I have to prove the following assertion:
If $f(z)$ is continuous in the sector $0<\vert z-a\vert \le r_0, 0\le arg(z-a)\le \alpha ,(0<\alpha \le 2\pi)$ and the limit $\lim_{z\to a}[(z-a)f(z)]=A$ exist, then $$\lim_{r\to 0}\int_{C_r} f(z)dz=iA\alpha,$$ where $C_r$ is the arc of the circle $\vert z-a \vert <r$ which is present in the given sector, traversed in the positive direction.
Proof (attempt):
Let $\epsilon>0, \epsilon_2>0.$ Then as $f$ is continuos in the whole sector $0<\vert z-a\vert \le r_0$ then $f$ is continuous in the particular point $a$, i.e. there exist $\delta>0$ such that if $\vert z-a\vert <\delta$ then $\vert f(z)-f(a)\vert<\epsilon$.
Also we have the existence of a limit, so there exist $\delta_2$ such that if $0<\vert z-a\vert <\delta_2$ then $\vert (z-a)f(z)-A \vert<\epsilon_2$.
And now I just need to show that $\vert \int_{C_r}f(z)dz-iA\alpha \vert<\epsilon$ i.e. $\vert \int_0^{2\pi}f(z)dz-i\alpha \lim_{z\to a}(z-a)f(z) \vert<\epsilon$.
How to continue the proof?
Can someone help me with this proof?
Thank you.
Set $z=a+r\mathrm{e}^{it}$, $t\in [0,\alpha]$. Then $$ \int_{C_r}f(z)\,dz=\int_0^\alpha f(a+r\mathrm{e}^{it})\,ir\mathrm{e}^{it}\,dt. $$ Meanwhile $$ (z-a)f(z)=A+o(|z-a|)\quad\text{or}\quad r\,\mathrm{e}^{it}f(a+r\mathrm{e}^{it})=A+o(r). $$ Hence $$ \int_{C_r}f(z)\,dz=\alpha A i+\alpha\cdot o(r). $$