\displaystyle \textbf{Question 1: } \text{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) }\left(i^{41}+\frac{1}{i^{257}}\right)^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}
\displaystyle \text{Answer:}

\displaystyle \text{i) }i^{457}=i^{4\times114+1}=(i^4)^{114}\times i=i

\displaystyle \text{ii) }i^{528}=i^{4\times132}=(i^4)^{132}=1

\displaystyle \text{iii) }\frac{1}{i^{58}}=\frac{1}{i^{4\times14+2}}=\frac{1}{(i^4)^{14}\times i^2}=-1

\displaystyle \text{iv) }i^{37}+\frac{1}{i^{67}}
\displaystyle =i^{4\times9+1}+\frac{1}{i^{4\times16+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) }\left(i^{41}+\frac{1}{i^{257}}\right)^9
\displaystyle =\left(i^{4\times10+1}+\frac{1}{i^{4\times64+1}}\right)^9=\left(i+\frac{1}{i}\right)^9=\left(\frac{i^2+1}{i}\right)^9=\left(\frac{-1+1}{i}\right)^9=0

\displaystyle \text{vi) }(i^{77}+i^{70}+i^{87}+i^{414})^3
\displaystyle =(i^{4\times19+1}+i^{4\times17+2}+i^{4\times21+3}+i^{4\times103+2})^3
\displaystyle =(i+i^2+i^3+i^2)^3=(-2+i+i^3)^3=(-2)^3=-8

\displaystyle \text{vii) }i^{30}+i^{40}+i^{60}
\displaystyle =i^{4\times7+2}+i^{4\times10}+i^{4\times15}=i^2+1+1=-1+1+1=1

\displaystyle \text{viii) }i^{49}+i^{68}+i^{89}+i^{110}
\displaystyle =i^{4\times12+1}+i^{4\times17}+i^{4\times22+1}+i^{4\times27+2}
\displaystyle =i+1+i+i^2=i+1+i-1=2i
\displaystyle \\

\displaystyle \textbf{Question 2: } \text{Show that }1+i^{10}+i^{20}+i^{30}\text{ is a real number.}
\displaystyle \text{Answer:}
\displaystyle 1+i^{10}+i^{20}+i^{30}
\displaystyle =1+i^{4\times2+2}+i^{4\times5}+i^{4\times7+2}
\displaystyle =1+i^2+1+i^2
\displaystyle =1-1+1-1=0
\displaystyle \therefore 1+i^{10}+i^{20}+i^{30}=0,\text{ which is a real number.}
\displaystyle \\

\displaystyle \textbf{Question 3: } \text{Evaluate the following:}
\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}}
\displaystyle \text{vi) }1+i^2+i^4+i^6+\ldots+i^{20}\hspace{1.0cm}\text{vii) }(1+i)^6+(1-i)^3
\displaystyle \text{Answer:}

\displaystyle \text{i) }i^{49}+i^{68}+i^{89}+i^{110}
\displaystyle =i^{4\times12+1}+i^{4\times17}+i^{4\times22+1}+i^{4\times27+2}
\displaystyle =i+1+i+i^2=i+1+i-1=2i

\displaystyle \text{ii) }i^{30}+i^{80}+i^{120}
\displaystyle =i^{4\times7+2}+i^{4\times20}+i^{4\times30}
\displaystyle =i^2+1+1=-1+1+1=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\times1+1}+i^{4\times2+2}+i^{4\times3+3}
\displaystyle =i+i^2+i^3=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\times148}+i^{4\times147+2}+i^{4\times147}+i^{4\times146+2}+i^{4\times146}}{i^{4\times145+2}+i^{4\times145}+i^{4\times144+2}+i^{4\times144}+i^{4\times143+2}}
\displaystyle =\frac{1+i^2+1+i^2+1}{i^2+1+i^2+1+i^2}
\displaystyle =\frac{1-1+1-1+1}{-1+1-1+1-1}=\frac{1}{-1}=-1

\displaystyle \text{vi) }1+i^2+i^4+i^6+\ldots+i^{20}
\displaystyle =(1+i^2)+(i^4+i^6)+(i^8+i^{10})+(i^{12}+i^{14})+(i^{16}+i^{18})+i^{20}
\displaystyle =(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+1=1

\displaystyle \text{vii) }(1+i)^6+(1-i)^3
\displaystyle =[(1+i)^2]^3+(1-i)^3
\displaystyle =(1+i^2+2i)^3+(1-3i+3i^2-i^3)
\displaystyle =(2i)^3+(1-3i-3+i)
\displaystyle =8i^3-2-2i
\displaystyle =-8i-2-2i=-2-10i
\displaystyle \\

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