Find the smallest positive integer value of n for which 1- i n -1- n n-2 is a real number-

Let z=(1+i)^n/(1 n)^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× i) (∵ 1+i/1 i=i, using problem 10 ) =i^n(1 1 2i) = 2i× i^n = ...

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Standard XII

Mathematics

Purely Real

Find the smal...

Question

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

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Solution

$L e t z = \frac{(1 + i)^{n}}{(1 - n)^{n - 2}} = \frac{(1 + i)^{n}}{(1 - i)^{n}} (1 - i)^{2} = {(\frac{1 + i}{1 - i})}^{n} \times (1 - i)^{2} = i^{n} (1 + i^{2} - 2 \times 1 \times i) (∵ \frac{1 + i}{1 - i} = i, using problem 10) = i^{n} (1 - 1 - 2 i) = - 2 i \times i^{n} = - 2 i^{n + 1} ∴ For n=1 z = - 2 i^{1 + 1} = - 2 i^{2} = 2, which is a real number$

$∴$ The smallest positive integer value of n is 1.

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

Let z=(1+i)n(1−n)n−2=(1+i)n(1−i)n(1−i)2=(1+i1−i)n×(1−i)2=in(1+i2−2×1×i) (∵ 1+i1−i=i, using problem 10)=in(1−1−2i)=−2i×in=−2in+1∴ For n=1z=−2i1+1=−2i2=2, which is a real number ∴ The smallest positive integer value of n is 1.

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