Range of real values of sin(z) [closed]

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.

closed as unclear what you're asking by Mark Viola, user21820, Brahadeesh, user91500, José Carlos SantosSep 17 '18 at 15:20

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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.

• 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 $z = \text{Arg} (t)$. – AlexanderJ93 Sep 17 '18 at 1:45
• @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. – egreg Sep 17 '18 at 8:30
• Oh yeah, not sure what I was thinking – 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 $\sin(\mathbb C)=\mathbb C$.

• Nice use of the sledgehammer! – egreg Sep 16 '18 at 19:54
• @egreg always fun to use it when (un)necessary ;) – Hirshy Sep 16 '18 at 19:56
• Do you mean "odd" instead of "uneven"? – Eric Duminil Sep 17 '18 at 12:27
• @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) – Hirshy Sep 17 '18 at 17:41
• @Hirshy Alles klar! – 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

$$ic+\sqrt{1-c^2}=0$$

$$ic=-\sqrt{1-c^2}$$

$$-c^2=1-c^2$$

$$0=1$$

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