I am self-taught, so apologies if I have missed something obvious.
I am trying to find more information about this curve I plotted in the complex plane.
$$z(t)=\left(1+\frac{1}{t+i}\right)^{t+i} \quad \forall t \in \mathbb{R}$$
You may notice that as $t \to \pm \infty, z(t) \to e$. (I came up with this idea when trying to consider complex inputs to the limit definition for $e$).
Curiously though, this curve looks circular and tangent to the real axis at $e$. If this is a circle, I would love to know either its radius or center. If it is a circle with radius $r$, I can see that the center should be $e + ri$. I have tried differentiating to find $dz/dt$ to find when Im$(dz/dt)$ is zero so that I can find the point $e+2ri$, but I began to think this method was useless.
I would greatly appreciate someone more experienced in this field guiding me to simpler cases of this problem, or giving me a hint that may lead me to an answer.
Thank you
P.S. If you were intrigued by this, I also found that you can create a circle that is "perpendicular" to the real axis at e that came in this form. Seeing how this one had two real values in the range, I found it equally interesting.
$$z(t)=\left(1+\frac{1}{1+ti}\right)^{1+ti} \quad \forall t \in \mathbb{R}$$
Your curve needs to be investigated for $t \to \infty$, which is a bit inconvenient. However, after the substitution $t \leftarrow - \frac{e}{2t}$ we can investigate near $t=0$. With a tool like Wolfram Alpha you will find $$\left(1+\frac1{\mathrm i - \frac{e}{2 t}}\right)^{\left(\mathrm i - \frac{e}{2 t}\right)} = e + t + \left(\frac{11}{6e}+\frac{2}{e}\mathrm i\right) t^2 + \ldots.$$
The radius of curvature at $t=0$ is then $r = \frac{e}{4}$. This is the reciprocal of twice the imaginary part of the second order coefficient (i.e. of $t^2$).