prove that the definition of $\log(z)$ is not everywhere continuous I would like to prove by contradiction that $\log(z) = \ln(r) + i\theta$ is not continuous at $\alpha = \theta$, for $r > 0$ and for $\alpha \le \theta \lt \alpha + 2\pi $. 
It seems pretty intuitive that $\alpha$ should not be equal to $\theta$ otherwise the function's derivative at that point would yield two results (i.e. one for $\alpha$ another for $\alpha + 2\pi$) in a neighbourhood around the point the derivative is taken and therefore violate the stipulation that the derivative must be consistent regardless of the path taken. 
However how can I prove this? 
I am assuming, (proof by contradiction) that $\log(z) = \ln(r) + i\theta$ is continuous at $\alpha = \theta$. But I don't know where to go from there. I am assuming that maybe the application of Cauchy-Riemann may be necessary. 
EDIT
Assuming that $\log(z) = \ln(r) + i\theta$ is continuous at $\alpha = \theta$, we know that the function is differentiable so long as the domain for $r$ avoids the singularity at $r = 0$, which it does. So we can say that $\log(z) = \ln(r) + i\theta$ is also complex differentiable. Because the function is differentiable it must satisfy Cauchy-Riemann:
C-R in complex form:
$$\begin{cases}
\frac{\partial}{\partial r} u(r, \theta) = \frac1r\frac{\partial}{\partial\theta}v(r, \theta) \\\\
\frac{\partial}{\partial r} v(r, \theta) = -\frac{1}{r}\frac{\partial}{\partial\theta}u(r, \theta)
\end{cases}
$$
we can manipulate our function into the form $\log(z) = u(r, \theta) + i v(r, \theta)$ such that:
$$\begin{cases}
u(r, \theta) = \ln(r)\\
v(r, \theta) = \theta
\end{cases}
$$
using C-R it can be shown that 
$1/r = 1/r$ and that $0=0$
which checks out. but if we replace $\theta$ by $\alpha$ we get 
$$\begin{cases}
\frac{\partial}{\partial r} \ln(r) = \frac{1}{r} \frac{\partial}{\partial \theta} \alpha\\\\
\frac{\partial}{\partial r} \alpha = -\frac{1}{r} \frac{\partial}{\partial \theta} \ln(r)
\end{cases}
$$
it is easy to show that these reduce to absurdities namely, 
$$\begin{cases}
\frac{1}{r} = 0\\\\
0=0
\end{cases}
$$
Is this a sufficient proof? 
 A: $\log$ is continuous iff it's sequentially continuous. But this is clearly not the case for $\theta = \alpha$: Let  $\alpha_n := \alpha+2\pi- \frac{1}{n}$. Then $$e^{\imath \, \alpha_n} \to e^{\imath \, \alpha} \qquad (n \to \infty)$$
but $$\log(e^{\imath \, \alpha_n}) = \imath \, \left(\alpha+2\pi- \frac{1}{n} \right) \to \imath \, (\alpha+2\pi) \not= \log(e^{\imath \, \alpha}) = \imath \, \alpha$$
Edit: No, your proof doesn't work, because you cannot simply replace $\theta$ by $\alpha$. If you write down the Cauchy-Riemann equations for $\theta = \alpha$, you obtain simply
$$\begin{cases}
\frac{\partial}{\partial r} u(r, \theta) \bigg|_{r=r_0,\theta=\alpha} = \frac1r\frac{\partial}{\partial\theta}v(r, \theta) \bigg|_{r=r_0,\theta=\alpha}  \\\\
\frac{\partial}{\partial r} v(r, \theta) \bigg|_{r=r_0,\theta=\alpha}  = -\frac{1}{r}\frac{\partial}{\partial\theta}u(r, \theta)\bigg|_{r=r_0,\theta=\alpha} 
\end{cases}
$$ i.e. you have to calculate the derivatives first and afterwards (!) evaluate the derivatives at the point $(r_0,\alpha)$. 
The point is that these derivatives don't exist at points of the form $(r_0,\alpha)$, since your functions are defined on $(0,\infty) \times [\alpha,\alpha+2\pi) $ and this domain isn't a neighboorhood of $(r_0,\alpha)$.
A: It's not quite clear what $\alpha$ means in your question, but anyway: Your problem has nothing to do with continuity; it's manifest before you have learned anything about continuity or limits, let alone CR-equations.
The central fact at stake here is that you cannot define the polar angle $\theta$ as a real valued function on $\dot{\mathbb R}^2$, resp. $\dot{\mathbb C}$. Whenever you admit going a full turn around the origin the polar angle will increase (or decrease) by the amount $2\pi$. This is a basic phenomenon (an "Urphänomen") of geometry that will only be obscured by taking care of it using CR-equations.
A: The conversion of $z$ to polar form yields $Re^{i \theta}$. If we let $z_0$ be some arbitrary point in the complex plane, it is the ray with length $R$ located at the angle $\alpha$ from 0.
If the function given is everywhere continuous over the interval $\alpha \le \theta < \alpha + 2 \pi$, where $\alpha$ is some arbitrary angle, then $\lim_{z \to z_0}f(z)$ must exist and its one-sided limits must also exist. So we show that:
$\lim_{z \to z_0^+}f(z) = \lim_{z \to z_0^-}f(z)$ (1)
(If we parametrize $f(z)$ with respect to $\theta$ we get: $\ln(R) + i \alpha$ and $ln(R) + i (\alpha + 2 \pi$) )
So we can rewrite (1) in polar form as:
$\lim_{\theta \to \alpha^c} \ln(R) + i \alpha = \lim_{\theta \to \alpha^{cc}} \ln(R) + i (\alpha + 2 \pi)$
where the script $c$ and $cc$ denotes clockwise and counter clockwise respectively.
as $\theta \to \alpha$, the clockwise limit goes to $i \alpha$ and the counter clockwise limit goes to $i (\alpha + 2 \pi)$. so no matter how small my value for $epsilon$ gets, the limits will always differ by a factor of $2 \pi i $. Because $\lim_{z \to z_0^+}f(z) \neq \lim_{z \to z_0^-}f(z)$, the function cannot be everywhere continuous over $\alpha \le \theta < \alpha + 2 \pi$.
