Question

If $z=r{e}^{i\theta }$, then $\left|e^{iz}\right|$ is equal to

• A. $e^{r\sin \theta }$
• B. $e^{-r\sin \theta }$
• C. $e^{-r\cos \theta }$
• D. $e^{r\cos \theta }$

Answer 1

The correct option is B

$e^{-r\sin \theta }$

Explanation for the correct option.

Find the value of the given expression.

It is known that $e^{i\theta }=\cos \theta +i\sin \theta$.

So the complex number $z$ is given as:

$\begin{array}{rcl}z& =& r{e}^{i\theta }\\ & =& r\left(\cos \theta +i\sin \theta \right)\end{array}$

Now multiply both sides by $i$ and find the complex number $iz$.

$\begin{array}{rcl}iz& =& ir\left(\cos \theta +i\sin \theta \right)\\ & =& r\left(i\cos \theta +i^2\sin \theta \right)\\ & =& r\left(i\cos \theta -\sin \theta \right)\left[i^2=-1\right]\\ & =& -r\sin \theta +ir\cos \theta \end{array}$

Now, $\left|e^{iz}\right|$ is given as:

$\begin{array}{rcl}\left|e^{iz}\right|& =& \left|e^{-r\sin \theta +ir\cos \theta }\right|\\ & =& \left|e^{-r\sin \theta }·e^{ir\cos \theta }\right|\left[a^{m+n}=a^m{a}^n\right]\\ & =& \left|e^{-r\sin \theta }\right|·\left|e^{ir\cos \theta }\right|\left[e^{iz}=1\right]\\ & =& e^{-r\sin \theta }\end{array}$

So, $\left|e^{iz}\right|$ is equal to $e^{-r\sin \theta }$.

Hence, the correct option is B.