A) \[{{n}_{1}}={{n}_{2}}+1\]
B) \[{{n}_{1}}={{n}_{2}}\]
C) \[{{n}_{1}},\,\,{{n}_{2}}\]are any two negative integers
D) \[{{n}_{1}},\,\,{{n}_{2}}\]are both any positive integers
Correct Answer: D
Solution :
Expression can be written as \[={{(1+i)}^{{{n}_{1}}}}+{{(1-i)}^{{{n}_{1}}}}+{{(1+i)}^{{{n}_{2}}}}+{{(1-i)}^{{{n}_{2}}}}\] \[={{2}^{{{n}_{1}}/2}}{{\left( \frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}} \right)}^{{{n}_{1}}}}+{{2}^{{{n}_{1}}/2}}{{\left( \frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}} \right)}^{{{n}_{1}}}}\] \[+{{2}^{{{n}_{2}}/2}}\,{{\left( \frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}} \right)}^{{{n}_{2}}}}+{{2}^{{{n}_{2}}/2}}\,{{\left( \frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}} \right)}^{{{n}_{2}}}}\] \[={{2}^{{{n}_{1}}/2}}\left\{ {{\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right)}^{{{n}_{1}}}}+{{\left( \cos \frac{\pi }{4}-i\sin \frac{\pi }{4} \right)}^{{{n}_{1}}}} \right\}\] \[+{{2}^{{{n}_{2}}/2}}\left\{ {{\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right)}^{{{n}_{2}}}}+{{\left( \cos \frac{\pi }{4}-i\sin \frac{\pi }{4} \right)}^{{{n}_{2}}}} \right\}\] \[={{n}^{{{n}_{1}}/2}}\left\{ \cos \frac{{{n}_{1}}\pi }{4}+i\sin \frac{{{n}_{1}}\pi }{4}+\cos \frac{{{n}_{1}}\pi }{4}-i\sin \frac{{{n}_{1}}\pi }{4} \right\}\] \[+{{2}^{{{n}_{2}}/2}}\left\{ \cos \frac{{{n}_{2}}\pi }{4}+i\sin \frac{{{n}_{2}}\pi }{4} \right.\]\[\left. +\cos \frac{{{n}_{2}}\pi }{4}-\sin \frac{{{n}_{2}}\pi }{4} \right\}\] \[={{2}^{{{n}_{1}}/2}}\cdot 2\cos \frac{{{n}_{1}}\pi }{4}+{{2}^{{{n}_{2}}/2}}\cdot 2\cos \frac{{{n}_{2}}\pi }{4}=\]real \[ie\],\[{{n}_{1}},\,\,{{n}_{2}}\]are any two positive integers.You need to login to perform this action.
You will be redirected in
3 sec