Prove that : `(i)" "i^(n) + i^(n+1) + i^(n+2) + i^(n+3) = 0` `(ii) (1+i)^(4) xx (1 + (1)/(i))^(4) = 16`.

Prove that : `(i)" "i^(n) + i^(n+1) + i^(n+2) + i^(n+3) = 0` `(ii) (1+

1.	Prove that : `(i)" "i^(n) + i^(n+1) + i^(n+2) + i^(n+3) = 0` `(ii) (1+i)^(4) xx (1 + (1)/(i))^(4) = 16`.
Answer» We have `(i)" "i^(n) + i^(n+1) + i^(n+2) + i^(n+3)` `= i^(n)(1 + i + i^(2) + i^(3))` `= i^(n)(1+i-1-i)=(i^(n)xx0)=0." "[because i^(2) = -1 and i^(3) = -i]` `(ii)" "i^(107)+i^(112)+i^(117)+i^(122)` `= i^(107)(1+i^(5)+i^(10)+i^(15))=i^(107)(1+i^(4)xx i + i^(8) xx i^(2) + i^(12) xx i^(3))` `=i^(107)(1 + i + i^(2) + i^(3))" "[because i^(4) = 1, i^(8) = 1 and i^(12) = 1]` `=i^(107)(1+i-1-i)=(i^(107)xx0)=0." "[because i^(2) = -1, i^(3) = -i]` `(1+i)^(4) xx (1 + (1)/(i))^(4)` `=(1+i)^(4) xx (1+(1)/(i)xx(i)/(i))^(4)(1+i)^(4)(1-i)^(4)" "[because i^(2) = -1]` `={(1+i)(1-i)}^(4) = (1-i^(2))^(4) = {1-(-1)}^(4) = 2^(4) = 16`.