Let $z_0 \neq z \in \mathbb C$ be two complex numbers for which $[z_0,z] = \{(z-z_0)t + z_0 : t \in [0,1]\}$ does not cross $0$. Let $\alpha \in \mathbb R \setminus \{-1\}$.
I want to upper bound the path-integral $$\left| \int_{[z_0,z]} |y|^{\alpha} dy \right| = |z-z_0| \int_0^1 |(z-z_0)t + z_0|^{\alpha}dt$$ I would expect something like $$\left| \int_{[z_0,z]} |y|^{\alpha} dy \right| \leq \frac{||z|^{\alpha + 1} - |z_0|^{\alpha+1}|}{|\alpha + 1|}$$ or at least $$\left| \int_{[z_0,z]} |y|^{\alpha} dy \right| \leq \frac{|z|^{\alpha + 1} + |z_0|^{\alpha+1}}{|\alpha + 1|}$$ to hold, up to some universal constant. The $(\alpha + 1)^{-1}$ factor (and the $\alpha$ dependence in general) is important. Otherwise, it's quite easy to estimate the integral as done in the Estimation Lemma.
I tried to do something simple as follows $$\int_0^1 |(z-z_0)t + z_0|^{\alpha}dt \leq \max\{2,2^{\alpha/2}\} \left( \int_0^1 |\Re(z-z_0)t + \Re(z_0) |^{\alpha} dt + \int_0^1 |\Im(z-z_0)t + \Im(z_0) |^{\alpha} dt \right)$$ but splitting the integral like this lead to singularities of type $\Re(z-z_0)^{-1}$, $\Im(z-z_0)^{-1}$ that I cannot get rid of. Moreover, the $2^{\alpha/2}$ factor is not ideal.
If we don't split the integral like I did, we are led to study an integral of the type $$\int_0^1 (at^2 + bt + c)^{\alpha/2}ds$$ for some real, positive second-order polynomial on $[0,1]$, but I don't know how to study such an integral. However, there might be some easier way to proceed since I'm not looking for an exact value.
The problem is symmetrical up to rotations so we can assume that $z, z_0$ have the same real part $x>0$ and write $z= x+ia, z_0= x+ib$. $x$ is actually the norm of the orthogonal projection of $0$ on $(z, z_0)$. The integral then rewrites $$I =\int_a^b |x+it|^\alpha dt$$ With a change of variables $u=t/x$: $$I=x^{\alpha+1} \int_{a/x}^{b/x} |1+iu|^\alpha dt$$ Name $J = I / x^{\alpha+1}$ that second integral. When $x \rightarrow \infty$, $J$ behaves like $\frac {b-a} x$ (because $a/x, b/x \rightarrow 0$ and $|1+iu|^\alpha \simeq 1$ for $u \simeq 0$. So for $x \rightarrow \infty$, $I \simeq (b-a)*x^\alpha$
For $x\rightarrow 0$, $J \le \int_{-\infty}^{\infty} |1+iu|^\alpha < \infty$ if $\alpha <0$ and $J \lesssim \max(|a/x|^\alpha,|b/x|^\alpha) * \frac {b-a} x$ so we get:
-$\alpha \ge 0$: $I \lesssim x^{\alpha} \cdot (b-a)$
For $\alpha \ge 0$ you can actually write from the beginning $I \le \max(|z|^\alpha, |z_0|^\alpha) *(|z-z_0|)$