Question 3, Exercise 1.1

Solutions of Question 3 of Exercise 1.1 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.

Simplify the following $\dfrac{(2+i)(3-2i)}{1+i}$

Solution. $$\begin{align}&\dfrac{(2+i)(3-2i)}{1+i}\\ =&\dfrac{6-2i^2+3i-4i}{1+i}\\ =&\dfrac{8-i}{1+i}\\ =&\dfrac{8-i}{1+i}\times \dfrac{1-i}{1-i}\\ =&\dfrac{8+i^2-8i-i}{1^2-i^2}\\ =&\dfrac{7-9i}{2}\\ =&\dfrac{7}{2}-\dfrac{9}{2}i\end{align}$$

Simplify the following $\dfrac{1+i}{(2+i)^2}$

Solution.

$$\begin{align}&\dfrac{1+i}{(2+i)^2}\\ =&\dfrac{1+i}{4+i^2+4i}\\ =&\dfrac{1+i}{3+4i}\\ =&\dfrac{1+i}{3+4i}\times \dfrac{3-4i}{3-4i}\\ =&\dfrac{3-4i^2-4i+3i}{9-16i^2}\\ =&\dfrac{7-i}{25}\\ =&\dfrac{7}{25}-\dfrac{1}{25}i. \end{align}$$

Simplify the following $\dfrac{1}{3+i}-\dfrac{1}{3-i}$

Solution.

$$\begin{align}&\dfrac{1}{3+i}-\dfrac{1}{3-i}\\ =&\dfrac{(3-i)-(3+i)}{3^2-i^2}\\ =&\dfrac{-2i}{9+1}\\ =&-\dfrac{2}{10}i\\ =&-\dfrac{1}{5}i\end{align}$$

Simplify the following $(1+i)^{-2}+(1-i)^{-2}$

Solution.

$$\begin{align}&(1+i)^{-2}+(1-i)^{-2}\\ =&\dfrac{1}{(1+i)^2}+\dfrac{1}{(1-i)^2}\\ =&\dfrac{1}{1+i^2+2i}+\dfrac{1}{1+i^2-2i}\\ =&\dfrac{1}{2i}+\dfrac{1}{-2i}\\ =&\dfrac{1}{2i}-\dfrac{1}{2i}\\ =&0\end{align}$$

Simplify: $(2+i)^2+\dfrac{7-4i}{2+i}$

Solution.

$$\begin{align}&(2+i)^2+\dfrac{7-4i}{2+i}\\ =&(4+i^2+4i)+\dfrac{7-4i}{2+i}\\ =&(3+4i)+\dfrac{7-4i}{2+i}\\ =&\dfrac{(3+4i)(2+i)+(7-4i)}{2+i}\\ =&\dfrac{(6+4i^2+8i+3i)+(7-4i)}{2+i}\\ =&\dfrac{2+11i+7-4i}{2+i}\\ =&\dfrac{9+7i}{2+i}\\ =&\dfrac{9+7i}{2+i}\times\dfrac{2-i}{2-i}\\ =&\dfrac{18-7i^2-9i+14i}{4-i^2}\\ =&\dfrac{25+5i}{5}\\ =&5+i. \end{align}$$

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