**38.
Supplementary remarks**** **We will solve
a number of mapping problems, not considered in **37.**

We will study several cases which have not been
worked out in **37. **

1) *One of the vertices of a polygon is the
image of the point at infinity. *In order to reduce this case
to one considered already, we perform the linear mapping *z *= -1/*z *+ *a*_{n}'
of the half-plane Im *z* > 0 onto the half-plane Im *z *> 0, moving the points *a*_{1},
*a*_{2}, ··· , *a*_{n}
= ¥ to the finite points *a'*_{1},
*a'*_{2}, ··· , *a*_{n}'
. Applying (7) of **37.**, we find

* If one of the points *a*_{k}
= 0, one must use *z *=
-1/(*z - a*) + *a*_{n}'
, where *a *differs from all the other *a*_{k}.

We reduce the expression in each
brackets to the common denominator and extract from each of them
the factor *a*_{n}' - *a*_{k}'
. (*k*= 1, 2, ··· , *n - *1) and find

where *a*_{k} = 1/(*a*_{n}'
- *a*_{k}') are some real constants and *C
*is a complex constant (all omitted factors are included in
it). Using elementary geometric assumptions relating to the sum
of angles of an *n*-fold polygon, according to which

we obtain finally

Thus, *if one of the vertices of the
polygon **D** coincides
with the point at infinity, then the factor belonging to this
point does not occur in the Schwarz-Christoffel formula.*

This circumstance is employed in practice to
simplify the Schwarz-Christoffel integral (cf. **39.** et seq.).

2) *One or several vertices of the polygon lie
at the point at infinity. *Let the vertex *A*_{k}
of the polygon *D *lie at the
point at infinity. We choose on the rays *A*_{k-1}*A*_{k}
and *A*_{k}*A*_{k+1 }arbitrarily
points *A*_{k}*' *and *A*_{k}",
join them by straight segments *A*_{k}*'
*and* A*_{k}" and consider the (*n*
+ 1)-polygon *D**' *obtained
(Fig. 81). The function, mapping the half-plane onto the -polygon
*D**' *, according to
above, is given by

where *a*_{k}*'** *and *a*_{k}*"
*are angles measured in fractions of *p
*of the vertices *A*_{k}*' *and*
A*_{k}" , *a*_{k}*'
*and* a*_{k}" are the points of
the *x*-axis*, *corresponding to these vertices.

Let the segment *A*_{k}*'
A*_{k}" be moved to infinity, remaining
parallel to itself; then the points *a*_{k}*'** *and *a*_{k}*"
*join at the single point *a*_{k}*,
c*orresponding to the vertex *A*_{k}*
*and in the limit the factors of (3), involving *a*_{k}*'
*and* a*_{k}", become We will
denote by *a*_{k}*p*, taken with a minus sign, the
angle of intersection of the rays *A*_{k-1}*A*_{k}
and *A*_{k-1}*A*_{k }at
the finite point* A*_{k}**. *We
then find from the triangle *A*_{k}*'A*_{k}*"A*_{k}***
that *a*_{k}*'** *+*a*_{k}*"
- **a*_{k}* = *1, i.e., *a*_{k}*'** *+*a*_{k}*"
- *2 = *a*_{k}*
- *1, and (3) ( it assumes the ordinary form

However, this reasoning may also be applied in the case when several vertices of the polygon lie at infinity.

Thus, *the formula of Schwarz-Christoffel
remains also true for multiply-connected domains, in which one or
several vertices lie at the point at infinity, if for this the
angle between the two lines with the vertex at infinity is
determined as the angle at the finite point of their
intersection, taken with the minus sign *(cf. **31.**).

For our definition of the angle at infinity,
there remains in force Relation (1) for the sum of the angles of
the polygon. In fact, for an (*n* + 1*)-*polygon*
**D*' with finite angles,
we have, on the basis of (1), that **S****
'** + *a*_{k}*'** *+*a*_{k}*"
= **p *- 1, where **S**** ' **denotes
the sum of all the angles *D**',
*but the angle for the vertex* A*_{k}*=
*¥ (we retain the notion adopted
above). Setting *a*_{k}*'*+*a*_{k}*"=**a*_{k}+1, we find (1) also for the polygon *D.*

3) *Mapping of the outside of a polygon. *This
case differs from the one in **37.****
**in that at some finite* point *a* of the upper
half-plane, corresponding to the point at infinity of the
polygon, the function *f*(*z*) has a first order
pole (presence of an upper order pole would contradict
single-sheetedness). It is shown, as in **37.**,
that at this point *g*(*z*) will have a first order
pole with residue -2. We will have the same also for the point of the
lower half-plane, because is a first order pole for the analytic
continuation of the function *f*(*z*). Thus, we
will have for *g*(*z*) the expansion

* If the point at infinity of the
polygon corresponds to the point at infinity of the *z*-plane,
this point is a boundary point for the polygon and we have the
case already treated in 2).

