A uniform bound on a family of complex integrals In my research, I have encountered the following integrals, which I want to bound. For $0 < t < 1$, let $C_t$ denote the portion of the complex unit circle consisting of $z$ for which $\Re (z) \ge t$ and define
$$
I_t (n,k) := \int_{C_t} \left( 
\frac{1-tz}{|1-tz|} \right)^{1+i+n} \frac{1}{(1-tz)^{1+i}} z^k  dz =  \int_{C_t} \left( 
\frac{1-tz}{|1-tz|} \right)^{n} \cdot z^k \cdot \frac{1}{|1-tz|^{1+i}} dz.
$$

I want to see that there exists $C>0$ such that $|I_t(n,k)| \leq C$ for all $1/2 < t < 1$ and $(n,k) \in \mathbb{Z}_{\ge 1} \times \mathbb{Z}_{\ge 1}$.

Any ideas will be appreciated.
 A: It seems to me that no such constant exists. Let me first give an estimate from above in terms of $t$, which might be useful if you restrict $t\in (1/2, 1-\epsilon)$ for fixed $\epsilon>0$. The fact that the estimate is sharp is readily seen from the estimate itself, but I will give another argument as well.
Upper estimate: Let $z=e^{is}$, and note that $C_t$ is parametrized by $s\in[-\arccos t,\arccos t]$. Then,
\begin{align*}
|I_t(n,k)|=\left|\int_{-\arccos t}^{\arccos t} \left(\frac{1-te^{is}}{|1-te^{is}|}\right)^n \frac{e^{iks}}{|1-te^{is}|^{1+i}} i e^{is} ds \right|
\end{align*}
We can estimate by passing $|\cdot|$ inside the line integral, and use $|z|^r=|z^r|$ (shown here). Then,
\begin{align*}
|I_t(n,k)|\leq \int_{-\arccos t}^{\arccos t} 1\cdot \frac{|e^{i(k+1)s}|}{\left||1-te^{is}|^{1+i}\right|} ds
\end{align*}
Now $\left||1-te^{is}|^{1+i}\right|=|e^{(1+i)\ln|1-te^{is}|}|=|1-te^{is}||e^{i\ln|1-te^{is}|}|=|1-te^{is}|$ (independently of the branch cut of $\ln$ chosen to be $\mathbb R^-_0$). Furthermore $|1-te^{is}|\geq 1-t$ (this is readily seen by a picture: the minimal distance between the circumference $te^{is}$ and $1$ is achieved by the right-most point of $te^{is}$, which is on the real axis).
Thus,
\begin{align*}
|I_t(n,k)|\leq \int_{-\arccos t}^{\arccos t} \frac{1}{1-t} ds \leq \frac{2\arccos t}{1-t} =: a_t.
\end{align*}
Note: $\lim_{t\to 1^-} a_t=+\infty$.
Sharpness:
The sought constant $C$ exists if and only if $|t^{k+1}I_t(n,k)|\leq C$ (we can even choose the same $C$ since $t<1$). Yet,
$$(1) \qquad\qquad t^{k+1}I_t(n,k)=\int_{C_t} \left( 
\frac{1-tz}{|1-tz|} \right)^{n} \cdot (tz)^k \cdot \frac{1}{|1-tz|^{1+i}} d(tz)=\int_{C_t} \left( 
\frac{1-\zeta}{|1-\zeta|} \right)^{n} \cdot \zeta^k \cdot \frac{1}{|1-\zeta|^{1+i}} d\zeta\ .$$
The divergence arises from the singular point $\zeta=1$ when $t=1$. The contour is immaterial in this respect: on any contour $C'_t$ arbitrarily close to $1$ as $t\to 1$ we would have the same problem. The last integrand in (1) is continuous, hence uniformly bounded, on $C_{t,\epsilon} := \{te^{is}: \Re(te^{is})\geq 1-\epsilon\}$. Here $C_{t,\epsilon}$ is an arc on the circumference of radius $t<1$, symmetric around the real axis, and of length $\asymp_{t\to 1} 2\arccos(1-\epsilon)\ll 1$.
