sin（z）的实际值范围-Rangeofrealvaluesofsin(z)

3  

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

等式$ \ sin z = w $为每个复数$ w $(特别是实数)提供解决方案。

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.

实际上,如果$ t = e ^ {iz} $,则等式变为$$ \ frac {tt ^ { - 1}} {2i} = w $$即$$ t ^ 2-2iwt-1 = 0 $$因此$ t = iw + \ sqrt {1-w ^ 2} $(或其倒数)。特别是$ 0 $永远不是解决方案,而等式$ e ^ {iz} = t $肯定有解决方案。

2  

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

基于我的评论,这是一种证明它而无需计算任何解决方案的方法:

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 uneven 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 $\sin(\mathbb C)=\mathbb C$.

as $ \ sin:\ mathbb C \ rightarrow \ mathbb C $是完全且非常数,我们可以应用Little Picard定理,因此我们有$ \ sin(\ mathbb C)= \ mathbb C $或$ \ sin( \ mathbb C)= \ mathbb C \ setminus \ {a \} $表示$ a \ in \ mathbb C $。让我们假设后者。因为$ \ sin $是一个不均匀的函数,这意味着$ -a \ notin \ sin(\ mathbb C)$。这只能是$ a = 0 $(否则我们会与Picard相矛盾),但我们都知道$ \ sin(0)= 0 $,因此$ \ sin(\ mathbb C)= \ mathbb C $。

0  

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

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

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 $\mathbb{R} \to [-1,1]$.

因此$ \ frac {\ pi} {2} + i \ mathbb {R} $的图像负责获得$ \ mathbb {R} _ {\ geq 1} $。类似地,$ \ frac { - \ pi} {2} + i \ mathbb {R} $的图像负责获得$ \ mathbb {R} _ {\ leq -1} $。 $ \ mathbb {R} $的图像负责获得$ [ - 1,1] $,$ \ sin $被视为通常的$ \ mathbb {R} \到[-1,1] $。