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.
Question 1(i)
Simplify ${{i}^{9}}+{{i}^{19}}$.
Solution
\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}
Question 1(ii)
Simplify ${{\left( -i \right)}^{23}}$.
Solution
\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}
Question 1(iii)
Simplify ${{\left( -1 \right)}^{\frac{-23}{2}}}$.
Solution
\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}
Question 1(iv)
Simplify ${{\left( -1 \right)}^{\frac{15}{2}}}$.
Solution
\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}
Go to