How can I show this statement. Show that there is no holomorphic fuction $f$ in the unit disc $D$ that  extends continuously to boundary of $D$ such that $f(z)=\frac{1}{z} ~for~ z\in \partial( D) $.


I tried to apply maximum principle but I couln't find the way to prove it.

Help me please.

I just update the full statement and I think it probably assume it is not constant fuction.

Thank you.
If such a function existed, then for all $r &lt; 1$ you'd have $\int_{|z| = r} f(z)\,dz = 0$ by Cauchy's Theorem. Taking limits as $r \rightarrow 1$ would give $\int_{|z| = 1} f(z)\,dz = 0$ as well, 
contradicting that  $\int_{|z| = 1} {1 \over z}\,dz = 2\pi i$.
By maximum principle we get that the function $g(z)=zf(z)-1$ can take the maximum value as $0$, but it can have the value less than $1$ just like in case if $z=0$, $g(0)=-1$, why it is a contradiction?