Question 1, Exercise 1.1

Solutions of Question 1 of Exercise 1.1 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.

Evaluate ${{i}^{31}}$.

Solution.

\begin{align}{{i}^{31}}&=i\cdot{{i}^{30}}\\ &=i\cdot{{\left( {{i}^{2}} \right)}^{15}}\\ &=i\cdot{{\left( -1 \right)}^{15}} \quad \because i^2=-1\\ &=i\cdot(-1)\\ &=-i.\end{align}

Evaulate ${{\left( -i \right)}^{6}}$.

Solution.

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

Evaluate ${{\left( -1 \right)}^{\frac{-13}{2}}}$. Solution.

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

Evaluate $\dfrac{2}{(-1)^{\frac{3}{2}}}$.

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 \cdot (i^2)^7 \\ &=i \cdot (-1)^7 \\ &= -i \end{align}

Evaluate $i^{23}+i^{58}+i^{21}$.

Solution.

\begin{align}i^{23}+i^{58}+i^{21} \\ =&i^{22}\cdot i+i^{58}+i^{20}\cdot i \\ =&(i^2)^{11}\cdot i+(i^2)^{29}+(i^2)^{10}\cdot i \\ =&(-1)^{11}\cdot i+(-1)^{29}+(-1)^{10}\cdot i \\ =& -i - i +i = -i \end{align}

Question 2>