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

what values of $z$ makes $g(z)$ real, imaginary, or modulus <1?

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

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

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

What about $z$ = ? =

  • $\begingroup$ What do you mean by "$z=?=$" $\endgroup$ – Samrat Mukhopadhyay Feb 11 '15 at 4:54
  • $\begingroup$ the question explicitly asks for values of $z$ where $g$ is real, imaginary, or modulus <1 $\endgroup$ – rebc Feb 11 '15 at 5:11
  • $\begingroup$ Ok, then you should make it clearer in the question. $\endgroup$ – Samrat Mukhopadhyay Feb 11 '15 at 5:23

$g(z)=e^{-2x}(\cos 2y-i\sin 2y)$. So $g(z)$ is real iff $e^{-2x}\sin 2y=0\implies y=\frac{n\pi}{2},n\in \mathbb{Z}$ which is a set of horizontal lines in complex plane. Similarly $g(z)$ is imaginary if $y=\frac{2n+1}{4}\pi,\ n\in \mathbb{Z}$. $|g(z)|=e^{-2x}\le 1\ \forall x\in \mathbb{R}^+\cup\{0\}$, i.e. $g(z)$ is inside the unit circle in complex plane $\forall z\in \mathbb{C}$.

  • $\begingroup$ Hi Samrat thanks for the help. however, how do we explicitly find the values of $z$ and not only $y$? $\endgroup$ – rebc Feb 12 '15 at 2:18
  • $\begingroup$ For the case, say $y=\frac{n\pi}{2}$, the $x$ value can be anything, i.e. $x\in \mathbb{R}$, so $z$ can be any one of the points on the line $y=\frac{n\pi}{2}$ $\endgroup$ – Samrat Mukhopadhyay Feb 12 '15 at 6:31
  • $\begingroup$ isnt $g(z) = e^{-2x}(cos(-2y)-isin(-2y)$) ? $\endgroup$ – rebc Feb 12 '15 at 20:50
  • 1
    $\begingroup$ @rebc $e^{-i2y}=\cos (-2y)+i\sin(-2y)=\cos(2y)-i\sin(2y)$ $\endgroup$ – Samrat Mukhopadhyay Feb 13 '15 at 8:54

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