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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!

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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

2 Answers 2

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: http://www.gratisu.org/.

Peace, Steve

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This is very helpful. Thank you very much! –  Whats My Name Sep 13 '13 at 23:36

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 $$\dfrac{u^2}{\cosh^2{c}}+\dfrac{v^2}{\sinh^2{c}}=1.$$

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