====== 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. =====Question 6===== Find conjugate of $\dfrac{1}{3+4i}$.\\ ====Solution==== \begin{align}\dfrac{1}{3+4i}&=\dfrac{1}{3+4i}\times \dfrac{3-4i}{3-4i}\\ &=\dfrac{3-4i}{9+16}\\ &=\dfrac{3-4i}{25}\\ &=\dfrac{3}{25}-\dfrac{4i}{25}\end{align} =====Question 7===== Find the multiplicative inverse of $\dfrac{3i+2}{3-2i}$.\\ ====Solution==== \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 $\dfrac{3i+2}{3-2i}=-i$ =====Question 8===== Find the quadrative equation $z+\dfrac{2}{z}=2.$\\ ====Solution==== \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} ====Go To==== [[fsc-part1-kpk:sol:unit01:review-ex-1-p3 |< Question 4 & 5]]