Let $f:\mathbb{C}\rightarrow\mathbb{C}$ satisfy these conditions:
(i) $f$ is holomorphic in $D=\{z\in \mathbb{C} \ |\ |z| \leq 1\}$
(ii) $f(z)=z²+2, \forall z\in \partial D$
(iii) $f(0)=2$
Does exist any function $f$ satisfying all these condition?
I worked through the problem and my answer is: Yes, that is.
We have $z=x+iy$, which implies that $z²=x²-y²+i(2xy)$
Working in $\partial D$ we know $|z|=1$, therefore $x²+y² = 1$
Plugging into the second condition, we get $f(z)=(x²-y²+2)+i(2xy)$
And when $z=0$, then $x=y=0$ and $f(0)=2$, so the third condition is fine.
If $f$ is holomorphic then it should attend Cauchy-Riemman conditions, and it does satisfy this condition (I derived it in my notebook).
Then all conditions is satisfied and our function is $f(z)=(x²-y²+2)+i(2xy)$ with $x,y \in \mathbb{R}$.
I wonder, however, if this is correct because it works (At least I think) when we're working in $\partial D$, but I'm not sure it works for all $|z| \leq 1$.
Is this right?
You need to have f at least holomorphic on the interior of D and continuous on D ;I guess holomorphic on D (a closed set ) implies that . Anyway the Cauchy integral formula applies here with inegration around the boundary of D and determines the values of the function at every point of the interior of D Obviously this function is f(z)= z^2 -2 which is holomorphic on the entire plane .Luckily you have f(0) =2 so this is the unique function required . If f(0) |=2 there would be no function satisfying your conditions ,Regards