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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.

$$z(t)=\left(1+\frac{1}{1+ti}\right)^{1+ti} \quad \forall t \in \mathbb{R}$$

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    $\begingroup$ Won't this function be multivalued? Which branch are you using? $\endgroup$ Commented Jan 9, 2020 at 18:23
  • $\begingroup$ I'm not sure what branch I am using. This is my first time plotting a curve in the complex plane as a "function" of t. I see what you mean though about there being multiple outputs, given the non-real power, but is there a way to ask more specifically a "principal power". I am truly sorry about my lack of appropriate vocabulary. $\endgroup$
    – Mr Manning
    Commented Jan 9, 2020 at 18:53
  • $\begingroup$ Complex analysis is not my strong suit either, so I don't know much of the appropriate vocabulary myself. The key is that $a^b \equiv e^{b\log a} = e^{b(\log |a| + i\arg(a) + 2\pi i k)}$ with $k \in \mathbb{Z}$. That is, each choice of $k$ gives a different answer. I think traditionally the "principal part" (I'm not sure this is the right term) is given by $k = 0$ with $\arg(a) \in [0, 2\pi)$. $\endgroup$ Commented Jan 9, 2020 at 19:56
  • $\begingroup$ The good news is that it seems to me that one branch will suffice: the curve $1 + \frac{1}{1 + ti}$ doesn't wrap around the origin. $\endgroup$ Commented Jan 9, 2020 at 20:07

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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 $t^2$).

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  • $\begingroup$ Amazing. Thank you. How did you figure to substitute that particular expression? Where can I read more about why the imaginary part of the 2nd order coefficient gives this answer? $\endgroup$
    – Mr Manning
    Commented Jan 9, 2020 at 19:20
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    $\begingroup$ I first tried $1/t$ and found that you get $e - \frac{e}{2}t + \ldots$. So with an extra fiddle factor of $-\frac{2}{e}$ it starts $e + t + \ldots$. As for the radius: it is easiest to remember that the curve $t\mapsto (t,t^2)$ has a curvature radius $\tfrac12$ at $t=0$ and scale from there. $\endgroup$
    – WimC
    Commented Jan 9, 2020 at 19:44
  • $\begingroup$ @WimC hi, I have just seen this answer,I wonder if you care to take a look at this post : math.stackexchange.com/questions/3742885/… $\endgroup$
    – dvd280
    Commented Jul 4, 2020 at 17:20

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