1. ## Complex Analysis

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

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

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

How would one go to tackle this problem?

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

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

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

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

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

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