Question

Solve the following complex expression
$\dfrac{{{\left( \cos \theta +i\sin \theta \right)}^{4}}}{{{\left( \sin \theta +i\cos \theta \right)}^{5}}}$
(a) $\cos 9\theta -i\sin 9\theta$
(b) $\cos 9\theta +i\sin 9\theta$
(c) $\sin 9\theta -i\cos 9\theta$
(d) $\sin 9\theta +i\cos 9\theta$

Answer 1

Hint: Any complex number that can be represented in the form $\cos \theta +i\sin \theta$ can be also written as ${{e}^{i\theta }}$. This form of the complex number is also called the euler form of the complex number. Using this euler form, we can solve this question.

Complete step-by-step answer:

Before proceeding with the question, we must know the concept and the formula that will be required to solve this question.
Any complex number that can be written in the form of $\cos \theta +i\sin \theta$ can be also expressed in the euler form. From the euler form, we can write the complex number $\cos \theta +i\sin \theta ={{e}^{i\theta }}$ . . . . . . . . (1)
Also, in the complex number, we have a formula ${{i}^{2}}=-1$ . . . . . . . . . (2).
In the question, we have to evaluate $\dfrac{{{\left( \cos \theta +i\sin \theta \right)}^{4}}}{{{\left( \sin \theta +i\cos \theta \right)}^{5}}}$.
$\dfrac{{{\left( \cos \theta +i\sin \theta \right)}^{4}}}{{{\left( \sin \theta +i\cos \theta \right)}^{5}}}$can be also written as,

& \dfrac{{{\left( \cos \theta +i\sin \theta \right)}^{4}}}{{{\left( i \right)}^{5}}{{\left( \dfrac{1}{i}\sin \theta +\cos \theta \right)}^{5}}} \\\ & \Rightarrow \dfrac{{{\left( \cos \theta +i\sin \theta \right)}^{4}}}{{{\left( i \right)}^{4}}\left( i \right){{\left( \dfrac{i}{{{i}^{2}}}\sin \theta +\cos \theta \right)}^{5}}} \\\ & \Rightarrow \dfrac{{{\left( \cos \theta +i\sin \theta \right)}^{4}}}{{{\left( {{i}^{2}} \right)}^{2}}\left( i \right){{\left( \dfrac{i}{{{i}^{2}}}\sin \theta +\cos \theta \right)}^{5}}} \\\ \end{aligned}$$ Using formula (2), we can write ${{i}^{2}}=-1$. So, we get, $$\begin{aligned} & \dfrac{{{\left( \cos \theta +i\sin \theta \right)}^{4}}}{{{\left( -1 \right)}^{2}}\left( i \right){{\left( -i\sin \theta +\cos \theta \right)}^{5}}} \\\ & \Rightarrow \dfrac{1}{i}\dfrac{{{\left( \cos \theta +i\sin \theta \right)}^{4}}}{{{\left( -i\sin \theta +\cos \theta \right)}^{5}}} \\\ & \Rightarrow \dfrac{i}{{{i}^{2}}}\dfrac{{{\left( \cos \theta +i\sin \theta \right)}^{4}}}{{{\left( -i\sin \theta +\cos \theta \right)}^{5}}} \\\ & \Rightarrow -i\dfrac{{{\left( \cos \theta +i\sin \theta \right)}^{4}}}{{{\left( \cos \theta -i\sin \theta \right)}^{5}}} \\\ \end{aligned}$$ From (1), we can write $\cos \theta +i\sin \theta ={{e}^{i\theta }}$ and $\cos \theta -i\sin \theta ={{e}^{-i\theta }}$. So, we get, $$\begin{aligned} & -i\dfrac{{{\left( {{e}^{i\theta }} \right)}^{4}}}{{{\left( {{e}^{-i\theta }} \right)}^{5}}} \\\ & \Rightarrow -i\dfrac{{{\left( {{e}^{i\theta }} \right)}^{4}}}{{{\left( {{e}^{i\theta }} \right)}^{-5}}} \\\ \end{aligned}$$ There is a rule of exponents from which we can say $\dfrac{{{a}^{b}}}{{{a}^{c}}}={{a}^{b-c}}$. Using this formula in the above expression, we get, $\begin{aligned} & -i{{\left( {{e}^{i\theta }} \right)}^{4-\left( -5 \right)}} \\\ & \Rightarrow -i{{\left( {{e}^{i\theta }} \right)}^{9}} \\\ & \Rightarrow -i\left( {{e}^{i9\theta }} \right) \\\ \end{aligned}$ Converting this euler form to the complex number in the form of $\cos \theta $ and $\sin \theta $, we can get, $$\begin{aligned} & -i\left( {{e}^{i9\theta }} \right)=-i\left( \cos 9\theta +i\sin 9\theta \right) \\\ & \Rightarrow -i\cos 9\theta -{{i}^{2}}\sin 9\theta \\\ & \Rightarrow -i\cos 9\theta -\left( -1 \right)\sin 9\theta \\\ & \Rightarrow \sin 9\theta -i\cos 9\theta \\\ \end{aligned}$$ Hence, the answer is option (c). Note: There is a possibility that one may commit a mistake while applying the formula $\dfrac{{{a}^{b}}}{{{a}^{c}}}={{a}^{b-c}}$ on the expression $$\dfrac{{{\left( {{e}^{i\theta }} \right)}^{4}}}{{{\left( {{e}^{i\theta }} \right)}^{-5}}}$$. It is possible that one may write the expression $$\dfrac{{{\left( {{e}^{i\theta }} \right)}^{4}}}{{{\left( {{e}^{i\theta }} \right)}^{-5}}}$$ as $${{\left( {{e}^{i\theta }} \right)}^{4-5}}$$ instead of $${{\left( {{e}^{i\theta }} \right)}^{4-\left( -5 \right)}}$$ which will lead us to an incorrect answer. So, one must be careful while applying such formulas specially on those terms which are having a negative exponent.