Gujarat BoardEnglish MediumSTD 11 ScienceMATHSComplex Numbers2 Marks
Question
Evaluate the following: $\frac{1}{\text{i}^{58}}$
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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 $ $\therefore \ \frac{1}{\text{i}^{58}}=\frac{1}{\text{i}^{4\times14}\times\text{i}^2}$ $=\frac{1}{1\times\text{i}^2}$ $=\frac{1}{-1} \ \big[\therefore \ \text{i}^2=-1\big]$ $=-1$
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