Continuity of $f(z) = \sin(\theta)$ - how to prove? If $f: \mathbb{C}\to\mathbb{C}$ is defined by $f[r(\cos(\theta)+i\sin(\theta)]=\sin\theta$ if $r>0$, and $f(0) = 0$, then how does one prove that $f$ is discontinuous at $0$ and continuous everywhere else? I've attempted to show continuity with $\varepsilon$-$\delta$, but failed. I have no idea how to apply this definition in this case (I was trying to show that the limit exists everywhere, unless $r=0$).
For the discontinuity case, can it be shown by considering $\lim\limits_{z\to z_0} f(z)=\lim\limits_{r\to 0}\frac{\sin\theta}{r(\cos\theta-i\sin\theta)}\to\infty$, and thus the limit does not exist, hence $f$ is not continuous at $0$?
 A: For continuity away from the origin, consider points $z = re^{i \theta}$ and $z_0 = r_0e^{i \theta_0}.$ 
Suppose $|z - z_0| < \delta.$ In this case, $z$ is in a disk of radius $\delta$ with center $z_0$. Using some geometry we find
$$|\theta - \theta_0| \leqslant \arcsin \frac{\delta}{r_0},$$
and
$$| \sin \theta - \sin \theta_0| = 2 \left|\sin \left(\frac{\theta - \theta_0}{2} \right)\right| \left|\cos \left(\frac{\theta + \theta_0}{2} \right)\right| \leqslant 2 \left|\sin \left(\frac{\theta - \theta_0}{2} \right)\right| \\ \leqslant |\theta - \theta_0|.$$
For any $\epsilon > 0$, choose  $\delta < r_0 \sin  \epsilon.$  
Then $|z - z_0| < \delta \implies | \sin \theta - \sin \theta_0| < \epsilon$
The function is discontinuous at $z = 0$, since $|\sin \theta -0|$ can assume non-zero values in any neighborhood regardless of the magnitude of $|z - 0| = r.$
A: For discontinuity at $0$, approach $0$ along the lines $\theta = \pi/3$ and $\theta = \pi/4$.
For continuity, let $\varepsilon > 0$ and $x = r_0(\cos \theta_0 + \mathrm{i} \sin \theta_0)$ be given.  Can you find a $\delta$ (depending on $x$ and $\varepsilon$) such that for any $y = r(\cos \theta + \mathrm{i} \sin \theta)\in \Bbb{C}$ with $|x-y| < \delta$, it must be the case that $|f(x) - f(y)| = |\sin \theta_0 - \sin \theta| < \varepsilon$?  (Note that $|x-y| < \delta$ places limits on how much $r$ and $r_0$ may differ and on how much $\theta$ and $\theta_0$ my vary.  Be careful with $x$ near zero since small neighborhoods of $x$ can contain points on the other side of the origin -- i.e., having very different $\theta$s.  You may want to make sure that $\delta$ is so small that you don't need to worry about including points on the far side of the origin.)
(The answer is clearly, "yes, you can" because the condition $|\sin \theta_0 - \sin \theta| < \varepsilon$ just means that $y$ is in a certain sector of the plane and no matter where $x$ is (as long as $x \neq 0$) you can always pick $\delta$ so small that the disk of radius $\delta$ around $x$ fits in that sector.  This argument is not a proof -- you need inequalities to show that this geometric description works.  In fact this is generic: did you draw a picture to help you understand what $f$ is doing?)
A: Using Euler's identity you can write a direct formula for $f$, which makes continuity obvious where polar coordinates are unique.
Geometrically speaking, you function works like this: draw a line $L_z$ through your given point $z$ and the origin, and then project the intersection of $L_z$ and the unit circle onto the y-axis.
In particular, $f$ is "projective", ie constant on lines through the origin. Now you can just approach $0$ along any line which isn't the x-axis to get a jump discontinuity at $0$.
edit: actually it's not projective, since it's only constant on half-lines through the origin. On the other half the sign changes.
