Question

The amplitude of ${{e}^{{{e}^{-i\theta }}}},\text{where }\theta \in R\text{ and }i=\sqrt{-1}$ is:
(a) $\sin \theta$
(b) $-\sin \theta$
(c) ${{e}^{\cos \theta }}$
(d) ${{e}^{\sin \theta }}$

Answer 1

First of all use ${{e}^{i\theta }}=\cos \theta +i\sin \theta$ in the given expression by replacing $\theta \text{ with }\left( -\theta \right)$. Now use ${{a}^{x+y}}={{a}^{x}}.{{a}^{y}}$ and separate the expression in two terms that is, ${{e}^{\cos \theta }}.{{e}^{i\sin \left( -\theta \right)}}$. Now, again use ${{e}^{i\theta }}=\cos \theta +i\sin \theta$ and then compare it with z = x + iy and use $\tan \alpha =\dfrac{y}{x}$ where $\alpha$ is the amplitude of the given expression.

Complete step-by-step answer:
In this question, we have to find the amplitude of ${{e}^{{{e}^{-i\theta }}}}$. First of all, we know that we can write any complex number z = x + iy as $r{{e}^{i\theta }}$ where r is the modulus of z and $\theta$ is the amplitude or argument of z. Also,
${{e}^{i\theta }}=\cos \theta +i\sin \theta ....\left( i \right)$
Let us consider the expression given in the question.
$E={{e}^{{{e}^{-i\theta }}}}$
By using equation (i) and replacing $\theta \text{ by }\left( -\theta \right)$ in it, we get,
$E={{e}^{\left[ \cos \left( -\theta \right)+i\sin \left( -\theta \right) \right]}}$
We know that ${{a}^{x+y}}={{a}^{x}}.{{a}^{y}}$. By using this in the above expression, we get,
$E={{e}^{\cos \left( -\theta \right)}}.{{e}^{i\sin \left( -\theta \right)}}$
We know that cos (– x) = cos x. By using this, we get,
$E={{e}^{\cos \theta }}.{{e}^{i\sin \left( -\theta \right)}}....\left( ii \right)$
By again using equation (i) and considering $\theta =\sin \left( -\theta \right)$ in it, we get,
$E={{e}^{\cos \theta }}\left[ \cos \left( \sin \left( -\theta \right) \right)+i\sin \left[ \sin \left( -\theta \right) \right] \right]$
$E={{e}^{\cos \theta }}\cos \left[ \sin \left( -\theta \right) \right]+i{{e}^{\cos \theta }}\sin \left[ \sin \left( -\theta \right) \right]$
Let us compare the above complex number with the general complex number z = x + iy. From this, we get,
$x={{e}^{\cos \theta }}\cos \left[ \sin \left( -\theta \right) \right]....\left( iii \right)$
$y={{e}^{\cos \theta }}\sin \left[ \sin \left( -\theta \right) \right]....\left( iv \right)$
We know that if the amplitude of z = x + iy is $\alpha$, then $\tan \alpha =\dfrac{y}{x}$. So, by substituting y and x from equation (iv) and (iii) respectively, we get,
$\tan \alpha =\dfrac{{{e}^{\cos }}\sin \left[ \sin \left( -\theta \right) \right]}{{{e}^{\cos \theta }}\cos \left[ \sin \left( -\theta \right) \right]}$
By canceling the like terms from the above equation, we get,
$\tan \alpha =\dfrac{\sin \left[ \sin \left( -\theta \right) \right]}{\cos \left[ \sin \left( -\theta \right) \right]}$
We know that $\dfrac{\sin x}{\cos x}=\tan x$. By using this in the above equation and considering $x=\sin \left( -\theta \right)$, we get,
$\tan \alpha =\tan \left[ \sin \left( -\theta \right) \right]$
By comparing the LHS and RHS of the above equation, we get,
$\alpha =\sin \left( -\theta \right)$
We know that sin (– x) = – sin x. By using this, we get,
$\alpha =-\sin \theta$
So, we get the amplitude of ${{e}^{{{e}^{-i\theta }}}}\text{ as }-\sin \theta$.
Hence, option (b) is the right answer.

Note: In this question, many students get confused between amplitude, argument, and modulus of complex numbers. So, they must note that in any general complex number of the form $r{{e}^{i\theta }}$, r is the modulus of a complex number and $\theta$ is the argument or amplitude of complex numbers. In the above question, students can also directly find the amplitude by comparing the expression of equation (ii) that is ${{e}^{\cos \theta }}.{{e}^{i\sin \left( -\theta \right)}}$ by $r{{e}^{i\theta }}$. Here, $r={{e}^{\cos \theta }}\text{ and }\theta =\sin \left( -\theta \right)$ which is our amplitude.