Question

For positive integers \[{{n}_{1}}\] and \[{{n}_{2}}\] the values of the expression,
\[{{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{3}} \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{5}} \right)}^{{{n}_{2}}}}+{{\left( 1+{{i}^{7}} \right)}^{{{n}_{2}}}}\], where \[i=\sqrt{-1}\]
Is real if and only if
(a). \[{{n}_{1}}={{n}_{2}}+1\]
(b). \[{{n}_{1}}={{n}_{2}}-1\]
(c). \[{{n}_{1}}={{n}_{2}}\]
(d). \[{{n}_{1}}>0,{{n}_{2}}>0\]

Answer 1

Hint: We will first convert all the complex numbers of the form \[a+ib\] to the polar form. Then we will use the law of indices which says \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\]. This would give us all the terms in the same format of \[{{e}^{i\theta }}\] which becomes easier to solve. Finally we simplify and get the answer.

Complete step-by-step answer:

Given, expression is \[{{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{3}} \right)}^{{{n}_{1}}}}+{{\left( 1+{{i}^{5}} \right)}^{{{n}_{2}}}}+{{\left( 1+{{i}^{7}} \right)}^{{{n}_{2}}}}\].
Now, \[{{i}^{3}}\] can be written as, \[{{i}^{2+1}}={{i}^{2}}\times i\].
Now, \[i=\sqrt{-1}\], thus, \[{{i}^{2}}=-1\].
Therefore, \[{{i}^{3}}=-i\].
Now, \[{{i}^{5}}\] can be written as, \[{{i}^{4+1}}={{i}^{4}}\times i\].
Now, \[{{i}^{4}}={{\left( {{i}^{2}} \right)}^{2}}\]
\[\begin{align}
  & {{i}^{4}}={{\left( -1 \right)}^{2}} \\
 & {{i}^{4}}=1 \\
\end{align}\]
Thus, \[{{i}^{5}}=1\times i\]
\[{{i}^{5}}=i\]
Now, \[{{i}^{7}}={{i}^{5+2}}\]
\[{{i}^{7}}={{i}^{5}}\times {{i}^{2}}\]
Now, \[{{i}^{5}}=i\] and \[{{i}^{2}}=-1\]. Thus, \[{{i}^{7}}=i\times -1\], which is \[{{i}^{7}}=-i\].
Thus, our expression now becomes,
\[{{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1-i \right)}^{{{n}_{1}}}}+{{\left( 1+i \right)}^{{{n}_{2}}}}+{{\left( 1-i \right)}^{{{n}_{2}}}}\]
Now, let us convert all the terms of the form \[a+ib\] to polar form \[r{{e}^{i\theta }}\].
We use, \[a+bi=\left( \sqrt{{{a}^{2}}+{{b}^{2}}} \right){{e}^{i\theta }}\], where \[\theta \] depends on the quadrant.
\[\theta ={{\tan }^{-1}}\left( \left| \dfrac{b}{a} \right| \right)\], if \[\left( a,b \right)\in \] I Quadrant.
\[\theta =\pi -{{\tan }^{-1}}\left( \left| \dfrac{b}{a} \right| \right)\], if \[\left( a,b \right)\in \] II Quadrant.
\[\theta =\pi +{{\tan }^{-1}}\left( \left| \dfrac{b}{a} \right| \right)\], if \[\left( a,b \right)\in \] III Quadrant.
\[\theta =-{{\tan }^{-1}}\left( \left| \dfrac{b}{a} \right| \right)\], if \[\left( a,b \right)\in \] IV Quadrant.
Thus, \[1+i\] is actually (1, 1) which belongs to I Quadrant.
Thus, \[1+i=\left( \sqrt{{{1}^{2}}+{{1}^{2}}} \right){{e}^{i\theta }},\theta ={{\tan }^{-1}}\left( \left| \dfrac{1}{1} \right| \right)\]
\[1+i=\sqrt{2}{{e}^{i\theta }},\theta ={{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4}\]
Therefore, \[1+i=\sqrt{2}{{e}^{i\dfrac{\pi }{4}}}\]
Similarly, \[1-i\] is actually (1, -1) which belongs to IV Quadrant.
Thus, \[1-i=\left( \sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}} \right){{e}^{i\theta }},\theta =-{{\tan }^{-1}}\left( \left| \dfrac{1}{-1} \right| \right)\]
\[1-i=\left( \sqrt{1+1} \right){{e}^{i\left( \dfrac{-\pi }{4} \right)}}\], since, \[\theta =-{{\tan }^{-1}}\left( 1 \right)=\left( \dfrac{-\pi }{4} \right)\]
 \[1-i=\sqrt{2}{{e}^{\dfrac{-i\pi }{4}}}\]
Thus, the expression \[{{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1-i \right)}^{{{n}_{1}}}}+{{\left( 1+i \right)}^{{{n}_{2}}}}+{{\left( 1-i \right)}^{{{n}_{2}}}}\] changes to be,
\[\begin{align}
  & {{\left( \sqrt{2}{{e}^{\dfrac{i\pi }{4}}} \right)}^{{{n}_{1}}}}+{{\left( \sqrt{2}{{e}^{\dfrac{-i\pi }{4}}} \right)}^{{{n}_{1}}}}+{{\left( \sqrt{2}{{e}^{\dfrac{i\pi }{4}}} \right)}^{{{n}_{2}}}}+{{\left( \sqrt{2}{{e}^{\dfrac{-i\pi }{4}}} \right)}^{{{n}_{2}}}} \\
 & ={{\left( \sqrt{2} \right)}^{{{n}_{1}}}}\left( {{e}^{\dfrac{{{n}_{1}}i\pi }{4}}} \right)+{{\left( \sqrt{2} \right)}^{{{n}_{1}}}}\left( {{e}^{\dfrac{-{{n}_{1}}i\pi }{4}}} \right)+{{\left( \sqrt{2} \right)}^{{{n}_{2}}}}\left( {{e}^{\dfrac{{{n}_{2}}i\pi }{4}}} \right)+{{\left( \sqrt{2} \right)}^{{{n}_{2}}}}\left( {{e}^{\dfrac{-{{n}_{2}}i\pi }{4}}} \right) \\
 & ={{\left( \sqrt{2} \right)}^{{{n}_{1}}}}\left[ {{e}^{\left( \dfrac{{{n}_{1}}\pi }{4} \right)i}}+{{e}^{-\left( \dfrac{{{n}_{1}}\pi }{4} \right)i}} \right]+{{\left( \sqrt{2} \right)}^{{{n}_{2}}}}\left[ {{e}^{\left( \dfrac{{{n}_{2}}\pi }{4} \right)i}}+{{e}^{-\left( \dfrac{{{n}_{2}}\pi }{4} \right)i}} \right] \\
\end{align}\]
(Taking out \[{{\left( \sqrt{2} \right)}^{{{n}_{1}}}}\] and \[{{\left( \sqrt{2} \right)}^{{{n}_{2}}}}\] as common)
Now, we know Euler’s formula as, \[{{e}^{i\theta }}=\cos \theta +i\sin \theta \] and \[{{e}^{-i\theta }}=\cos \theta -i\sin \theta \].
Adding \[{{e}^{i\theta }}\] and \[{{e}^{-i\theta }}\], we get,
\[\begin{align}
  & {{e}^{i\theta }}+{{e}^{-i\theta }}=\cos \theta +i\sin \theta +\cos \theta -i\sin \theta \\
 & \Rightarrow {{e}^{i\theta }}+{{e}^{-i\theta }}=2\cos \theta \\
\end{align}\]
Thus in the expression we have,
\[{{e}^{i\left( \dfrac{{{n}_{1}}\pi }{4} \right)}}+{{e}^{-i\left( \dfrac{{{n}_{1}}\pi }{4} \right)}}\] and \[{{e}^{i\left( \dfrac{{{n}_{2}}\pi }{4} \right)}}+{{e}^{-i\left( \dfrac{{{n}_{2}}\pi }{4} \right)}}\]
Replacing them with \[2\cos \left( \dfrac{{{n}_{1}}\pi }{4} \right)\] and \[2\cos \left( \dfrac{{{n}_{2}}\pi }{4} \right)\], respectively we get,
\[={{\left( \sqrt{2} \right)}^{{{n}_{1}}}}\left( 2\cos \left( \dfrac{{{n}_{1}}\pi }{4} \right) \right)+{{\left( \sqrt{2} \right)}^{{{n}_{2}}}}\left( 2\cos \left( \dfrac{{{n}_{2}}\pi }{4} \right) \right)\]
This expression is always real irrespective of what \[{{n}_{1}}\] and \[{{n}_{2}}\] values are.
But since the question says positive values of \[{{n}_{1}}\] and \[{{n}_{2}}\], we go with option (d).
Thus the correct option is option (d).

