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Complex Numbers — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsComplex Numbers3 Marks
Question
Evaluate the following:
$\big(\text{i}^{77}+\text{i}^{70}+\text{i}^{87}+\text{i}^{414}\big)^3$
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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 $
$\big(\text{i}^{77}+\text{i}^{70}+\text{i}^{87}+\text{i}^{414}\big)^3=\big(\text{i}^{4\times9}\times\text{i}^{1}+\text{i}^{4\times17}\times\text{i}^{2}+\text{i}^{4\times21}\times\text{i}^3+\text{i}^{4\times103}\times\text{i}^2\big)^3$
$=\big(1\times\text{i}+1\times\text{i}^2+1\times\text{i}^3+1\times\text{i}^2\big)^3$
$=(\text{i}-1-\text{i}-1)^3$
$=(-2)^3$
$=-8$
$\big(\therefore(\text{i}^{77}+\text{i}^{70}+\text{i}^{87}+\text{i}^{414}\big)^3=-8$

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