Hence we will have for the
function, realizing *a conformal mapping of the upper
half-plane onto the outside of a polygon*

Here, *a*_{k}*
*are the angles of the polygon measured in fractions of *p*, *a*_{k}*
*the points of the real axis corresponding to its vertices, *a*
is the point of the upper half-plane, corresponding to the point
at infinity of the polygon, *z*_{0}, *C*, *C*_{1}*
*are certain constants.

4) *Mapping of the inside *(*outside*)
*of the unit circle onto the inside* (*outside*) *of
the polygon *represented by the function

Here, *a*_{k}*
*are the internal (external) angles of the polygon, measured
in fractions of *p*, |*a*_{k}| =1 are the points of
the unit circle corresponding to to its vertices, *C* and *C*_{1}*
*are certain constants. For the mapping of the outside of the
circle onto the outside of the polygon, it is assumed, in
addition, that the points at infinity of the *z*- and *w-*planes
correspond to each other.

For the mapping of the *inside
of the unit circle onto the outside of the polygon*, we have

where the notation is the same as in (6) and it is assumed that the point at infinity of the polygon corresponds to the centre of the circle.

Formulae (6) and (7) reduce to the preceding
one with the aid of a supplementary bi-linear mapping of the *z*-plane
such as this is done at the start of **38.** for the
derivation of (2).

5) *The inverse problem. *Now, let there
be given an arbitrary set of real numbers *a*_{k}
and *a*_{k}, satisfying the conditions

and the arbitrary complex numbers *C*
and *C*_{1}. We construct with their help the
Schwarz-Christoffel integral

It turns out that, under these conditions, *the
Schwarz-Christoffel function determines the function realizing a
conformal mapping of the upper half-plane onto a certain polygon
with the vertex angles **a*_{k}*p**.*

In fact, the argument of the derivative of (8)

remains constant on any of the
segments (*a*_{k}, *a*_{k+1}),
*k* = 1, 2, ··· , *n* - 1 of the real axis *,
and the same derivative *dw*/*dz* does not vanish
inside such a segment. Consequently, the function (8) maps
mutually-single-valuedly the segment (*a*_{k},
*a*_{k+1}) onto some straight segment (*A*_{k},
*A*_{k+1}). The same is also true for the
segment (*a*_{n}, *a*_{1})
of the real axis, which contains *z* = ¥. In fact, firstly, by the condition **S***a*_{k }**= ***n* -2*, *the segments ( *a*_{n},
¥) and ( -¥,
*a*_{1}) are turned by this
transformation by the same angle, secondly, the integral (8) goes
to the point *z* = ¥ **; consequently,
the function *w* tends to one and the same limit as z ® ± ¥.

* We assume that arg (*x - **a*_{k})
is 0 or *p *depending of whether *x >** a*_{k }or*
x <** a*_{k }, whence on every segment* *(*a*_{k},
*a*_{k+1}) all terms of the sum (9) are
constant.

** The first statement follows
since *dw*/*dz *equals arg *C *on the
segment (*a*_{n},¥) and on the segment (-¥, *a*_{1}) arg *C*+(S*a*_{k}-*n*)*p* =arg*C*-2*p*; the second statement that the principal term of the
integrand near the point *z* = ¥ has the form

Thus, the function (8) establishes a
mutually-single-valued relationship between the real axis and the
contour of some polygon *A*_{1},
A_{2}*, *··· A_{n}. Note that certain of the
vertices of this polygon may lie at infinity - it will be the
vertex A_{k} for which *a*_{k} £ 0 (in fact, for approachment to the
corresponding points *a*_{k },
the function *w *® ¥, because the integral (8) diverges, since
the order of infinity of the integrand ³1).
However, also in this case, we apply the boundary correspondence
principle and may assert that this function (7) realizes the
conformal mapping of the upper *z - *half-plane onto the
inside of the polygon *A*_{1},
A_{2}*, *··· A_{n}. The angle for the vertex A_{k} of this polygon equals *a*_{k}*p**, *because,
by (9), arg* dw*/*dz *changes for the passage
through each point *a*_{k }in the
direction from the left to the right by -*p*(*a*_{k}
- 1) = *p -a*_{k}*p*, whence the corresponding segment
*A*_{k-1} A_{k}*, *rotates by the
angle *p -a*_{k}*p* counter-clockwise. This proves
completely the assertion.