Note: A simple observation can solve this question in very less time.
Consider, \[{{\left( 1+i \right)}^{1}}+{{\left( 1-i \right)}^{1}}\].
This becomes \[1+i+1-i=2\], which is real.
Now, consider, \[{{\left( 1+i \right)}^{2}}+{{\left( 1-i \right)}^{2}}\].
This becomes, \[1+2i+{{i}^{2}}+1-2i+{{i}^{2}}\], which is \[2+2{{i}^{2}}\]. Since, \[{{i}^{2}}=-1\], \[2+2{{i}^{2}}=0\], which is real.
Now, consider, \[{{\left( 1+i \right)}^{3}}+{{\left( 1-i \right)}^{3}}\], which becomes, \[1+3i+3{{i}^{2}}+{{i}^{3}}+1-3i+3{{i}^{2}}-{{i}^{3}}\], which is \[2+6{{i}^{2}}\].
Since, \[{{i}^{2}}=-1\], \[2+6{{i}^{2}}=-4\], which is real.
Thus, \[{{\left( 1+i \right)}^{{{n}_{1}}}}+{{\left( 1-i \right)}^{{{n}_{1}}}}+{{\left( 1+i \right)}^{{{n}_{2}}}}+{{\left( 1-i \right)}^{{{n}_{2}}}}\] is independent of \[{{n}_{1}}\] and \[{{n}_{2}}\].
Thus we got with option (d) as the question says positive integers.