# Is this holomorphic function injective? How can we find its image?

Let's consider the function: $f(z)= z+2z^2+3z^3+4z^4+...$ defined on the unit disk D equal to ${|z|<1}$. How can I prove that the function $f$ is injective? And how can I find $f(D)$?

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Integrate the power series expansion of $\frac{f(z)}{z}$ term-wise to discover a power series expansion of a function you are familiar with to obtain a more explicit description of $f$. – levap Oct 29 '12 at 18:48

Many power series can be summed by termwise differentiation and integration. If $|z|<1$ then $$f(z)=z(1+2z+3z^2+\cdots)=z{{\rm d}\over {\rm d}z}(z+z^2+z^3+\cdots)\\ =z{{\rm d}\over {\rm d}z}\Bigl({1\over 1-z}-1\Bigr)={z\over(1-z)^2}$$ If $f(z)=f(w)$ then $$z(1-w)^2=w(1-z)^2$$ A little algebra converts this into $$(z-w)(1-zw)=0$$ If both $|z|<1$ and $|w|<1$ then $1-zw\neq 0$, so we must have $z-w=0$. Hence $f$ is injective.
To find $f(D)$, consider $$f({\rm e}^{{\rm i}\theta})={{\rm e}^{{\rm i}\theta}\over (1-{\rm e}^{{\rm i}\theta})^2}={{\rm e}^{{\rm i}\theta}(1-{\rm e}^{-{\rm i}\theta})^2 \over (1-{\rm e}^{{\rm i}\theta})^2(1-{\rm e}^{-{\rm i}\theta})^2} ={{\rm e}^{{\rm i}\theta}-2+{\rm e}^{-{\rm i}\theta} \over (1-{\rm e}^{{\rm i}\theta})^2(1-{\rm e}^{-{\rm i}\theta})^2}$$ In the last expression, both the nominator and the denominator are real, and that the nominator is negative. We substitute ${\rm e}^{{\rm i}\theta}=\cos(\theta)+{\rm i}\sin(\theta)$ and simplify to get $$f({\rm e}^{{\rm i}\theta})={2\cos(\theta)-2 \over (2-2\cos(\theta))^2}=-{1\over 2(1-\cos(\theta))}$$ From this expression, it is easy to see that $f({\rm e}^{{\rm i}\theta})$ will run along the real axis from $-\infty$ up to $-{1\over 4}$ as $\theta$ runs from $0$ to $\pi$, and then back down to $-\infty$ as $\theta$ continues from $\pi$ to $2\pi$. Hence $$f(D)=\Bbb C \setminus \bigl(-\infty,-{1\over 4}\bigr]= \Bigl\{x+{\rm i}y : y\neq 0 {\rm \ or\ }\Bigl(x>-{1\over 4} {\rm \ and\ }y=0\Bigr)\Bigr\}$$
But $f(0)=0$ hence $0$ is in $f(D)$... – Did Oct 29 '12 at 20:06