Consider the function $f (z) = z +2z^2 +3z^3 +··· = \sum_{n≥0} nz^n$ defined on the open disk $\{z \vert|z| < 1\}$. Choose the correct option:
(a) f is not injective but attains every complex value at least once.
(b) f is injective but does not attain every complex value.
(c) f is injective and attains every complex value.
(d) None of the above.
Attempt: $f(z)=z +2z^2 +3z^3 +··· =z(1+2z+3z^2+...)=\frac{z}{(1-z)^2}.$ Now, for injectivity: $f(z_1)=f(z_2)\implies z_1(1-z_2)^2=z_2(1-z_1)^2\implies (z_1-z_2)(1-z_1z_2)=0$. Since in the open unit disk $z_1z_2\neq1$, so we have $z_1=z_2$ and hence the function is injective.
I looked the function $\frac{z}{(1-z)^2}$ known as the Koebe function and wikipedia says $f(z)$ can never achieve $\frac{-1}{4}$ as $f(z)=\frac{-1}{4} \iff z=-1$ but, $z=-1$ is not in the unit disk, I wish to see a different approach on the surjectivity part. Can someone please help me with the onto (surjective) part?
Let's just do it in the straightforward way. We solve $$\frac z{(1-z)^2} = c.$$ This simplifies to $cz^2 - (2c + 1)z + c = 0$. If $c = 0$ then $z = 0$, which is in the open disk. Otherwise, $z^2 - \frac{2c+1}c z + 1 = 0.$ So the two solutions $z_1$ and $z_2$ will always have product $1$, from Vieta's theorem. Therefore, to find a $c$ such that no $z$ in the open unit disk satisfies the equation, you have to let $|z_i| = 1 \quad (i = 1, 2)$. So we substitute $z_1 = e^{i\theta}$ and $z_2 = e^{-i\theta}$, which gets you $$\frac{2c+1}{c} = 2\cos \theta,$$ by Vieta's theorem again. Note that $\theta$ is real while $c$ may be complex.
After solving this, $c = \frac{1}{2\cos\theta - 2} \in \left(-\infty, -\frac14\right]$. And any $c$ in this interval is not obtainable.