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, ${{i}^{4}}={{\left( {{i}^{2}} \right)}^{2}}$

& {{i}^{4}}={{\left( -1 \right)}^{2}} \\\ & {{i}^{4}}=1 \\\ \end{aligned}$$ Thus, $${{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{aligned} & {{\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{aligned}$$ (Taking out $${{\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{aligned} & {{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{aligned}$$ Thus in the expression we have, $${{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.