Question

If $z=\cos \theta +\sin \theta$ , then the value of ${{z}^{n}}+\dfrac{1}{{{z}^{n}}}$ will be:
a)$\sin 2n\theta$
b)$2\sin n\theta$
c)$2\cos n\theta$
d)$\cos 2n\theta$

Answer 1

Hint: Consider the fact that ${{e}^{i\theta }}$ can also be written as $\cos \theta +i\sin \theta$.
Then $z={{e}^{i\theta }}$ which can also be written as ${{z}^{n}}={{e}^{in\theta }},{{z}^{-n}}={{e}^{-i\theta }}$ and then use the form that ${{e}^{in\theta }}$ can be written as $\cos \left( n\theta \right)+i\sin \left( n\theta \right),{{e}^{-in\theta }}$ can be written as $\cos n-i\sin \theta$.

Complete step-by-step answer:
After that add them and get the desired result.

In the question we are given the value of function z which is written as,
$z=\cos \theta +i\sin \theta$
As we know that ${{e}^{i\theta }}=\cos \theta +i\sin \theta ,{{e}^{-i\theta }}=\cos \theta -i\sin \theta$ we will first prove it use it in the solution,
By using Taylor’s series we can write $\sin \theta$ as,
$\sin \theta =\theta -\dfrac{{{\theta }^{3}}}{3!}+\dfrac{{{\theta }^{5}}}{5!}-\dfrac{{{\theta }^{7}}}{7!}+...........$
By using Taylor’s series we can write $\cos \theta$ as,
$\cos \theta =1-\dfrac{{{\theta }^{2}}}{2!}+\dfrac{{{\theta }^{4}}}{4!}-\dfrac{{{\theta }^{6}}}{6!}+...........$
By using Taylor’s series we can write ${{e}^{\theta }}$ as ,
${{e}^{\theta }}=1+\theta +\dfrac{{{\theta }^{2}}}{2!}+\dfrac{{{\theta }^{3}}}{3!}+...........$
By using Taylor’s series we can write ${{e}^{-\theta }}$ as,
${{e}^{-\theta }}=1+\left( -\theta \right)+\dfrac{{{\left( -\theta \right)}^{2}}}{2!}+\dfrac{{{\left( -\theta \right)}^{3}}}{3!}+...........$
${{e}^{-\theta }}=1-\theta +\dfrac{{{\theta }^{2}}}{2!}-\dfrac{{{\theta }^{3}}}{3!}+...........$
In the series of ${{e}^{\theta }}$ we will just replace $\theta$ by $i\theta$ then we get,
${{e}^{i\theta }}=1+\left( i\theta \right)+\dfrac{{{\left( i\theta \right)}^{2}}}{2!}+\dfrac{{{\left( i\theta \right)}^{3}}}{3!}+...........$
Which can be simplified and written by using fact that
${{i}^{2}}=-1,{{i}^{3}}=-i.$
So, we get
${{e}^{i\theta }}=1+i\theta -\dfrac{{{\theta }^{2}}}{2!}-\dfrac{{{\theta }^{3}}}{3!}...........$
Now separating terms which are with ‘i’ and which are without ‘i’ so we get,
${{e}^{i\theta }}=\left( 1-\dfrac{{{\theta }^{2}}}{2!}+........ \right)+i\left( \theta -\dfrac{{{\theta }^{3}}}{3!}+........ \right)$
Now we found that,
$\begin{aligned} & \cos \theta =1-\dfrac{{{\theta }^{2}}}{2!}+......... \\\ & sin\theta =\theta -\dfrac{{{\theta }^{3}}}{3!}+......... \\\ \end{aligned}$
We get,
${{e}^{i\theta }}=\cos \theta +i\sin \theta$
In the question we were given,
$z=\cos \theta +i\sin \theta$
So we can also write
$z={{e}^{i\theta }}$
So if,$z={{e}^{i\theta }}$ then ${{z}^{n}}={{e}^{in\theta }}$ .
${{e}^{in\theta }}=\cos \theta +i\sin \left( n\theta \right)$
Then we can say,
${{z}^{n}}=\cos \theta +i\sin \theta \left( n\theta \right)$
Now if $z=\cos \theta +i\sin \theta$ then,
$\dfrac{1}{z}=\dfrac{1}{\cos \theta +i\sin \theta }$
Now rationalizing or multiplying $\cos \theta -i\sin \theta$ to numerator and denominator we get,
$\dfrac{1}{z}=\dfrac{\cos \theta -i\sin \theta }{\left( \cos \theta +i\sin \theta \right)\left( \cos \theta -i\sin \theta \right)}$
Now by using formula
$\left(a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
We will simplify it as,
$\dfrac{1}{z}=\dfrac{\cos \theta -i\sin \theta }{{{\cos }^{2}}\theta -{{i}^{2}}\sin \theta }$
The value of ${{i}^{2}}=-1$ so we can write it as,
$\dfrac{1}{z}=\dfrac{\cos \theta -i\sin \theta }{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }$
Now by using identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ we get,
$\dfrac{1}{z}=\cos \theta -i\sin \theta$
We can write it as,
$\dfrac{1}{z}=\cos \left( -\theta \right)+i\sin \left( -\theta \right)$
So if $\cos \theta +\sin \theta ={{e}^{i\theta }}$
Then we can say,
$\cos \left( -\theta \right)+i\left( \sin \left( -\theta \right) \right)={{e}^{i\left( -\theta \right)}}$
So,
$\begin{aligned} & \cos \theta +i\sin \theta ={{e}^{-i\theta }} \\\ & \dfrac{1}{z}={{e}^{-i\theta }} \\\ \end{aligned}$
Then we can say,
$\dfrac{1}{{{z}^{n}}}={{e}^{-in\theta }}$
We can express ${{e}^{-in\theta }}$ as,
$\begin{aligned} & {{e}^{-in\theta }}=\cos \left( -n\theta \right)+i\sin \left( -n\theta \right) \\\ & =\cos \left( n\theta \right)-i\sin \left( n\theta \right) \\\ \end{aligned}$
Hence we can tell,
$\dfrac{1}{z}=\cos n\theta -i\sin n\theta$
We have been asked to find ${{z}^{n}}+\dfrac{1}{{{z}^{n}}}$ so,
$\begin{aligned} & {{z}^{n}}+\dfrac{1}{{{z}^{n}}}=\cos \left( n\theta \right)+i\sin \left( n\theta \right)+\cos n\theta -i\sin \left( n\theta \right) \\\ & {{z}^{n}}+\dfrac{1}{{{z}^{n}}}=2\cos \theta \\\ \end{aligned}$
The correct option is ‘c’.

Note: Students generally have confusion how to find ${{z}^{n}}$ if $z=\cos \theta +\sin \theta$ as they forget that $\cos \theta +i\sin \theta$ can be represented as ${{e}^{i\theta }},{{e}^{in\theta }}$ can also be represented as $\cos n\theta +i\sin n\theta$ that is why they should well versed with all the forms to use them easily.