Math Help - Complex Analysis

1. Complex Analysis

Let f be a continuous function on $D=\left\{z:|z|=1\right\}$. Laat $\gamma:[0,2\pi]\to D$, defined by $\gamma(t)=e^{it}$.

Define the function $F$ as follows:
$F(z) = f(z)$ if $|z|=1$
$F(z) = \frac{1}{2\pi i}\int_\gamma \frac{f(\zeta)}{\zeta-z}d\zeta$ if $|z|< 1$

Is $F$ continuous on $\overline{D}(0,1)$ [Hint:Consider $f(z)=\overline{z}$]

How would one go to tackle this problem?

2. Originally Posted by Dinkydoe
Let f be a continuous function on $D=\left\{z:|z|=1\right\}$. Laat $\gamma:[0,2\pi]\to D$, defined by $\gamma(t)=e^{it}$.

Define the function $F$ as follows:
$F(z) = f(z)$ if $|z|=1$
$F(z) = \frac{1}{2\pi i}\int_\gamma \frac{f(\zeta)}{\zeta-z}d\zeta$ if $|z|< 1$

Is $F$ continuous on $\overline{D}(0,1)$ [Hint:Consider $f(z)=\overline{z}$]

How would one go to tackle this problem?
Do what the hint says, take $f(z) = \overline{z}$. If |z| = 1 then $F(z) = \overline{z}$.

If |z| < 1 then $F(z) = \frac{1}{2\pi i}\int_\gamma \frac{\overline{\zeta}}{\zeta-z}d\zeta$. But when $|\zeta|=1$, $\overline{\zeta} = \zeta^{-1}$. So we can write F(z) as $\frac{1}{2\pi i}\int_\gamma \frac1{\zeta(\zeta-z)}d\zeta = -\frac1z+\frac1z=0$ (using the residue theorem to evaluate the integral, since the integrand is now analytic except for poles at $\zeta=0$ and $\zeta=z$).

That shows that F(z) is not continuous anywhere on $\gamma$.