Question

The value of ${{\left( 1+i \right)}^{8}}+{{\left( 1-i \right)}^{8}}$ is:
a). 16
b). -16
c). 32
d). -32

Answer 1

Hint: In this problem, first we will use the formula which is used to convert complex numbers of the form \[a+bi\] to the form of $r{{e}^{i\theta }}$ . Then will apply the law of indices to simplify the terms. Adding both the terms in the final step will yield us the answer.

Complete step-by-step solution -
The formula we are going to use is $a+bi=\left( \sqrt{{{a}^{2}}+{{b}^{2}}} \right){{e}^{i\theta }}$ , where $\theta ={{\tan }^{-1}}\left( \dfrac{b}{a} \right)$ .
Thus, applying the formula to the first term in the question, we get,
$1+i=\left( \sqrt{{{1}^{2}}+{{1}^{2}}} \right){{e}^{i\theta }}$ , where $\theta ={{\tan }^{-1}}\left( \dfrac{1}{1} \right)$ .
Solving further we get, $1+i=\sqrt{1+1}{{e}^{i\dfrac{\pi }{4}}}$ , Since $\theta ={{\tan }^{-1}}\left( 1 \right)$ which is $\left( \dfrac{\pi }{4} \right)$
$\therefore 1+i=\sqrt{2}{{e}^{\dfrac{i\pi }{4}}}$ ……………………. (i)
Similarly let us apply the same formula to the second term $\left( 1-i \right)$ .
Thus, applying the formula, we get,
$1-i=\left( \sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}} \right){{e}^{i\theta }}$ , where $\theta ={{\tan }^{-1}}\left( \left| \dfrac{-1}{1} \right| \right)+\pi $
Solving further we get,
$1-i=\sqrt{1+1}{{e}^{i\left( \dfrac{5\pi }{4} \right)}}$ since $\theta ={{\tan }^{-1}}\left( \left| -1 \right| \right)+\pi $ which is $\left( \dfrac{5\pi }{4} \right)$
$\therefore 1-i=\sqrt{2}{{e}^{\dfrac{5i\pi }{4}}}$ …………………. (ii)
Now, the equation says,
${{\left( 1+i \right)}^{8}}+{{\left( 1-i \right)}^{8}}={{\left( \sqrt{2}{{e}^{\dfrac{i\pi }{4}}} \right)}^{8}}+{{\left( \sqrt{2}{{e}^{\dfrac{5i\pi }{4}}} \right)}^{8}}$ …………. Substituting from (i) and (ii).
Now applying the law of indices which says ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ , we get,
${{\left( \sqrt{2} \right)}^{8}}\left( {{e}^{\dfrac{i\pi }{4}\times 8}} \right)+{{\left( \sqrt{2} \right)}^{8}}\left( {{e}^{\dfrac{5i\pi }{4}\times 8}} \right)$
$=\left( {{2}^{\dfrac{1}{2}\times 8}} \right)\left( {{e}^{i\pi \times 2}} \right)+\left( {{2}^{\dfrac{1}{2}\times 8}} \right)\left( {{e}^{i\pi \times 10}} \right)$…………………. Since $\sqrt{2}={{2}^{\dfrac{1}{2}}}$
$\left( {{2}^{4}} \right)\left( {{e}^{2\pi i}} \right)+\left( {{2}^{4}} \right)\left( {{e}^{10\pi i}} \right)$ ………………… (iii)
Now we know the Euler’s formula as ${{e}^{i\theta }}=\cos \theta +i\sin \theta $ .
Now let us put $\theta $ as $2\pi $ and as $10\pi $ one at a time.
$\theta =2\pi \Rightarrow {{e}^{2\pi i}}=\cos \left( 2\pi \right)+i\sin \left( 2\pi \right)$
$\Rightarrow {{e}^{2\pi i}}=1$ ……….. (iv)
$\theta = 10\pi \Rightarrow {{e}^{10\pi i}}=\cos \left( 10\pi \right)+i\sin \left( 10\pi \right)$
$\Rightarrow {{e}^{10\pi i}}=1$ …………… (v)
Substituting (iv) and (v) in (iii) we get,
$\begin{align}
  & \left( {{2}^{4}} \right)\left( 1 \right)+\left( {{2}^{4}} \right)\left( 1 \right) \\
 & ={{2}^{4}}+{{2}^{4}} \\
 & =16+16 \\
 & 32 \\
\end{align}$
Thus, ${{\left( 1+i \right)}^{8}}+{{\left( 1-i \right)}^{8}}=32$ .
Therefore, option (c) is correct.

Note: An important point we need to remember is the formula for angle $\theta ={{\tan }^{-1}}\left( \dfrac{b}{a} \right)$ .
The formula actually has four parts,
i) $\theta ={{\tan }^{-1}}\left( \left| \dfrac{b}{a} \right| \right)$ , if $\left( a,b \right)\in I$ quadrant.
ii) $\theta =\pi -{{\tan }^{-1}}\left( \left| \dfrac{b}{a} \right| \right)$ , if $\left( a,b \right)\in II$ quadrant.
iii) $\theta =\pi +{{\tan }^{-1}}\left( \left| \dfrac{b}{a} \right| \right)$ , if $\left( a,b \right)\in III$ quadrant.
iv) $\theta =-{{\tan }^{-1}}\left( \left| \dfrac{b}{a} \right| \right)$ , if $\left( a,b \right)\in IV$ quadrant.
Applying the wrong formula would result in a wrong calculation of angle and thus would give you a wrong answer.