{(1+i)(1-i)} \\
&=\dfrac{3+1+i(-3+1)}{1+1} \\
&=\dfrac{4-2i}{2} = 2-i.
\end{align}
Now
\begin{align}&\left|\dfrac{z_{1}}{z_{2}}+\lambda\right|=\sqrt{\lambda+2}\\
\implies & |2-i+\lambda|=\sqrt{\lambda+2} \\
\implies & |2+\lambda-i|^2=\lambda+2 \\
\implies &(2+\lambda)^2+1=\lambda+2 \\
\implies &4+4\lambda+\lambda^2+1-\lambda-2=0 \\
\implies &\lambda^2+3\lambda+3=0.
\end{align}
By using quadratic formula, we have 
\begin{align}
\lambda &=\dfrac{-3\pm\sqrt{9-4(1)(3)}}{2(1)} \\
& =\dfrac{-3\pm\sqrt{-3}}{2} \\
&=\dfrac{-3\pm i\sqrt{3}}{2} 
\end{align}
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