Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have just started studying some complex functions and I do not understand the concept very well (I cannot visualize them easily). So I would appreciate some help for the following problem:

Say we have the complex function f(z) = cosh(z) . I would like to know how the f-plane looks like when x=const. and respectively y=const.

Thank you very much!

share|cite|improve this question
When you can visualize complex functions easily, let me know. I still have a hard time, too :) – Clayton Sep 13 '13 at 20:29
Ok, I see that the "easily" is rather stupid but what I wanted to say is that the transition from real to complex functions is not very smooth in my head.:) – Whats My Name Sep 13 '13 at 20:45

You may be interested in this paper regarding visualizing complex functions:

"Phase Plots of Complex Functions: A Journey in Illustration" by Elias Wegert and Gunter Semmler.

Jim Fowler and I (Steve Gubkin) wrote this webpage to help visualization of complex functions:

We hope to produce an online complex analysis course soon which employs a diverse set of visualization methods. For example, we will use webGL to produce 3d plots of modulus colored by phase, phase plots on the riemann sphere, interactive geometry tools for the poincare metric on the disk, etc. We should also have a lot of computational exercises which are verified directly by a server running an instance of sage. We are still developing back end stuff right now, but we should have something up and running within the year. We do not have anything up there now, but watch this space:

Peace, Steve

share|cite|improve this answer
This is very helpful. Thank you very much! – Whats My Name Sep 13 '13 at 23:36
I'm interested by such a course. I already illustrated a whole course of mathematics for bachelor using complex functions on my website: Students appreciated it. – Mikaël Mayer May 5 '15 at 13:25
@MikaëlMayer Ya, I still have plans to do this, but we got sort of distracted by producing calculus content (also, I defend my thesis tomorrow). I cannot promise when, but I definitely intend to build this course! – Steven Gubkin May 5 '15 at 18:31

Let $z=x+iy.$ Separating the real and imaginary parts of $f,$ we have $$\begin{align} f(z)=f(x+iy)=\cosh(z)=\dfrac{e^{x+iy}+e^{-x-iy}}{2}\\ =\dfrac{1}{2}[e^x(\cos{y}+i\sin{y} )+e^{-x}(\cos{y}-i\sin{y})] \\=\cos{y}\cdot\dfrac{e^x+e^{-x}}{2}+i\sin{y}\cdot\dfrac{e^x-e^{-x}}{2}. \end{align}$$ Denote $$ u(x+iy)=\Re{f(x+iy)}=\cos{y}\cdot\dfrac{e^x+e^{-x}}{2}=\cosh{x}\cdot\cos{y},\\ v(x+iy)=\Im{f(c+iy)}=\sin{y}\cdot\dfrac{e^x-e^{-x}}{2}=\sinh{x}\cdot\sin{y}$$ For fixed $x=c$

$$\dfrac{u(c+iy)}{\cosh{c}}=\cos{y},\\ \dfrac{v(c+iy)}{\sinh{c}}=\sin{y}.$$ Squaring last equalities, we have ellipse $$\dfrac{u^2}{\cosh^2{c}}+\dfrac{v^2}{\sinh^2{c}}=1.$$

share|cite|improve this answer

Although you asked this question a while ago, I have a new answer. You may be interested by the website in "expert" mode. For example, the $cosh$ function is the following:

$z\rightarrow cosh(z)$ (click on the picture to visit the website and enter other formulas)


Although the axis are not shown, the window is between $-4-3i$ and $4+3i$. The zero is black and white is infinite. 1 is red and -1 is cyan. This representation is quasi-invariant if you take the negative of the picture, it will represent the inverse function.

To analyze this picture, you can consider that the values on the horizontal axis (the real axis) are red and darkest a zero, which means that $cosh(x) > cosh(0)$. The black spot are zeroes of the function corresponding to $+/- \pi/2$; Colored points denote complex functions. $i$ is greenish whereas $-i$ is purplish.

By clicking on the image above, you can modify the formula and try out other ones. I feel it is a very good way to represent complex functions.

To answer your original question, I wrote an algorithm usable on the website to compute the f-plane, for example for x constant. You can modify the value of the parameter c (constant for $x$). @M.Strochyk was right, these are ellipses.

let c = 1; set k = 0; let min = -4; let max = 4; let n = 100; func f = cosh(z); set result = 0; let threshold = 0.1; repeat n in set theta = ((max-min)*k/n+min)*i+c; set result = if(abs(f(theta)-z) < threshold, theta, result); set k = k + 1; result

f-plane of cosh for $x=1$

Click on the picture to be able to modify it by entering different values.

DISCLAIMER: I am one of the co-authors of this website.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.