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I've managed to prove that
$$\tan(z)=\tan(x+iy)=\frac{\sin 2x}{\cos 2x+\cosh 2y}+i\,\frac{\sinh 2y}{\cos 2x+\cosh 2y}$$

Now, I've also been able to use this identity to show that the vertical line $x=\pi/4$ gets mapped onto a portion of a circle.

However, I am wondering if there is a way to use the tangent identity above to show that if $x=a$ is a vertical line, where $-\pi/4<a<\pi/4$, then $\tan(z)$ maps this line onto a portion of a circle.

Update Thanks to Maxim's comment below, solving the first of the two equations below

$$u=\frac{\sin 2a}{\cos 2a+\cosh 2y}\qquad\text{and}\qquad v=\frac{\sinh 2y}{\cos 2a+\cosh 2y}$$

gave:

$$\cosh 2y=\frac{\sin 2a-u\cos 2a}{u}$$

Substituting this into the second equation and solving, gave:

$$\sinh 2y=\frac{v\sin 2a}{u}$$

Substituting each of these into $\cosh^22y-\sinh^22y=1$ and expanding and manipulating and completing the square led to the result

$$\left(u+\frac{\cos 2a}{\sin 2a}\right)^2+v^2=1+\frac{\cos^22a}{\sin^22a}$$

Substituting $a=\pi/8$ gives the equation $(u+1)^2+v^2=2$. Here is the Matlab code and image, which shows that his suggestion worked.

figure('Units','Normalized','Position',[0.1,0.1,0.8,0.5])
y=linspace(-5,5); x=pi/8*ones(size(y));
subplot(1,2,1), plot(x,y,'LineWidth',2,'Color','r'), hold on
grid on, axis([-3,3,-3,3]), xticks(-3*pi/4:pi/4:3*pi/4)
xticklabels({'-3\pi/4','-\pi/2','-\pi/4','0','pi/4','\pi/2','3\pi/4'})
xlabel('x'),ylabel('y'), ax=gca; ax.FontSize=14;
ax=gca; ax.XAxisLocation='origin'; ax.YAxisLocation='origin';
f=@(z) tan(z);
z=complex(x,y); w=f(z); 
subplot(1,2,2)
plot(w,'LineWidth',2,'Color','b'), hold on
g=@(u,v) (u+1).^2+v.^2-2;
fimplicit(g)
grid on, axis([-3,3,-3,3]), 
xlabel('u'),ylabel('v'), ax=gca; ax.FontSize=14;
ax.XAxisLocation='origin'; ax.YAxisLocation='origin';

enter image description here

Note that the blue part in the graph on the right is image of the vertical line under the transformation $f(z)=\tan(z)$. It is only a portion of the full circle, but exactly what I wanted.

Arashrostami
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David
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    Let $\tan z = w = u + i v$. We want to eliminate $y$ from the equations $$(u, v) = \left( \frac {\sin 2 a} {\cos 2 a + \cosh 2 y}, \frac {\sinh 2 y} {\cos 2 a + \cosh 2 y} \right).$$ Solving the equations for $\sinh 2 y$ and $\cosh 2 y$ and substituting the result into $\cosh^2 2 y - \sinh^2 2 y = 1$ gives an equation of a circle in variables $u$ and $v$. – Maxim Nov 03 '19 at 00:34
  • @Maxim Your suggestion worked. See my update above. Thanks for a nice suggestion. – David Nov 03 '19 at 04:49
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    Have a look at https://math.stackexchange.com/a/1072348/42969: $\tan$ can be written a the composition of three “simpler” mappings, and that allows to determine the image of the vertical line. – Martin R Nov 03 '19 at 07:04
  • I think each horizontal line also gets mapped onto a portion of a circle. Can you prove it? – Arashrostami Jun 24 '22 at 12:33

1 Answers1

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We will first assume that the parametric equations are governed by that of a circle, in order to determine its centre $c$ and radius $r$. We will then verify that the circle is the correct model through substitution of the parametric equations.

It is clear that for $a\in\left(-\frac\pi4,\frac\pi4\right)$ and $z=a+iy$, we have $|\Im\tan z|\le1$ in $\hat{\Bbb C}=\Bbb C\cup\{\infty\}$ with equality as $y=\pm\infty$. Symmetry of the maximum and minimum infers that $\Im c=0$.

In turn, this implies that $m=\max\Re\tan z$ occurs at $\Im\tan z=0$, so $m=\frac{\sin2a}{1+\cos2a}>0$. The Euclidean distances (in $\hat{\Bbb C}$) between the two points and the centre $c$ can now be equated to give $$\sqrt{1+(\Re c)^2}=m-\Re c\implies2\Re c=m-\frac1m.$$ Since $m-\Re c=r$, the circle sought has equation $$\left(u-\frac m2+\frac1{2m}\right)^2+v^2=\frac14\left(m+\frac1m\right)^2$$ in the $w$-plane, with $w=u+iv$.

The verification part should now be straightforward, on substituting $u=\Re\tan z$ and $v=\Im\tan z$.

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