Evaluate the following: i^ 49 +i^ 68 +i^ 89 +i^ 110

Question

Evaluate the following:
$\text{i}^{49}+\text{i}^{68}+\text{i}^{89}+\text{i}^{110}$

✓

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 $
$\text{i}^{49}+\text{i}^{68}+\text{i}^{89}+\text{i}^{110}=\text{i}^{4\times12}\times\text{i}^{1}+\text{i}^{4\times17}+\text{i}^{4\times22}\times\text{i}^1+\text{i}^{4\times27}\times\text{i}^2$
$=1\times\text{i}+1+1\times\text{i}+1\times\text{i}^2$
$=\text{i}+1+\text{i}-1$
$=2\text{i}$
$\therefore\text{i}^{49}+\text{i}^{68}+\text{i}^{89}+\text{i}^{110}=2\text{i}$

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