I want to show that $f(z) = \frac{1}{2}(\frac{z}{(1-z)^2}+\frac{z}{(1+z)^2})$ is not injective in the unit disk.
What I have done so far:
Simplify $f(z)$, $f(z) = \frac{z^3+z}{(z^2-1)^2}$
Setting $f(z) = f(w)$
$\frac{z^3+z}{(z^2-1)^2} = \frac{w^3+w}{(w^2-1)^2}$
$(z^3+z)(w^2-1)^2-(w^3+w)(z^2-1)^2=0$
Factor the expression to get: $(w-z)(zw-1)((zw+1)^2+(z+w)^2) = 0$
So I really need to show there is a pair of $z, w \in \mathbb{D}$ that satisfies $(zw+1)^2+(z+w)^2= 0$. I have trouble proceed from here. In addition, is there an easier way to do this? I thought of Rouche's Theorem but didn't get anything meaningful.
You could try the obvious route. $(zw + 1)^2 = -(z+w)^2$ is implied by $zw+1 = i(z+w)$, so collecting coefficients it would suffice for $z(w-i) - iw + 1 = 0$ or
$$z = \frac{iw - 1}{w - i} = i \left(\frac{w + i}{w - i}\right).$$
Try $w = it$ to obtain
$$z = i\left(\frac{t + 1}{t - 1}\right)$$
and for $-1 < t < 0$ we have $|z|< 1$.
Explicitly, you can pick $w = -i/2$ and $z = -i/3$. We have $f(z) = f(w) = -12i/50$.