Find the smallest positive integer value of n for which (1+i)^n (1-i)^ n-2 is a real number.

Let z= (1+i)^n (1-i)^ n-2 = (1+i)^n (1-i)^ n (1-i)^2 = ( 1+i 1-i )^n (1-i)^2 =i^n (1+i^2-2 1 ) ( 1+i 1-i =i,using problem 10 ) =i^n(1-1-2i) =-2i ^n =-2i^ n+1 For n = 1 z=-2i^ 1+1 =-2i^2 =2, which is real number The smallest positive integer value of n is 1.

Find the smallest positive integer value of n for which (1+i)^n (…

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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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