Gujarat BoardEnglish MediumSTD 11 ScienceMATHSComplex Numbers3 Marks
Question
Evaluate the following: $\Big(\text{i}^{41}+\frac{1}{\text{i}^{257}}\Big)^9$
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Answer
We know that $\text{i}=\sqrt{-1}$ $\text{i}^2 = -1$ $\text{i}^3 = -\text{i}$ $\text{i}^4 = 1$ In order to find $i^n$ Where n > 4, we divide n by 4 to get quotient p and remainder q, So that $\text{n} = 4\text{p} + \text{q}, \ 0\leq\text{q}<4 $ Then $\text{i}^\text{n} =\text{i}^{4\text{p}+\text{q}}$ $=\text{i}^{4\text{p}}\times\text{i}^\text{q}$ $=\big(\text{i}^{4}\big)^{\text{p}}\times\text{i}^\text{q}$ $=\text{i}^{\text{p}}\times\text{i}^\text{q}$ $=\text{i}^\text{q} \ \big[\therefore \ 1^{\text{p}-1}\big]$ Hence $\text{i}^\text{n} =\text{i}^\text{q},$ where $0\leq\text{q}<4 $ $\Big(\text{i}^{41}+\frac{1}{\text{i}^{257}}\Big)^9=\Bigg(\text{i}^{4\times10}\times\text{i}^1+\frac{1}{\text{i}^{4\times64}\times\text{i}^1}\Bigg)^9$ $=\Big(1\times\text{i}+\frac{1}{1\times\text{i}}\Big)^9$ $=\Big(1+\frac{1}{\text{i}}\Big)^9$ $=\Big(\text{i}+\frac{1}{\text{i}\times\text{i}}\times\text{i}\Big)^9$ $=\Big(\text{i}+\frac{\text{i}}{-1}\Big)^9$ $=(\text{i}-\text{i})^9$ $=0$
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