Consider the function $f (z) = z +2z^2 +3z^3 +··· = \sum_{n≥0} nz^n$ defined on the open disk $\{z ||z| < 1\}$. 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.
My Try: $f(z)=z +2z^2 +3z^3 +··· =z(1+2z+3z^2+...)=z\frac{d}{dz}\frac{1}{1-z}=\frac{z}{(1-z)^2}.$ For injectivity, Consider the equation. $f(a)=f(b)\implies a(1-b)^2=b(1-a)^2\implies (a-b)(ab-1)=0$. So, when the product of two numbers is 1. then also, above equation satisfies. So, It is not injective. For surjectivity, I consider the equation $c\in \mathbb C$ $c=\frac{z}{(1-z)^2} $. I got $z$ in terms of $c$. So, (a) is the apt answer for this question. Am I judging correctly?
This is the famous Koebe function which is the maximal univalent function - or Riemann map if one wishes - on the disc in many ways (under the usual normalizations, $f(0)=0, f'(0)=1$) - eg for coefficients (Bieberbach conjecture/de Branges Theorem), for integral means (Baernstein theorem) etc; its image omits the part of the negative axis that starts at $-\frac{1}{4}$ and goes to $-\infty$ which can be seen by looking at the image of the unit circle under it or even simpler, noting that $4K(z)=(\frac{1+z}{1-z})^2-1$ and then the function inside the square root is the Mobius transform of the unit disc onto the right half-plane, squaring means the image of the disc omits only the negative axis, then subtract $1$, we omit the part from $-1$ to $-\infty$ and then dividing by $4$ etc; injectivity on the unit disc follows trivially from this representation too