# 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.

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*}

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}

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*}

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*}