# Finding properties of curves in the complex plane.

I am self-taught, so apologies if I have missed something obvious.

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

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.

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

• Won't this function be multivalued? Which branch are you using? Commented Jan 9, 2020 at 18:23
• 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. Commented Jan 9, 2020 at 18:53
• 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 $\arg(a) \in [0, 2\pi)$. Commented Jan 9, 2020 at 19:56
• 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. Commented Jan 9, 2020 at 20:07

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 $$t^2$$).
• 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.