Question 1: Evaluate the following:

i) i^{457}          ii) i^{528}          iii) \frac{1}{i^{58}}           iv) i^{37} + \frac{1}{i^{67}}           v) \Big( i^{41}+ \frac{1}{i^{257}} \Big)^9

vi) ( i^{77} + i^{70} + i^{87} + i^{414} )^3   vii) i^{30} + i^{40} + i^{60}   viii) i^{49} + i^{68} + i^{89} + i^{110}

Answer:

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

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

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

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

=  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

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

= \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

vi) ( i^{77} + i^{70} + i^{87} + i^{414} )^3

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

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

vii) i^{30} + i^{40} + i^{60}

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

viii) i^{49} + i^{68} + i^{89} + i^{110}

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

= i + 1 + i + i^2 = i + 1 + i - 1 = 2i

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

Answer:

1 + i^{10}+ i^{20}+i^{30}

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

= 1 + i^2 + 1 + i^2

= 1 - 1 + 1 - 1 = 0 . This is a real number.

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Question 3:

i) i^{49} + i^{68} + i^{89} + i^{110}           ii) i^{30} + i^{80} + i^{120}

iii) i + i^{2} + i^{3} + i^{4}           iv) i^{5} + i^{10} + i^{15}

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

vii) (1+i)^6+(1-i)^3

Answer:

i) i^{49} + i^{68} + i^{89} + i^{110}

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

= i + 1 + i + i^2 = 2i

ii) i^{30} + i^{80} + i^{120}

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

= i^2 + 1 + 1 = -1 + 2 = 1

iii) i + i^{2} + i^{3} + i^{4}

= i - 1 - i + 1 = 0

iv) i^{5} + i^{10} + i^{15}

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

= i + i^2 + i^3

= i - 1 - i = - 1

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

= \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}}

= \frac{1+i^2+1+i^2+1}{i^2+1+i^2+1+i^2} = \frac{1}{-1} = -1

vi) 1 + i^{2} + i^{4} + i^{6}+ \ldots + i^{20}

= [ 1+(-1)] + [ 1+(-1)] + [ 1+(-1)] + \ldots + 1 = 1

vii) (1+i)^6+(1-i)^3

= [(1+i)^3]^3 + ( 1 - i)^3

= ( 1 + i^2 + 2i)^3 + ( 1 - i^3 + 3i^2 - 3i)

= 8i^3 - 2 - 2i

= - 8i - 2 - 2i

= -2 - 10i