This expression $$\tan^{-1}\left(e^{\pi/c}\right)+\cot^{-1}\left(e^{\pi /c}\right)$$ is equal to $\frac{\pi}{2}$ for real $c$.

But when $c\in\mathbb C$, it gives either $\frac{\pi}{2}$ or $-\frac{\pi}{2}$.

For an example, for $c=\frac{1}{2}-\frac{i}{2}$, it gives $- \frac{\pi}{2}$; and for $c=1-i$, it gives $\frac{\pi}{2}$.

I am trying to find the ranges for values that give two of these answers. How do I find that?

  • $\begingroup$ As pointed out in the link given by @TheSimpliFire the result strongly depends on the definition of $\arctan(z)$ and $\operatorname{arccot}(z)$. $\endgroup$
    – user
    May 17 '21 at 7:34

This really is quite an interesting problem. MATLAB has a pretty good plotting feature that can plot complex numbers. I used the following MATLAB script to produce 100,000 random complex numbers where both the real and imaginary parts have magnitude $\le 2.5$ (complex numbers with either magnitude exceeding $2.5$ produce $\pi/2$, as you will see in a second).

close all
clear all

%Number of randomly generated points
N = 100000;
%Produce a random matrix with 2N values; the first N will be the real part of z, 
%while the last N will contribute to the imaginary part of z
randmat = -2.5 + (2.5+2.5).*rand(2*N,1);

negnums = zeros(N,1);
posnums = zeros(N,1);

for n=1:N
    %Generate a random complex number
    z(n) = randmat(n,1) + randmat(N+n,1)*1i;
    num(n) = atan(exp(pi/z(n))) + acot(exp(pi/z(n)));
    num(n) = real(num(n));
    if num(n)>0
        posnums(j) = z(n);
        negnums(k) = z(n);

negnums = nonzeros(negnums);
posnums = nonzeros(posnums);
hold on
legend('Produce pi/2','Produce -pi/2')

I then plot each complex number and colored it green if it produced $\pi/2$ and blue if it produced $-\pi/2$. That graphic is as follows:

A plot of complex values

When zooming in, our picture becomes even more fascinating:

A zoomed in plot of complex values

I don't have the slightest clue why this is happening, but perhaps you can use this as a stepping stone to investigate further.

  • $\begingroup$ Are those circles in your first image (in blue)? They look pretty close $\endgroup$
    – QC_QAOA
    May 16 '21 at 5:06
  • $\begingroup$ I believe so. They appear to have radii of 1. $\endgroup$
    – BSplitter
    May 16 '21 at 8:56
  • $\begingroup$ I've provided closed forms for these equations. $\endgroup$
    – TheSimpliFire
    May 16 '21 at 10:21

Claim. Let $k\in[1/(4r-1),1/(4r+1)]$ and $\ell\in[1/(4r+1),1/(4r+3)]$ where $r\in\Bbb Z$. Then\begin{align}\arctan e^{\pi/c}+\operatorname{arccot}e^{\pi/c}=\begin{cases}\pi/2,&\quad|c-ik|=k\\-\pi/2,&\quad|c-i\ell|=\ell\end{cases}.\end{align}

Proof. It is easy to show that for principal values, \begin{align}\arctan z+\operatorname{arccot}z=\begin{cases}\pi/2,&\quad\Re z\ge0\\-\pi/2,&\quad\Re z<0\end{cases}.\end{align} Let $c=x+iy$. Then $\arctan e^{\pi/c}+\operatorname{arccot}e^{\pi/c}=\pi/2$ if and only if $$\Re\exp\left(\frac{\pi(x-iy)}{x^2+y^2}\right)=\exp\left(\frac{\pi x}{x^2+y^2}\right)\cos\frac{\pi y}{x^2+y^2}\ge0.$$ Let $r\in\Bbb Z$ so the inequality is equivalent to $$\frac y{x^2+y^2}\in2r+\left[-\frac12,\frac12\right].$$ Writing $y=n(x^2+y^2)$ where $n\in2r+[-1/2,1/2]$, we see that this is the equation of a circle $$x^2+(y-k)^2=k^2$$ centred at $(0,k)$ of radius $k$ where $k=1/(2n)\in[1/(4r-1),1/(4r+1)]$.

Similarly, $\arctan e^{\pi/c}+\operatorname{arccot}e^{\pi/c}=-\pi/2$ if and only if $$x^2+(y-\ell)^2=\ell^2$$ where $\ell\in(1/(4r+1),1/(4r+3)).$


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