I don't know how to use the condition of jacobian of f(z). I want to start with the definition of open set .let z ∈D f(z)∈I'm(f) i want to prove there is a neighborhood of f(z) contained in Im(f).If f'(z)≠0 i want to use f is conformal mapping to prove the image under f of disk is a disk but then i think this idea is wrong.
Hint for $9$ : Write $f(z)=(z-z_0)^Nh(z)$ near with $h(z_0)\not=0$, by continuity of $h$ at $z_0$ there is a simply connected nbd of $z_0$ in which $h$ never vanishes. Now, every nonvanishing function defined on a simply connected open set has a continuous, and hence holomorphic $N$-th root. So, we can write $f(z)=(z-z_0)^N \big(H(z)\big)^N=\big((z-z_0)H(z)\big)^N$, where $\big(H(z)\big)^N=h(z)$, near a simply connected nbd of $z_0$ for some holomorphic $H$.
Hint for $11$: When $f'(z_0)=0$, the write $f(z)=(z-z_0)^N \big(H(z)\big)^N=\big((z-z_0)H(z)\big)^N$, near a simply connected nbd of $z_0$ as above, where $N\geq 2$. Now, $G(z):=(z-z_0)H(z)$ defined near a nbd of $z_0$ has the property that $G'(z)=H(z)+(z-z_0)H'(z)$ near $z_0$, hence, $G'(z_0)=H(z_0)\not=0$. So, $G$ maps every small enough nbd of $z_0$ onto an open set of $\Bbb C$, by considering Jacobian, as said in the question. Therefore, $f=G^N$ also maps every small enough nbd of $z_0$ onto an open set of $\Bbb C$. So, we are done.
How to use Jacobian: Let $f:\Omega\to\Bbb C$ be holomorphic and $f'(z_0)\not=0$, write $f=u+i v$, as real and imaginary parts. Then, $$f'(z_0)\not=0\implies\det\begin{bmatrix}u_x(z_0) & u_y(z_0)\\ v_x(z_0) & v_y(z_0)\end{bmatrix}\not=0.$$ So, inverse mapping theorem gives that $f$ maps every small enough nbd of $z_0$ onto an open set of $\Bbb C$.
