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

Hint: 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.