Gujarat BoardEnglish MediumSTD 11 ScienceMATHSComplex Numbers2 Marks
Question
Evaluate the following: $\text{i}^{37}+\frac{1}{\text{i}^{67}}$
✓
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 $
$\therefore \ \text{i}^{37}+\frac{1}{\text{i}^{67}}=\text{i}^{4\times9}\times\text{i}^1+\frac{1}{\text{i}^{4\times16}\times\text{i}^3}$ $=1\times\text{i}^1+\frac{1}{1\times\text{i}^3}$
$=\text{i}+\frac{1}{\text{i}^3\times\text{i}}\times\text{i}$ $=\text{i}+\frac{\text{i}}{\text{i}^4}$
$=\text{i}+\frac{\text{i}}{1} \ \big[\therefore \ \text{i}^4=1\big]$ $=2\text{i}$ $\therefore\text{i}^{37}+\frac{1}{\text{i}^{67}}=2\text{i}$
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