Gujarat 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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