====== 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.
=====Question 9=====
Find the conjugate of   $\dfrac{\left( 3-2i \right)\left( 2+3i \right)}{\left( 1+2i \right)\left( 2-i \right)}$.
====Solution====
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}


=====Question 10=====
Evalute   ${{\left[ {{i}^{18}}+{{\left( \dfrac{1}{i} \right)}^{25}} \right]}^{3}}$.
====Solution====
\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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