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Prove. Consider a polygon in the w plane having vertices at w_{1}, w_{2}, ... , w_{n} with corresponding
interior angles α_{1}, α_{2}, ... , α_{n} respectively. See Fig. 1b.

A transformation which maps the upper half plane R of the z plane onto the interior R' of the polygon and the real axis of the z plane onto the boundary of the polygon is given by

or

where the points x_{1}, x_{2}, ... , x_{n} on the real axis of the z plane are mapped respectively into the
points w_{1}, w_{2}, ... , w_{n} of the polygon and A and B are complex constants.

Proof. The following will be more demonstration than proof. It will give intuitive insight into the transformation.

First observe that from 1] we obtain

3] arg dw = arg dz + arg A + (α_{1}/π -1) arg(z - x_{1}) + (α_{2}/π -1) arg (z - x_{2}) + ... + (α_{n}/π -1)
arg(z - x_{n})

We direct our attention to what occurs as z moves along the real axis from the left, tracing out the
polygon as it moves. As z passes each of the zeros at x_{1}, x_{2}, ... , x_{n} an event occurs. There is an
abrupt change in argument at that point. And this abrupt change in argument causes an abrupt
change arg dw which causes w to make an abrupt change in direction at that point i.e. it turns one
of the corners of the polygon.

As z moves along the real
axis from the left toward x_{1},
let us assume that w moves
along a side of the polygon
toward w_{1}. When z crosses
from the left side of x_{1} to
the right side, θ_{1} = arg(z -
z_{1}) changes from π to 0
while all other terms in 3]
stay constant. Fig. 2a shows
the vector z - a_{1} before z
reaches point x_{1}. Fig. 2b
shows the

vector z - a_{1} after
z has passed x_{1}. We see
that vector z - a_{1} changes
from a left pointing vector
to a right pointing vector
i.e. changes in direction
from π to 0. Consequently
arg dw decreases by

(α_{1}/π -1) arg(z - x_{1}) = (α_{1}/π -1)π = α_{1} - π

or, what is the same thing, increases by π - α_{1} (an increase being in the counterclockwise
direction). As a consequence of this when w reaches point w_{1}, it turns through the angle π - α_{1}
and then moves along the side w_{1}w_{2} of the
polygon. See Fig. 3.

When z moves through point x_{2} the same thing
happens. Arg (z - x_{2}) changes from π to 0 while
all other terms in 3] stay constant. Thus another
turn through an angle of π - α_{2} at point w_{2} is
made. We thus see that as z moves along the x
axis w traces out the polygon.

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