MCQ
The value of $\frac{(\text{i}^5+\text{i}^6+\text{i}^7+\text{i}^8+\text{i}^9)}{(1+\text{i})}$ is:
- A$\frac{1}{2}(1+\text{i})$
- B$\frac{1}{2}(1-\text{i})$
- C$1$
- D$\frac{1}{2}$
Solution:
$\frac{(\text{i}^5+\text{i}^6+\text{i}^7+\text{i}^8+\text{i}^9)}{(1+\text{i})}$
$=\frac{\text{i}-1-\text{i}+1+\text{i}}{1+\text{i}} \ [$
As, $\text{i}^5=\text{i},\text{i}^6=-1,\text{i}^7=-\text{i},\text{i}^8=1,\text{i}^9=\text{i}]$$=\frac{\text{i}}{\text{i}+1}$
$=\frac{\text{i}}{\text{i}+1}\times\frac{\text{i}-1}{\text{i}-1}$
$=\frac{\text{i}(\text{i}-1)}{\text{i}^2-1}$
$=\frac{\text{i}^2-\text{i}}{-2}$
$=\frac{1}{2}(1+\text{i})$
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