If we have $f(z) = e^{iz} $, what values of $z$ for which $f(z)$ is real, imaginary, or of modulus <1?

We know that $e^{iz} = cos(z)+isin(z)$,

hence, $f(z)$ is real if $sin(z)=0$, imaginary if $cos (z)=0$, right?

What values of $z$ though? real: $z=sin^{-1}0 = 0$ ? I'm confused!

Also, what about if $f$ is of modulus $<1$ ?

Also, how do we go about if the function is $g(z) = e^{-2z}$ ?

$g(z) = e^{-2(x+iy)} $ = $e^{-2x-i2y} $

= $e^{-2x}e^{i(-2y)} $ =$e^{-2x} [cos(-2x)+isin(-2x)]$

Hence, real if $cos(-2x)=0$ ?


$f(z)$ is real if and only if $z=\pi k + yi$ for some integer $k$ and real $y$.

$f(z)$ is imaginary if and only if $z=\pi (k +\frac12)+ yi$ for some integer $k$ and real $y$.

$|f(z)|<1$ if and only if $z = x+yi$ for some real $x$ and some positive $y$.

These three conditions can be stated more succinctly as $\operatorname{Re} z =\pi k$, $\operatorname{Re} z =\pi (k+\frac12)$, and $\operatorname{Im} z >0$ where $k$ is integral.

  • $\begingroup$ could you also help me out with $g(z)$ ? $\endgroup$ – rebc Feb 11 '15 at 3:53
  • 1
    $\begingroup$ You need to think about why what I've said is true. Then you can try to write down similar conditions for $g$. $\endgroup$ – MPW Feb 11 '15 at 3:56

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