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The Schwarz Christoffel mapping is given by

$$f(\zeta) = \int_{\zeta_0}^\zeta \frac{1}{(w-z_1)^{1-(\alpha_1/\pi)}\cdots (w-z_n)^{1-(\alpha_n/\pi)} } \,\mathrm{d}w $$

where $z_i$ are complex points, and $\alpha_i \in (-\pi, \pi)$ are the corresponding angles.

To make plots of $f$ we need to evaluate that integral numerically as there is no closed form for general $z_i$ and $\alpha_i$.

What methods are there to evaluate this integral numerically?

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    If you use Matlab, you can integrate any function $Z=f(z)$. (https://fr.mathworks.com/help/matlab/ref/integral.html) under (almost) any (approximation of) path but maybe you want to do it by yourself... – 2017-02-06
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    Does that mean that we can apply our standard numerical quadrature formulas also in the complex case? – 2017-02-06
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    using "integral(...,...,...)" the detail of the method is hidden. Matlab surely optimizes: I have had opportunities to compute special functions using complex integral representation. The result is always very accurate, even with tricky functions with cuts, etc. – 2017-02-06
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    Have you tried ? Does it provide good results ? – 2017-02-08
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    I was not able to get `integral` to work, but I used `quad` instead, see e.g. [here](https://i.stack.imgur.com/5zObC.jpg) or [here](https://i.stack.imgur.com/kwJz3.jpg). – 2017-02-09

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Here is a way to use "integral" or "quad" to get numerical values with Matlab.

Let us take the example of function

$$\tag{1}f(z)=\int_{0}^z \dfrac{dw}{\sqrt{w^2-1}}$$

which is in fact identical to arcosh$(z)-i\pi/2$, or more exactly to $s$(arcosh$(z)-i\pi/2)$ where $s$ is the sign of $\Im(z)$, the imaginary part of $z$.

[It is the example given by Wikipedia in (https://en.wikipedia.org/wiki/Schwarz%E2%80%93Christoffel_mapping)] with a rectification.

Here is the corresponding program :

  clear all;close all;hold on;
  f=@(w)(1./((w-1).^(1/2).*(w+1).^(1/2)));
  r=-2:0.051:2;S=length(r);%range
  [X,Y]=meshgrid(r,r);
  for K=1:S
     for L=1:S
        z=X(K,L)+i*Y(K,L);
        t(K,L)=integral(f,0,z);% or "quad"
        s=sign(imag(z));
        t1(K,L)=s*acosh(z)-i*pi/2;
        %t1(K,L)=s*i*(acos(z)-pi/2);%identical to previous expression
     end;
  end;
  subplot(1,2,1);
  surf(X,Y,abs(t));view([19,32]);shading interp
  subplot(1,2,2);
  surf(X,Y,abs(t1));view([19,32]);shading interp

enter image description here

Fig. 1: (left) values of $\left|\displaystyle\int_{0}^z \dfrac{dw}{\sqrt{w^2-1}}\right|$ ; (right) values of $|\text{sign}(\Im(z))(\text{arcosh}(z)-i\pi/2|.$