Schwarz-Christoffel transformation help

I'm struggling to find the 2 angles from the image to compute the integral.

Does any1 have a method for this?

Travelling anti-clockwise, define the interior angle of the polygon to be $\alpha_{j} \pi$ and the exterior angle to be $\beta_{j} \pi = \pi - \alpha_{j} \pi$. This means that at a corner when we turn left, $\beta_{j}$ will be positive, and $\beta_{j}$ will be negative when we turn right at a corner.

In your question, we want to map (from the z plane to the w plane), z = -1 to w = -1 and z = 1 to w = 0. So the Schwarz Christoffel mapping will be of the form
$f(z) = A + C\int_{}^{z} (t+1)^{-\beta_{1}}(t-1)^{-\beta_{2}} \mathrm d x$,
where I've left the betas, A and C for you to determine. [Hint:].

Technically speaking, we need a third point (by the Riemann Mapping Theorem, we can fix the pre-images of 3 boundary points, i.e. 3 of the xj, and therefore a third angle (in order to close the polygon, imagine going all the way back round at (0, -infty) and (-1, infty)). But we can choose $x_{3} = \infty$ and "ignore" this point in the formula.

Is this Complex Analysis?

It looks so interesting... what is the pre requisites

Travelling anti-clockwise, define the interior angle of the polygon to be $\alpha_{j} \pi$ and the exterior angle to be $\beta_{j} \pi = \pi - \alpha_{j} \pi$. This means that at a corner when we turn left, $\beta_{j}$ will be positive, and $\beta_{j}$ will be negative when we turn right at a corner.

In your question, we want to map (from the z plane to the w plane), z = -1 to w = -1 and z = 1 to w = 0. So the Schwarz Christoffel mapping will be of the form
$f(z) = A + C\int_{}^{z} (t+1)^{-\beta_{1}}(t-1)^{-\beta_{2}} \mathrm d t$,
where I've left the betas, A and C for you to determine. [Hint:].

Technically speaking, we need a third point (by the Riemann Mapping Theorem, we can fix the pre-images of 3 boundary points, i.e. 3 of the xj, and therefore a third angle (in order to close the polygon, imagine going all the way back round at (0, -infty) and (-1, infty)). But we can choose $x_{3} = \infty$ and "ignore" this point in the formula.

I think i understand what you're trying to say.

If i start from the top of the diagram and as i approach u=-1 i can go clockwise which is -pi/2 the interior angle or i can go anti-clockwise which is pi/2 and the exterior angle, is this correct?

And as i approach u=0 i can again take the interior angle which is -3*pi/2 (
clockwise) or i can take the exterior angle which is pi/2 (anti-clockwise), is this correct?

In this Q should both the angles be pi/2?

I'm starting from the left and approaching W1 when I get to w1 I turn pi/2 anti-clockwise to continue and then when I get to w2 I turn by pi/2 anti-clockwise to continue.

Essentially, you should think about going around the polygon in an anti-clockwise direction. As you approach (u,v)= (-1,0) from the top, you turn left (left in the sense that if you are on the line going around the polygon in an anticlockwise direction) by pi/2 and then you turn right by pi/2 at the origin. In general, turning left gives a positive beta and turning right gives a negative one.

Perhaps this wasn't the wisest way of explaining this. I'll elaborate with the mathematical definition I posted earlier for.

At (u,v) = (-1,0), the interior angle is $\frac{\pi}{2}$, so $\beta_{1}\pi = \pi - \frac{\pi}{2}$, so $\beta_{1} = \frac{1}{2}$. At the origin, the interior angle is $\frac{3\pi}{2}$, so $\beta_{2} = -\frac{1}{2}$. Note that my exterior angle is positive when I turn left and negative when I turn right.

We then use the SCF which is:
$f(z) = A + C\int_{}^{z} \prod_{j}^{} (t-x_{j})^{-\beta_{j}} \mathrm d t$
where x_i are the points on the UHP you are mapping. You plug in x1 = -1, beta 1 = 1/2, etc and solve the integral.



See above - try and identify the exterior angles.

Okay i understand what you're saying. I imagine that in walking on the line starting from the top and if I turn left my angle will be positive and if I turn right my angle will be negative.

So for the first case the exterior angle is pi/2 and because I'm turning left this will be positive.

And when I approach the origin the origin the exterior angle is again pi/2 but because I'm turnin right this angle will be negative. (

Is that correct?

Also if I decide to start at the other end of the graph (not starting at the top). Then the sign of my angle will change, is this correct?