Question 9 & 10, Exercise 1.1

Solutions of Question 9 & 10 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.

Find the conjugate of $\dfrac{\left( 3-2i \right)\left( 2+3i \right)}{\left( 1+2i \right)\left( 2-i \right)}$.

Let \begin{align}z&=\dfrac{\left( 3-2i \right)\left( 2+3i \right)}{\left( 1+2i \right)\left( 2-i \right)}\\ &=\dfrac{6+6+9i-4i}{2+2+4i-i}\\ &=\dfrac{12+5i}{4+3i}\end{align} Now \begin{align}\bar{z}&=\dfrac{12-5i}{4-3i}\\ &=\dfrac{12-5i}{4-3i}\times \dfrac{4+3i}{4+3i}\\ &=\dfrac{48+15+36i-20i}{16+9+12i-12i}\\ &=\dfrac{63+16i}{25}\\ &=\dfrac{63}{25}+\dfrac{16}{25}i\end{align}

Evalute ${{\left[ {{i}^{18}}+{{\left( \dfrac{1}{i} \right)}^{25}} \right]}^{3}}$.

\begin{align}i^{18}+\left(\dfrac{1}{i}\right)^{25} &=i^{18}+\dfrac{1}{i^{25}}\\ &={{\left( {{i}^{2}} \right)}^{9}}+\dfrac{1}{i.{{\left( {{i}^{2}} \right)}^{12}}}\\ &={{\left( -1 \right)}^{9}}+\dfrac{1}{i.{{\left( -1 \right)}^{12}}}\\ &=-1+\dfrac{1}{i}\\ &=-1-i \quad \because \dfrac{1}{i}=-i \end{align} Now \begin{align} {{\left[ {{i}^{18}}+{{\left( \dfrac{1}{i} \right)}^{25}} \right]}^{3}}&={{\left( -1-i \right)}^{3}}\\ &=-{{\left( 1+i \right)}^{3}}\\ &=-\left( {{1}^{3}}+{{i}^{3}}+3\left( 1 \right)\left( i \right)\left( 1+i \right) \right) \\ &=-\left( 1-i+3i\left( 1+i \right) \right)\\ &=-\left( 1-i+3i-3 \right)\\ &=-\left( 2i-2 \right)\\ &=2-2i\end{align}

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