Evaluate the following: (i^ 41 +1 i^ 257 )^9

Question

Evaluate the following:
$\Big(\text{i}^{41}+\frac{1}{\text{i}^{257}}\Big)^9$

✓

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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