Question 1, Exercise 1.1

Solutions of Question 1 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.

Simplify ${{i}^{9}}+{{i}^{19}}$.

\begin{align}{{i}^{9}}+{{i}^{19}}&=i\cdot{{i}^{8}}+i\cdot{{i}^{18}}\\ &=i\cdot{{\left( {{i}^{2}} \right)}^{4}}+i\cdot{{\left( {{i}^{2}} \right)}^{9}}\\ &=i\cdot{{\left( -1 \right)}^{4}}+i\cdot{{\left( -1 \right)}^{9}}\\ &=i\cdot1+i\cdot\left( -1 \right)\\ &=i-i\\ &=0.\end{align}

Simplify ${{\left( -i \right)}^{23}}$.

\begin{align}{{\left( -i \right)}^{23}}&={{\left( -1 \right)}^{23}}{{i}^{23}}\\ &=( -1 ) \cdot {i}{\cdot} i^{22}\\ &=\left( -1 \right)\cdot i\cdot{{\left( {{i}^{2}} \right)}^{11}}\\ &=-i\cdot{{\left( {{i}^{2}} \right)}^{11}}\\ &=-i\cdot{{\left( -1 \right)}^{11}}\\ &=-i\cdot\left( -1 \right)\\ &=i\end{align}

Simplify ${{\left( -1 \right)}^{\frac{-23}{2}}}$.

\begin{align}{{\left( -1 \right)}^{\frac{-23}{2}}}&={{\left( \sqrt{-1} \right)}^{-23}}\\ &={{i}^{-23}} =\frac{1}{i{{\left( {{i}^{2}} \right)}^{11}}}\\ &=\frac{1}{i{{\left( -1 \right)}^{11}}} =\frac{1}{-i}\\ &=\frac{1}{-i}\times \frac{i}{i}=\frac{i}{-\left( -1 \right)}\\ &=i\end{align}

Simplify ${{\left( -1 \right)}^{\frac{15}{2}}}$.

\begin{align}{{\left( -1 \right)}^{\frac{15}{2}}}&={{\left[ {{\left( -1 \right)}^{\frac{1}{2}}} \right]}^{15}}\\ &=i^{15} = i \cdot i^{14} \\ &=i{{\left( {{i}^{2}} \right)}^{7}}\\ &=i{{\left( -1 \right)}^{7}}\\ &=-i\end{align}

Question 2 & 3 >