A) 3
B) 4
C) 1
D) 2
Correct Answer: C
Solution :
\[\frac{{{(1+i)}^{n}}}{{{(1-i)}^{n-2}}}={{\left( \frac{1+i}{1-i} \right)}^{n}}{{(1-i)}^{2}}\] \[=(1+{{i}^{2}}-2i){{\left( \frac{1+i}{1-i} \right)}^{n}}\] \[=(1-1-2i){{\left( \frac{1+i}{1-i} \right)}^{n}}\] \[=(-2i){{\left( \frac{1+i}{1-i} \right)}^{n}}\] \[=(-2i){{\left\{ \frac{(1+i)(1+i)}{(1-i)(1+i)} \right\}}^{n}}\] \[=(-2i){{\left\{ \frac{2i}{2} \right\}}^{n}}\] \[(\because \,\,{{i}^{2}}=-1)\] \[=(-2i)\,{{(i)}^{n}}\] \[=(-2i)\,{{i}^{n+1}}\] Put \[n=1,\] we get \[=(-2){{i}^{2}}=(-2)(-1)\] \[=2\Rightarrow \] positive integer. Hence, the least positive integer value of \[n=1\] for which the given expression have positive value.
You need to login to perform this action.
You will be redirected in
3 sec
