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.

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}