A) 4
B) 8
C) 2
D) 12
Correct Answer: A
Solution :
Given, \[{{(1+i)}^{2n}}={{(1-i)}^{2n}}\] \[\Rightarrow \] \[{{\left( \frac{1+i}{1-i} \right)}^{2n}}=1\] \[\Rightarrow \] \[{{\left[ \frac{{{(1+i)}^{2}}}{{{(1-i)}^{2}}} \right]}^{n}}=1\] \[\Rightarrow \] \[{{\left[ \frac{1+{{i}^{2}}+2i}{1+{{i}^{2}}-2i} \right]}^{n}}=1\] \[\Rightarrow \] \[{{\left[ \frac{1-1+2i}{1-1-2i} \right]}^{n}}=1\] \[[\because {{i}^{2}}=-1]\] \[\Rightarrow \] \[{{[-1]}^{n}}=1\] \[\Rightarrow \] \[n=2\] \[[\because {{(-1)}^{2}}=1]\]
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