Thus, for some complex constant $K$, Taylor-expanding around $\epsilon=0$,
\begin{align*}
t^{k+1}I_t(n,k)=& \ K+\int_{C_{t,\epsilon}} \left( 
\frac{1-\zeta}{|1-\zeta|} \right)^{n} \cdot \zeta^k \cdot \frac{1}{|1-\zeta|^{1+i}} d\zeta \underset{\substack{ t\to 1\\ \epsilon\to 0}}{\asymp} K+\int_{-t\sqrt{\epsilon}}^{t\sqrt{\epsilon}} \left(1+O(\epsilon^n)\right) \left(1+O(t^k\epsilon^k)\right) \frac{1}{t\epsilon} ds
\\
&\underset{\epsilon\to 0}{\asymp} 2\epsilon^{-1}\sqrt{\epsilon} \to \infty
\end{align*}
where we used that $\arccos(s)\asymp \sqrt{1-s}$ as $s\to 1$.
Edit: explicit computations.
I am adding some more explicit computations, to clarify the divergence of $I_t(n,k)$ as $t\to 1$.
Firstly,
$$I_t(n,k)\underset{t\to 1}{\asymp} \int_{C_t} \frac{dz}{|1-tz|^{i+1}}$$
since the other terms in the integrand are only a phase, for every $n,k\geq 1$.
Now, parametrize $C_t$ by $z=e^{i\theta}$ with $\theta\in [-\arccos t,\arccos t]$.
By some basic trigonometry
$$|1-tz|=\sqrt{(t-t\cos\theta+1-t)^2+(t\sin\theta)^2}=\sqrt{1+t^2-2t\cos\theta}.$$
Again up to the phase $i e^{i\theta}$ coming from the change of variable $z=e^{i\theta}$,
$$I_t(n,k)\underset{t\to 1}{\asymp} \int_{-\arccos t}^{\arccos t} \frac{d\theta}{\left(\sqrt{1+t^2-2t\cos\theta}\right)^{i+1}}$$
For real $r>0$ we have that $r^{1+i}=r(\cos\ln r+i\sin\ln r)$. Thus,
$$\frac{1}{r^{1+i}}=\frac{1}{r(\cos\ln r+i\sin\ln r)}=\frac{\cos\ln r-i\sin\ln r}{r(\cos^2\ln r+\sin^2\ln r)}=\frac{(\cos\ln r-i\sin\ln r)}{r}.$$
Applying this to the previous integral,
$$I_t(n,k)\underset{t\to 1}{\asymp} \int_{-\arccos t}^{\arccos t} \frac{\cos\left(\frac{1}{2}\ln(1+t^2-2t\cos\theta)\right)-i\sin\left(\frac{1}{2}\ln(1+t^2-2t\cos\theta)\right)}{\sqrt{1+t^2-2t\cos\theta}} d\theta .$$
Taylor-expanding the arguments of the logarithms at $1$ (regarded as functions of $\theta$, for fixed $t$),
$$I_t(n,k)\underset{t\to 1}{\asymp} \int_{-\arccos t}^{\arccos t} \frac{\cos\left(\frac{1}{2}(t^2-2t\cos\theta)\right)-i\sin\left(\frac{1}{2}(t^2-2t\cos\theta)\right)}{\sqrt{1+t^2-2t\cos\theta}} d\theta.$$
By additivity of the integral it suffices to show that
$$\int_{-\arccos t}^{\arccos t} \frac{\cos\left(\frac{1}{2}(t^2-2t\cos\theta)\right)}{\sqrt{1+t^2-2t\cos\theta}} d\theta$$
diverges as $t\to 1$. Since the numerator is uniformly bounded away from $0$ in a neighborhood of $\theta=0$ for all $t\in (0,1)$,
\begin{align*}
\int_{-\arccos t}^{\arccos t} \frac{\cos\left(\frac{1}{2}(t^2-2t\cos\theta)\right)}{\sqrt{1+t^2-2t\cos\theta}} d\theta &\underset{t\to 1}{\asymp} \int_{-\arccos t}^{\arccos t} \frac{1}{\sqrt{1+t^2-2t\cos\theta}} d\theta
\\
&\underset{t\to 1}{\asymp} \int_{-\sqrt{1-t}}^{\sqrt{1-t}} \frac{1}{\sqrt{1+t^2-2t\cos\theta}} d\theta
\end{align*}
which is readily seen to diverge.
