Given $f(z) = \sin(z)$ such that $z$ is an element of the complex numbers is the range of the real part of $f(z)$ all the reals? Is the range of the real part of $f(z)$ all reals given that the imaginary part is zero?

Sorry for the formatting, I’m on mobile.


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The equation $\sin z=w$ has solution for every complex $w$ (in particular real).

Indeed, if $t=e^{iz}$, the equation becomes $$ \frac{t-t^{-1}}{2i}=w $$ that is $$ t^2-2iwt-1=0 $$ Therefore $t=iw+\sqrt{1-w^2}$ (or its inverse). In particular $0$ is never a solution and the equation $e^{iz}=t$ certainly has solution.

  • $\begingroup$ Maybe it's obvious, but the reason that $e^{iz} = t$ has a solution is because $|t| = \sqrt{(w)^2 + (\sqrt{1-w^2})^2} = 1$, so $z = \text{Arg} (t)$. $\endgroup$ – AlexanderJ93 Sep 17 '18 at 1:45
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    $\begingroup$ @AlexanderJ93 That only holds for real $w$ such that $-1\le w\le 1$ (which is not surprising). The real reason is that $t=0$ is not a solution for $t^2-2iwt-1=0$ and that the complex exponential function only misses $0$ as value. $\endgroup$ – egreg Sep 17 '18 at 8:30
  • $\begingroup$ Oh yeah, not sure what I was thinking $\endgroup$ – AlexanderJ93 Sep 17 '18 at 17:27

Building on my comment here is a way to prove it without needing to calculate any solutions:

as $\sin:\mathbb C\rightarrow\mathbb C$ is entire and non constant, we can apply the Little Picard Theorem and therefore we have that $\sin(\mathbb C)=\mathbb C$ or $\sin(\mathbb C)=\mathbb C\setminus\{a\}$ for one $a\in\mathbb C$. Let us assume the later. Because $\sin$ is an odd function this would mean that also $-a\notin\sin(\mathbb C)$. This can only be true if $a=0$ (else we would contradict Picard), but we all know that $\sin(0)=0$ and therefor $\sin(\mathbb C)=\mathbb C$.

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    $\begingroup$ Nice use of the sledgehammer! $\endgroup$ – egreg Sep 16 '18 at 19:54
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    $\begingroup$ @egreg always fun to use it when (un)necessary ;) $\endgroup$ – Hirshy Sep 16 '18 at 19:56
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    $\begingroup$ Do you mean "odd" instead of "uneven"? $\endgroup$ – Eric Duminil Sep 17 '18 at 12:27
  • $\begingroup$ @EricDuminil in fact I do, thank you! (In german it is called "even" and "un-even", so I guess that's where it came from) $\endgroup$ – Hirshy Sep 17 '18 at 17:41
  • $\begingroup$ @Hirshy Alles klar! $\endgroup$ – Eric Duminil Sep 17 '18 at 17:42

$$ \sin (i z + \frac{\pi}{2}) = \cosh z \\ \sin (i z - \frac{\pi}{2}) = - \cosh z\\ $$

So the image of $\frac{\pi}{2}+i\mathbb{R}$ takes care of getting $\mathbb{R}_{\geq 1}$. Similarly the image of $\frac{-\pi}{2}+i\mathbb{R}$ takes care of getting $\mathbb{R}_{\leq -1}$. The image of $\mathbb{R}$ takes care of getting $[-1,1]$ with $\sin$ being viewed as a usual $\mathbb{R} \to [-1,1]$.


Using Taylor series, it can be seen that $\sin(ix)=i\sinh(x)$, and $\cos(ix)=\cosh(x)$. Combine this with the angle addition formula, and we have

$$\sin(x+iy) = \sin(x)\cosh(y) + i\cos(x)\sinh(y).$$

For this to be a real number, we need

$$\cos(x)\sinh(y) = 0$$

$$\cos(x) = 0 \qquad\text{or}\qquad \sinh(y) = 0$$

For the right-hand possibility, the only solution to $\sinh(y)=0$ is $y=0$ . This makes the original expression $\sin(x+iy)=\sin(x)$ cover the interval $[-1,1]$ as $x$ varies.

For the left-hand possibility, the only solutions to $\cos(x)=0$ are $x=\frac\pi2+n\pi$ . This makes $\sin(x)=(-1)^n$, so the original expression $\sin(x+iy)=(-1)^n\cosh(y)$ covers the interval $[1,\infty)$ or $(-\infty,-1]$, depending on whether $n$ is even or odd.

Thus the range of real values is

$$(-\infty,-1]\cup[-1,1]\cup[1,\infty) = \mathbb R.$$

More generally, the equation

$$\sin(z) = c$$

for any given $c\in\mathbb C$, can always be solved. Of course, the solution isn't unique; it can be expressed with the multi-valued complex logarithm. Using $\sin(z)=-i\sinh(iz)$ , and $\text{arsinh}(z)=\sinh^{-1}(z)=\ln(z+\sqrt{z^2+1})$ , produces

$$z = -i\,\text{arsinh}(ic)$$

$$= -i\ln(ic+\sqrt{1-c^2})$$

The logarithm is defined everywhere except 0 . So the only possibility for non-existence of a solution is





which is not a possibility. Thus the above expression for $z$ always works.


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