Question

Find: ${{\left( 1+i \right)}^{6}}+{{\left( 1-i \right)}^{6}}$
(a) 1
(b) 0
(c) – 1
(d) 2

Answer 1

Hint : First, use the formula $z=a+ib$ and find the value of ‘a’ and ‘b’ and then using them find $r=\sqrt{{{\text{a}}^{2}}+{{\text{b}}^{2}}}$, Now use the general form of complex number in the terms of $\cos \theta$ and $\sin \theta$ , $z=r\left( \cos \theta +i\sin \theta \right)$ , then find the value of $\theta$ from $\tan \theta$ , when we have $\cos \theta =1$ and $\sin \theta =1$ . Now, Substitute the required obtained values in $z=r\left( \cos \theta +i\sin \theta \right)$ and by using De Moivre’s theorem find the value of ${{\left( 1+i \right)}^{6}}$ . Similarly, find the value of ${{\left( 1-i \right)}^{6}}$ and add them up and find the final result.

Complete step-by-step answer :
In this question, we need to use the De Moivre’s theorem, to solve the following, let us solve separately for ${{\left( 1+i \right)}^{6}}$ and ${{\left( 1-i \right)}^{6}}$ .
We have the expression,
${{\left( 1+i \right)}^{6}}+{{\left( 1-i \right)}^{6}}$ ........................ (i)
Now, from this above equation, let us solve for ${{\left( 1+i \right)}^{6}}$ .
We know,
$z=a+ib$ and $z=1+i$ .............. (ii)
Hence, when we compare both the equations, we get a = 1 and b = 1
We know, $r=\sqrt{{{\text{a}}^{2}}+{{\text{b}}^{2}}}$
$\begin{aligned} & =\sqrt{{{1}^{2}}+{{1}^{2}}} \\\ & =\sqrt{2} \end{aligned}$
We also know that $z=r\left( \cos \theta +i\sin \theta \right)$ .............. (iii)
Compare the above equation with the equation (ii), we get,
$\cos \theta =1$ , $\sin \theta =1$ therefore, $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
Hence, $\tan \theta =1$
$\begin{aligned} & \theta ={{\tan }^{-1}}\left( 1 \right) \\\ & =\dfrac{\pi }{4} \end{aligned}$
Now, let us substitute the equation (ii) the value of $\theta$ and the value of $r$ in the equation (iii), we get
$1+i=\sqrt{2}\left[ \cos \left( \dfrac{\pi }{4} \right)+i\sin \left( \dfrac{\pi }{4} \right) \right]$
By De Moivre’s theorem, we know that
$\begin{aligned} & {{z}^{n}}={{\left[ r\left( \cos \theta +i\sin \theta \right) \right]}^{n}} \\\ & ={{r}^{n}}\left[ \cos \left( n\theta \right)+i\sin \left( n\theta \right) \right] \end{aligned}$
Using the above the theorem, we get
$\begin{aligned} & {{\left( 1+i \right)}^{6}}={{\left( \sqrt{2} \right)}^{6}}\left[ \cos \left( \dfrac{6\pi }{4} \right)+i\sin \left( \dfrac{6\pi }{4} \right) \right] \\\ & ={{\left( {{2}^{\dfrac{1}{2}}} \right)}^{6}}\left[ \cos \left( \dfrac{3\pi }{2} \right)+i\sin \left( \dfrac{3\pi }{2} \right) \right] \end{aligned}$
Simplify further to find the value of ${{\left( 1+i \right)}^{6}}$ , we get
${{\left( 1+i \right)}^{6}}={{\left( 2 \right)}^{3}}\left[ \cos \left( \dfrac{3\pi }{2} \right)+i\sin \left( \dfrac{3\pi }{2} \right) \right]$
Since, we know that $\dfrac{3\pi }{2}=\left( \pi +\dfrac{\pi }{2} \right)$ , we can expand the following.
${{\left( 1+i \right)}^{6}}=8\left[ \cos \left( \pi +\dfrac{\pi }{2} \right)+i\sin \left( \pi +\dfrac{\pi }{2} \right) \right]$
Now, according to the quadrant rules, we know, in the third quadrant in which we have $\left( \pi +\theta \right)$ , only ‘tan’ and ‘cot’ functions are positive. Hence, here, the values after solving for ‘cos’ and ‘sin’ the value we will get is negative.
${{\left( 1+i \right)}^{6}}=8\left[ -\cos \dfrac{\pi }{2}+i\left( -\sin \dfrac{\pi }{2} \right) \right]$
$\dfrac{\pi }{2}=90{}^\circ$ , therefore $\cos 90{}^\circ =0$ and $\sin 90{}^\circ =1$ , we get
$\begin{aligned} & {{\left( 1+i \right)}^{6}}=8\left[ -\left( 0 \right)+i\left( -1 \right) \right] \\\ & =8\left( 0-i \right) \\\ & =8\left( -i \right) \\\ & =-8i \end{aligned}$
Therefore, we got, ${{\left( 1+i \right)}^{6}}=-8i$ ....................... (iv)
Similarly, if we solve, for the next expression, which is ${{\left( 1-i \right)}^{6}}$ by solving with the help of De Moivre’s theorem, we will get ${{\left( 1-i \right)}^{6}}=8i$ ............ (v)
From, the equations, (i), (iv) and (v), we get
$\begin{aligned} & {{\left( 1+i \right)}^{6}}+{{\left( 1-i \right)}^{6}}=-8i+8i \\\ & =0 \end{aligned}$
So, the correct answer is “Option B”.

Note : We could also use Binomial theorem and expand the given expression but it would be a very long and tedious process due to which we can make many mistakes and can cause deduction in marks. In the first quadrant, all the trigonometric functions are positive, in the second quadrant, only sine and cosine functions are positive and in the fourth quadrant, only cosine and secant functions are positive.