====== Question 2, Review Exercise ======
Solutions of Question 2 of Review Exercise 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. 

===== Question 2(i) =====
Find the value of the following: $i^{2}+i^{4}+i^{6}+\cdots+i^{100}$\\

** Solution. **

\begin{align*}
& i^{2}+i^{4}+i^{6}+\ldots+i^{100} \\
=& i^2 + (i^2)^2 + (i^2)^3 + (i^2)^4 + \ldots +(i^2)^{49} +(i^2)^{50} \\
=& -1 + (-1)^2 + (-1)^3 + (-1)^4 + \ldots + (-1)^{49}+(-1)^{50} \\
=& -1+1-1+1- \ldots -1+1 \\
=& 0.
\end{align*}


===== Question 2(ii) =====
 Find the value of the following: $\left|\dfrac{(3-2 i)(1+i)}{2-3 i}\right|$

** Solution. **

\begin{align}
&\left|\dfrac{(3-2i)(1+i)}{2-3i}\right|\\
=& \dfrac{|3-2i||1+i|}{|2-3i|}\\
=&\dfrac{ \sqrt{9 + 4}\sqrt{1 + 1}}{\sqrt{4 + 9}}\\
=& \dfrac{\sqrt{13} \cdot \sqrt{2}}{\sqrt{13}}\\
=& \sqrt{2}\end{align}

===== Question 2(iii) =====
 Find the value of the following: $|\overline{(3-2 i)(4-i)}|$

** Solution. **

\begin{align*}
& |\overline{(3-2 i)(4-i)}|\\
=&|(3-2 i)(4-i)|\quad \because \,\, |\bar{z}|=|z|\\
=&|3-2 i||4-i|\\
=&\sqrt{9+4}\sqrt{16+1}\\
=&\sqrt{13}\sqrt{17}=\sqrt{221}.
\end{align*}
===== Question 2(iv) =====
Find the value of the following: $\left(\dfrac{3+5 i}{2-3 i}\right)^{-1}$
 
** Solution. **

\begin{align*}
& \left(\dfrac{3+5 i}{2-3 i}\right)^{-1}\\
=&\dfrac{2-3 i}{3+5 i}\\
=&\dfrac{(2-3 i)(3-5i)}{(3+5 i)(3-5i)}\\
=&\dfrac{6-15-10i-9i}{9+25}\\
=&\dfrac{-9-19i}{34}\\
=&-\dfrac{9}{34}-\dfrac{19}{34}i.
\end{align*}




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