The statement remains valid also in the case when there does not apply the condition

In this case,only there appears an additional (*n
*+ 1)-the vertex of the polygon corresponding to the point *z*
= ¥. The
reader will verify that this vertex will be finite, if

and moved to infinity if

6) *Mapping of the outside of a
"star". *In conclusion*, *following
L.M.Lakhtin, we find the general formula for the mapping of the
outside of the circle |*z*| > 1 onto the outside of the
"star:, formed by *n* straight segments *L*_{k},
extending from the origin of co-ordinates (Fig. 82). We denote
the angle between *L*_{k }and *L*_{k+1
}by *a*_{k}*p*, the point of the circle, corresponding to the vertex
of this angle and the end *L*_{k}, by *a*_{k
}and *b*_{k }, respectively (*k*
= 1, 2, ··· , *n*;* L*_{n+1 }= *L*_{n
}).

We consider first the mapping of
the upper half-plane *z *onto the given domain. Let *a*_{k}'
and *b*_{k}' be the points of the real
axis of the *z*-plane, corresponding to the vertices of the angles *LkLk*_{+1}and
the ends of the segments *L*_{k}, and *a*_{0}
the point in the upper half-plane corresponding to *w* = ¥. Obviously, the mapping function *w*
= *f*(*z*) must have in
the neighbourhood of the point *z*=*a*_{k}' the form

and, in the neighbourhood of *z*=*b*_{k}
, the form

where *j*_{k}(*z*) and *y*_{k}(*z*) are
regular in the neighbourhoods of the functions mentioned above.
Finally, at the points *a* and , it must have first
order poles and must be regular at the remaining points of the *z*-plane.
Hence, as also above, we conclude that the logarithmic derivative
of the mapping function must have first order poles with residues
*a*_{k} at the points *z *= *a*_{k}',
first order poles with residues -1 at the points *z *= *a *and
*z *= , and must be regular at the points *z *= *b*_{k}'
and the remaining points of the *z*-plane. Thus,

and, consequently, after integration and raising it to the exponential power, we obtain

where *C*' is a certain
constant.

Completing the supplementary mapping

of the upper half-plane onto the
outside of the unit circle |*z*| < 1, we find the
required mapping in the form

where *C *and *a*_{k}
(| *a*_{k}| = 1) are certain constants*.

* We have

(*a*_{k}" - constants),

Side by side with Lakhtin;s formula (10), one must employ the usual formula for the mapping of the outside of a circle onto the outside of a polygon, which in this case has the form

Joined consideration of (10) and (11) permits avoidance of tiresome integrations.

As an example, we consider the particular case *n*
= 2, when the polygon is the outside of two segments,
intersecting at the origin of co-ordinates by the angle *a*_{1} = *a*. We have *a*_{2}
= 2 - *a*_{ }and (10)
and (11) assume the forms

Equating the derivatives after simple transformations, we find

whence we find for the
determination of the constants *b*_{1}* *and *b*_{2} the equation

For example, one may set *b*_{1}* = *1, when (*a* - 1)(*a*_{2} - *a*_{1}) = 1 - *a*_{1}*a*_{2} and *b*_{2} = - .*a*_{1}*a*_{2}. Equation
(12) contains four real parameters by choosing them, one
may choose the correspondence of the points *b*_{1}* , b*_{2} and the
ends* L*_{1}* , L*_{2}.

We present next several examples of applications of the Schwarz-Christoffel formula to the mapping of polygons.

**39. Examples **1) *Mapping of the upper half-plane *Im *z*
> 0 onto the quadrangle *A*_{1}*A*_{2}*A*_{3}*A*_{4}* *(Fig.
83) Consider to start with the mapping of the first quadrant of
the *z*-plane onto the right half of *OA*_{1}*A*_{2}*B *of
the given rectangle with the correspondence of points: *O ***«** 0, *A*_{1}«1,* B* «
¥ . We denote the prototype of the
point *A*_{2 }by 1/*k*,
where 0 < *k* < 1. We may consider the required
mapping as the continuation of this mapping by the symmetry
principle of the positive semi-axis *y, *whence we may
assume that* A*_{3 }« 1/*k*. Thus, the required mapping
can be written in the form

(the constant *C*_{1}* *= 0, by the correspondence of the points *O
***«** 0).
For the determination of the constants *C *and *k, *we
use the correspondence of the points *A*_{1}« 1:

and likewise of the points *A*_{2
}«1/*k*:

(we have subdivided the integral from 0 to 1/*k*
into the two from 0 to 1 and from 1 to 1/*k* and employed
(10), whence

We will assume as given the constant *k*
(0 < *k* < 1) and the dimensions of the rectangles *K*
and *K' *so that the constants *C* in (1) and (2)
are 1:

Then the mapping of the half-plane onto our rectangle will be given by the function

This integral cannot be expressed in terms of
elementary functions; it belongs to the class of the so-called *elliptic
functions*, and its inverse (i.e., the function yielding the
mapping of a rectangle onto the half-plane) to the *elliptic
functions of Jacobi*. It has the special notation

and is called the *elliptic sine*. We will
learn its details in **102.;** we may consider the
solution of the problem of mapping of an arbitrary rectangle onto
a half-plane to have been solved (and not with fixed dimensions
as here).

