Question 11, Exercise 1.1

Solutions of Question 11 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.

Let ${{z}_{1}}=2-i$, ${{z}_{2}}=-2+i$. Find ${\rm Re}\left( \dfrac{{{z}_{1}}{{z}_{2}}}{\overline{{{z}_{1}}}} \right)$.

Given $z_1=2-i$ and $z_2=-2+i$, then $\overline{z_1}=2+i$. \begin{align} z_1 z_2&=(2-i)(-2+i)\\ &=-4+1+2i+2i\\ &=-3+4i \end{align} Now we take \begin{align} \dfrac{z_1 z_2}{\overline{{z_1}}}&=\dfrac{-3+4i}{2+i}\\ &=\dfrac{-3+4i}{2+i}\times \dfrac{2-i}{2-i}\\ &=\dfrac{-6+4+8i+3i}{4+1}\\ &=\dfrac{-2+11i}{5}\\ &=\dfrac{-2}{5}+\dfrac{11i}{5} \end{align} Hence, we have $${\rm Re}\left( \dfrac{{{z}_{1}}{{z}_{2}}}{\overline{{{z}_{1}}}} \right)=\dfrac{-2}{5}.$$

Let $z_1=2-i$. Find ${\rm Im}\left( \dfrac{1}{{{z}_{1}}\overline{{{z}_{1}}}} \right)$.

Given $z_1=2-i$, then $\overline{z_1}=2+i$. \begin{align} z_1\overline{z_1}&=(2-i)(2+i)\\ &=4+1+2i-2i=5.\end{align} This gives $$\dfrac{1}{z_1\overline{z_1}}=\dfrac{1}{5}.$$ Hence, we have $${\rm Im}\left(\dfrac{1}{z_1\overline{z_1}}\right)=0.$$

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