Evaluate the following: 1 i^ 58

Question

Evaluate the following:
$\frac{1}{\text{i}^{58}}$

✓

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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