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Complex Numbers — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsComplex Numbers5 Marks
Question
Find the smallest positive integer value of n for which $\frac{(1+\text{i})^\text{n}}{(1-\text{i})^{\text{n}-2}}$ is a real number.
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Answer

Let $\text{z}=\frac{(1+\text{i})^\text{n}}{(1-\text{i})^{\text{n}-2}}$
$=\frac{(1+\text{i})^\text{n}}{(1-\text{i})^{\text{n}}}(1-\text{i})^2$
$=\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^\text{n}\times(1-\text{i})^2$
$=\text{i}^\text{n}\big(1+\text{i}^2-2\times1\times\text{i}\big) \ \Big(\because\frac{1+\text{i}}{1-\text{i}}=\text{i},\text{using problem 10}\Big)$
$=\text{i}^\text{n}(1-1-2\text{i})$
$=-2\text{i}\times\text{i}^\text{n}$
$=-2\text{i}^{\text{n}+1}$
$\therefore$ For n = 1
$\text{z}=-2\text{i}^{1+1}$
$=-2\text{i}^2$
$=2,$ which is real number
$\therefore$ The smallest positive integer value of n is 1.

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