# Question 2 & 3, Exercise 1.1

Solutions of Question 2 & 3 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.

Prove that ${{i}^{107}}+{{i}^{112}}+{{i}^{122}}+{{i}^{153}}=0$.

\begin{align}L.H.S.&={{i}^{107}}+{{i}^{112}}+{{i}^{122}}+{{i}^{153}}\\ &=i\cdot i^{106}+i^{112}+i^{122}+i\cdot i^{152}\\ &=i.{{\left( {{i}^{2}} \right)}^{53}}+{{\left( {{i}^{2}} \right)}^{56}}+{{\left( {{i}^{2}} \right)}^{61}}+i.{{\left( {{i}^{2}} \right)}^{76}}\\ &=i.{{\left( -1 \right)}^{53}}+{{\left( -1 \right)}^{56}}+{{\left( -1 \right)}^{61}}+i.{{\left( -1 \right)}^{76}}\\ &=-i+1-1+i\\ &=0=R.H.S.\end{align}

Add the complex numbers $3\left( 1+2i \right),-2\left( 1-3i \right)$.

\begin{align}& 3\left( 1+2i \right)+-2\left( 1-3i \right)\\ &=\left( 3+6i \right)+\left( -2+6i \right)\\ &=\left( 3-2 \right)+\left( 6+6 \right)i\\ &=1+12i\end{align}

Add the complex numbers $\dfrac{1}{2}-\dfrac{2}{3}i,\dfrac{1}{4}-\dfrac{1}{3}i$.

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

Add the complex numbers $\left( \sqrt{2},1 \right),\left( 1,\sqrt{2} \right)$.

\begin{align}&\left( \sqrt{2},1 \right)+\left( 1,\sqrt{2} \right)\\ &=\left( \sqrt{2}+i \right)+\left( 1+\sqrt{2}i \right)\\ &=\left( \sqrt{2}+1 \right)+\left( 1+\sqrt{2} \right)i\end{align}