Question

If $z=re^{i\theta}$ , then $\left|e^{iz}\right|=$

• A. $1$
• B. $e^{2rsin\theta}$
• C. $e^{rsin\theta}$
• D. $e^{-rsin\theta}$

Answer 1

Answer: (D)

$z=re^{i\theta} =r\left(cos\,\theta+i\, sin\,\theta\right)$
$\therefore \, iz=ir\left(cos\,\theta+i\,sin\,\theta\right)=ir \, cos\,\theta+i^{2}\, r\, sin\,\theta$
$=-r sin\,\theta+ir\, cos\,\theta$
$\left|e^{iz}\right|=\left|e^{\left(-r\,sin\,\theta +ir\,cos\,\theta\right)}\right|=\left|e^{-r\,sin\,\theta}\right|\cdot\left|e^{ir\,cos\,\theta}\right|$
$=e^{-r\,sin\,\theta} \left[\sqrt{cos^{2}\left(r\,cos\,\theta\right)+sin^{2}\left(r\,cos\,\theta\right)}\right]$
$=e^{-r\, sin\,\theta} \times1=e^{-r\,sin\,\theta}$