Uniform convergence of complex exponent derivative I'm trying to prove the following:

Let $\Re z > 0$. Then $$\lim_{\varepsilon \to 0} \frac{t^{z + \varepsilon} - t^z}{\varepsilon} = t^z \log t$$ uniformly in $t \in [0,1]$.

I've tried to bound it in a straight forward way to no avail. Also, assuming the convergence is among real positive epsilons, I could reduce it to the purely real case ($\Im z$ does not matter) and apply Dini's theorem. However, it does not seem to work well for arbitary $\varepsilon$.
How to prove it for the general complex case?
 A: I assume we use $t^w = e^{w\log t}$ with the real-valued logarithm of $t$ for $t > 0$ and $t^w \equiv 0$ for $t = 0$ and $\Re w > 0$ as well as $0^z\log 0 = 0$, as suggested by continuity.
Then we have
$$\begin{align}
\left\lvert \frac{t^{z+\varepsilon}-t^z}{\varepsilon} - t^z\log t \right\rvert
&= t^{\Re z} \left\lvert\frac{t^\varepsilon-1}{\varepsilon}-\log t\right\rvert\\
&= t^{\Re z}\left\lvert\sum_{k=1}^\infty \frac{\varepsilon^{k-1}(\log t)^k}{k!} - \log t\right\rvert\\
&\leqslant t^{\Re z}\lvert\log t\rvert\sum_{k=1}^\infty \frac{\lvert\varepsilon\log t\rvert^{k}}{(k+1)!},\tag{1}
\end{align}$$
and equality holds when $\varepsilon < 0$.
So for every $0 < \lvert\varepsilon \rvert < \Re z$, the function
$$f_\varepsilon \colon t \mapsto \frac{t^{z+\varepsilon}-t^z}{\varepsilon} - t^z\log t$$
is continuous on $[0,1]$ (since $t^{z}\log t\to 0$ for $t\to 0$ and $\Re z > 0$), we have $\lvert f_\varepsilon(t)\rvert \leqslant \lvert f_{-\lvert\varepsilon\rvert}(t)\rvert$ for all $\varepsilon$ and $t$ under consideration, and from $(1)$ it is immediate that $0 < r_1 < r_2$ implies $\lvert f_{-r_1}(t)\rvert \leqslant \lvert f_{-r_2}(t)\rvert$ for all $t$.
Dini's theorem now asserts that
$$\lim_{r\searrow 0} \lvert f_{-r}(t)\rvert = 0$$
uniformly on $[0,1]$, and $\lvert f_\varepsilon\rvert \leqslant \lvert f_{-\lvert\varepsilon\rvert}\rvert$ shows that
$$\lim_{\varepsilon\to 0} \frac{t^{z+\varepsilon}-t^z}{\varepsilon} - t^z\log t = 0$$
uniformly on $[0,1]$ too.
