Find the amplitude of (e)^(e)^-i theta.
There is no typing mistake, power is actually two times. Answer is -sin theta. Experts pls help. Thanks

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} {\mathrm{e}}^{{\mathrm{e}}^{-\mathrm{i\theta }}} \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ \mathrm{z}=\left|\mathrm{z}\right|{\mathrm{e}}^{\mathrm{i}.\mathrm{sin\alpha }}=\left|\mathrm{z}\right|\left(\mathrm{cos\alpha }+\mathrm{isin\alpha }\right) ;\mathrm{equation}\left(\mathrm{i}\right) \\ \mathrm{where} \left|\mathrm{z}\right| \mathrm{is} \mathrm{modulus} \mathrm{of} \mathrm{z} \mathrm{and} \mathrm{\alpha } \mathrm{is} \mathrm{its} \mathrm{amplitude}. \\ {\mathrm{e}}^{{\mathrm{e}}^{-\mathrm{i\theta }}}= {\mathrm{e}}^{{\mathrm{e}}^{\mathrm{i}\left(-\mathrm{\theta }\right)}} \\ = {\mathrm{e}}^{\cos \left(-\mathrm{\theta }\right)+\mathrm{isin}\left(-\mathrm{\theta }\right)} \left[\mathrm{By} \mathrm{equation}\left(\mathrm{i}\right)\right] \\ ={\mathrm{e}}^{\mathrm{cos\theta }-\mathrm{isin\theta }} \\ ={\mathrm{e}}^{\mathrm{cos\theta }}.{\mathrm{e}}^{-\mathrm{isin\theta }} \\ ={\mathrm{e}}^{\mathrm{cos\theta }}.{\mathrm{e}}^{\mathrm{i}\left(-\mathrm{sin\theta }\right)} \\ \mathrm{On} \mathrm{comparing} \mathrm{it} \mathrm{with} \mathrm{equation}\left(\mathrm{i}\right), \mathrm{we} \mathrm{get}, \\ \mathbf{mod}\left(\mathbf{ }{\mathbf{e}}^{{\mathbf{e}}^{\mathbf{-}\mathbf{i\theta }}}\right)\mathbf{=}{\mathbf{e}}^{\mathbf{cos\theta }} \\ \mathbf{amplitude}\left({\mathbf{e}}^{{\mathbf{e}}^{\mathbf{-}\mathbf{i\theta }}}\right)\mathbf{=}\mathbf{-}\mathbf{sin\theta } \end{gathered}$$

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