# Complex values of z for real and imaginary part

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