(C) Given,$(1+i)^n=(1-i)^n$ $\Rightarrow \frac{(1+i)^n}{(1-i)^n}=1$ $\Rightarrow \left[\frac{(1+i)(1+i)}{(1-i)(1+i)}\right]^n=1$ $\Rightarrow \left[\f

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(C) Given,$(1+i)^n=(1-i)^n$ $\Rightarrow \frac{(1+i)^n}{(1-i)^n}=1$ $\Rightarrow \left[\frac{(1+i)(1+i)}{(1-i)(1+i)}\right]^n=1$ $\Rightarrow \left[\frac{1+i^2+2i}{1-i^2}\right]^n=1$ $\Rightarrow \left[\frac{1-1+2i}{1+1}\right]^n=1$ $\Rightarrow \left(\frac{2i}{2}\right)^n=1$ $\Rightarrow i^n=1$ Since the smallest positive integer $n$ for which $i^n=1$ is $4$,therefore $n=4$.

Class 11 Mathematics 4-1.Complex numbers Integral power of iota, Algebraic operations and Equality of complex numbers

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The least positive integer $n$ for which $(1+i)^n=(1-i)^n$ is

TS EAMCET 2014 All TS EAMCET

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A
$8$
B
$2$
C
$4$
D
$6$

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