Class 11: Complex Numbers – Exercise 13.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Evaluate the following:

$\displaystyle \text{i) } i^{457} \hspace{1.0cm} \text{ii) } i^{528} \hspace{1.0cm} \text{iii) } \frac{1}{i^{58}} \hspace{1.0cm} \text{iv) } i^{37} + \frac{1}{i^{67}}$ $\displaystyle \text{v) } \Big( i^{41}+ \frac{1}{i^{257}} \Big)^9$

$\displaystyle \text{vi) } ( i^{77} + i^{70} + i^{87} + i^{414} )^3 \hspace{1.0cm} \text{vii) } i^{30} + i^{40} + i^{60} \hspace{1.0cm} \text{viii) } i^{49} + i^{68} + i^{89} + i^{110}$

Answer:

$\displaystyle \text{i) } i^{457} = i^{4 \times 114 + 1} = {(i^4)}^{114} \times i = i$

$\displaystyle \text{ii) } i^{528} = i^{4 \times 132 } = {(i^4)}^{132} = 1$

$\displaystyle \text{iii) } \frac{1}{i^{58}} = \frac{1}{i^{4 \times 14+2}} = \frac{1}{(i^4)^{14} \times i^2} = -1$

$\displaystyle \text{iv) } i^{37} + \frac{1}{i^{67}}$

$\displaystyle = i^{4 \times 9 + 1}+ \frac{1}{i^{4 \times 15 + 3}} = (i^4)^9 \cdot i + \frac{1}{(i^4)^{16}\cdot i^3} = i - \frac{1}{i} = \frac{i^2 - 1}{i} = \frac{-2}{i} \times \frac{i}{i} = \frac{-2i}{i^2} = 2i$

$\displaystyle \text{v) } \Big( i^{41}+ \frac{1}{i^{257}} \Big)^9$

$\displaystyle = \Big( i^{4 \times 10 + 1}+ \frac{1}{i^{4 \times 64 + 1}} \Big)^9 = \Big( i+ \frac{1}{i} \Big)^9 = \Big( \frac{i^2+1}{i} \Big)^9 = \Big( \frac{-1+1}{i} \Big)^9 = 0$

$\displaystyle \text{vi) } ( i^{77} + i^{70} + i^{87} + i^{414} )^3$

$\displaystyle = ( i^{4 \times 19 + 1} + i^{4 \times 17 + 2} + i^{4 \times 21 + 3} + i^{4 \times 103 + 2} )^3$

$\displaystyle = ( i + i^{2} + i^{3} + i^{2} )^3 = ( - 2 + i + i^3)^3 = -8$

$\displaystyle \text{vii) } i^{30} + i^{40} + i^{60}$

$\displaystyle = i^{4 \times 7 + 2} + i^{4 \times 10} + i^{4 \times 15 } = i^2 + 1 + 1 = - 1 + 1 +1 = 1$

$\displaystyle \text{viii) } i^{49} + i^{68} + i^{89} + i^{110}$

$\displaystyle = i^{4 \times 12 + 1} + i^{4 \times 17 } + i^{4 \times 22 + 1} + i^{4 \times 27 + 2}$

$\displaystyle = i + 1 + i + i^2 = i + 1 + i - 1 = 2i$

$\displaystyle \\$

Question 2: Show that $\displaystyle 1 + i^{10}+ i^{20}+i^{30}$ is a real number

Answer:

$\displaystyle 1 + i^{10}+ i^{20}+i^{30}$

$\displaystyle = 1+ i^{4 \times 2 + 2}+ i^{4 \times 5 }+i^{4 \times 7 + 2}$

$\displaystyle = 1 + i^2 + 1 + i^2$

$\displaystyle = 1 - 1 + 1 - 1 = 0$ . This is a real number.

$\displaystyle \\$

Question 3:

$\displaystyle \text{i) } i^{49} + i^{68} + i^{89} + i^{110} \hspace{1.0cm} \text{ii) } i^{30} + i^{80} + i^{120}$

$\displaystyle \text{iii) } i + i^{2} + i^{3} + i^{4} \hspace{1.0cm} \text{iv) } i^{5} + i^{10} + i^{15}$

$\displaystyle \text{v) } \frac{i^{592} + i^{590} + i^{588} + i^{586} + i^{584}}{i^{582} + i^{580} + i^{578} + i^{576} + i^{574}} \hspace{1.0cm} \text{vi) } 1 + i^{2} + i^{4} + i^{6}+ \ldots + i^{20}$

$\displaystyle \text{vii) } (1+i)^6+(1-i)^3$

Answer:

$\displaystyle \text{i) } i^{49} + i^{68} + i^{89} + i^{110}$

$\displaystyle = i^{4 \times 12 + 1} + i^{4 \times 17} + i^{4 \times 22 + 1} + i^{4 \times 27 + 2}$

$\displaystyle = i + 1 + i + i^2 = 2i$

$\displaystyle \text{ii) } i^{30} + i^{80} + i^{120}$

$\displaystyle = i^{4 \times 7 + 2} + i^{4 \times 20} + i^{4 \times 30}$

$\displaystyle = i^2 + 1 + 1 = -1 + 2 = 1$

$\displaystyle \text{iii) } i + i^{2} + i^{3} + i^{4}$

$\displaystyle = i - 1 - i + 1 = 0$

$\displaystyle \text{iv) } i^{5} + i^{10} + i^{15}$

$\displaystyle = i^{4 \times 1 + 1} + i^{4 \times 2 + 2} + i^{4 \times 3 + 3}$

$\displaystyle = i + i^2 + i^3$

$\displaystyle = i - 1 - i = - 1$

$\displaystyle \text{v) } \frac{i^{592} + i^{590} + i^{588} + i^{586} + i^{584}}{i^{582} + i^{580} + i^{578} + i^{576} + i^{574}}$

$\displaystyle = \frac{i^{4 \times 148} + i^{4 \times 147 + 2} + i^{4 \times 147} + i^{4 \times 146 + 2} + i^{4 \times 146}}{i^{4 \times 145 + 2} + i^{4 \times 145} + i^{4 \times 144 + 2} + i^{4 \times 144} + i^{4 \times 143 + 2}}$