Question

Prove that:
(i) $i^n+i^{n+1}+i^{n+2}+i^{n+3}=0$
(ii) $i^{107}+i^{112}+i^{117}+i^{122}=0$
(iii) $(1+i{)}^4\times {(1+\frac{1}{i})}^4=16$

Answer 1

We have
(i) $i^n+i^{n+1}+i^{n+2}+i^{n+3}$
$=i^n(1+o+i^2+i^3)$
$=i^n(1+i-1-i)=(i^n\times 0)=0.$ $[\because i^2=-1\text{and}{i}^3=-i]$
(ii) $i^{107}+i^{112}+i^{117}+i^{112}$ $=i^{107}(1+i^5+i^{10}+i^{15})=i^{107}(1+i^4\times i+i^8\times i^2+i^{12}\times i^3)$ $=i^{107}(1+i+i^2+i^3)$ $[\because i^2=-1,i^3=-i]$
(iii) $(1+i{)}^4\times {(1+\frac{1}{i})}^4$ $=(1+i{)}^4\times {(1+\frac{1}{i}\times \frac{i}{i})}^4=(1+i{)}^4(1-i{)}^4$ $[\because i^2=-1]$ $=\left[\right(1+i\left)\right(1-i){]}^4=(1-i^2{)}^4=[1-(-1){]}^4=2^4=16$.