What is $\vert e^{(iz)} \vert$?

Question

$\vert z \vert = R$ is a given circle

$$z = Re^{(it)}$$

Therefore, $z = R \cos t + i R \sin t$

Now how do I calculate $\vert e^{(iz)} \vert$ in terms of "$\sin t$" only by substituting the value of $z$ in it?

Answer 1

By definition, if $z = a + ib$, then $|z| = \sqrt{a^2 + b^2}$. Thus,

$$ |e^{iz}| = |e^{ia}\cdot e^{-b}| = e^{-b} | \cos a + i \sin a| = e^{-b} $$

Answer 2

With

$z = R\cos t + iR \sin t, \tag 1$

we have

$iz = -R\sin t + iR\cos t; \tag 2$

thus

$e^{(iz)} = e^{-R\sin t + iR\cos t} = e^{-R\sin t}e^{iR \cos t}; \tag 3$

therefore,

$\vert e^{(iz)} \vert = \vert e^{-R\sin t} \vert \vert e^{iR \cos t} \vert = e^{-R\sin t}, \tag 4$

since

$\vert e^{iR \cos t} \vert = 1 \tag 5$

and

$\vert e^{-R \sin t}\vert = e^{-R\sin t}, \tag 6$

by virtue of the fact that $e^{-\sin t}$ is positive real.