Question 6, 7 & 8, Review Exercise 1

Solutions of Question 6, 7 & 8 of Review Exercise 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.

Find conjugate of $\dfrac{1}{3+4i}$.

Suppose $$z=\dfrac{1}{3+4i}.$$ Then $$\begin{align}z&=\dfrac{1}{3+4i}\times \dfrac{3-4i}{3-4i}\\ &=\dfrac{3-4i}{9+16}\\ &=\dfrac{3-4i}{25}\\ &=\dfrac{3}{25}-\dfrac{4}{25}i\end{align}$$ Thus $$\bar{z}=\dfrac{3}{25}+\dfrac{4}{25}i.$$

Find the multiplicative inverse of $\dfrac{3i+2}{3-2i}$.

$$\begin{align}\dfrac{3i+2}{3-2i}\\ \dfrac{3i+2}{3-2i}&=\dfrac{3i+2}{3-2i}\times \dfrac{3+2i}{3+2i}\\ &=\dfrac{9i-6+6+4i}{9+4}\\ &=\dfrac{13i}{13}\\ &=i\\ a&=0,\quad\quad b=1\\ \sqrt{{{a}^{2}}+{{b}^{2}}}&=1\end{align}$$ multiplicative inverse is
$$\dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}}-\dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$$ multiplicative inverse of $\dfrac{3i+2}{3-2i}=-i$

Find the quadrative equation $z+\dfrac{2}{z}=2.$

$$\begin{align}z+\dfrac{2}{z}&=2\\ {{z}^{2}}+2&=2z\\ {{z}^{2}}-2z+2&=0\end{align}$$ According to the quadratic formula, we have
$$a=1,\quad b=-2$$ and $$c=2$$
Quadratic formula is
$$\begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ z&=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( 2 \right)}}{2\left( 1 \right)}\\ z&=\dfrac{2\pm \sqrt{4-8}}{2}\\ z&=\dfrac{2\pm \sqrt{-4}}{2}\\ z&=\dfrac{2\pm 2i}{2}\\ z&=1\pm i\end{align}$$

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