We will only note here an interesting property of
the function sn *z* *: It turns out to be a meromorphic
function with two periods the ratio of which is purely imaginary.

* We change the notation of the variables.

In
order to prove this assertion, we denote our rectangle's width by
*I *and its sides by *I, II, II, IV*, as is shown
in Fig. 84. We continue the function *w=*sn *z*,
initially defined in the rectangle (*I*), by the symmetry
principle through the side *I* into the rectangles (2).
This continuation realizes the mapping (2) onto the lower
half-plane. Continuing this mapping through the side *II' *of
the rectangle (2), we find that *w* = sn *z* yields
the mapping of the rectangle (3) again onto the upper half-plane,
etc. (the rectangles, shaded in Fig. 84, are mapped onto the
upper half-plane , those not shaded onto the lower half-plane).

Thus, we continue the function *w*
= sn* z *into the entire *z*-plane. For this, the
function turns out to be single-valued, because, if for passage
along any closed contour we meet again, for example, rectangle (*I*),
then the new value of sn *z* will coincide with the old
ones (they map (*I*) onto the half-plane with the same
normalization as the earlier one). Moreover, this function is
regular inside the rectangles and everywhere on their boundaries,
except the point *iK* ' and all points , corresponding to
it for the continuations (they are indicated in Fig. 63 by
crosses), where it has poles, because these points go to *w*
= ¥ during the conformal
mapping. Thus, sn *z* is a meromorphic function.

Moreover, in Fig. 84, small black circles denote
the arbitrary point *z* of the rectangle and all points
which are obtained from it by an even number of continuations**.
At all these points, having the form *z* + 4*nK* +
2 *n'K;i*, where *n, n' = *0, ±1, ±2, ··· ,
the functions sn assume the same values:

This property also indicates that sn *z*
has the two periods *t* = 4*K
*and *t*' = 2*K*'*.*

** By white circles are denoted the points at which our function assumes the value

It also follows from the
continuation of sn *z* that this function is odd

2)* Strip with a horizontal cut
*(Fig. 85). This represents a four-angle figure, the three
vertices *A*_{1}, *A*_{2}*, A*_{3}*
*which lie at infinity and the angles *a*_{1} = *a*_{2}*
*=* **a*_{3 }= 0. We will assume that *a*_{4 }=
0, *a*_{1 }= 1, *a*_{2 }= ¥ and we also take *z*_{0
}= 0 (when we find immediately from the correspondence of
the points *a*_{4} and *A*_{4 }that
*C*_{1 }= 0). Consequently, there corresponds to
the point *A*_{3 }a certain point *a*_{3
}= -*a*_{4 }of the negative strip. The
Schwarz-Christoffel integral assumes the form:

(the
factor corresponding to the point *A*_{2 }vanishes;
1) in **38.**). We employ for. the determination of the constants *C*
' and *a *the following reasoning: When the point *z*
circles the point *a*_{1 }= 1 along half the
circle *c *(i.e., when the vector 1-*z*=*re*
rotates, changing it argument from 0 to -*p* ), then the
corresponding point *w* must travel with the ray *A*_{4}*A*_{1}*
*on *A*_{1}*A*_{2} and the
increase of *w* must differ little from -*ih*_{1}:

where *O*(*r*) is
infinitely small as *r *®
0. This reasoning is justified by the fact that
the image of the semi-circle *c*_{r} for
small *r *differs little from the* *segment of the
straight line, joining the rays *A*_{4}*A*_{1}*
*and *A*_{1}*A*_{2} and
perpendicular to it.

On the other hand, for such a
small increment *D**z*, the increase of the second term in the curly
brackets of (1) will likewise be small, because this term is
continuous at the point *z* = 1. However, the increase of
the first term ln (1 - *z*) = ln *r* + *i**j *equals*
-i**p, *whence

Comparing the expressions obtained
for *D**w* and going to the limit *r* ® 0, we find

Analogously, when the point *z*
moves around the point *a*_{3} = - *a*
along the circle *z* + *a = re*^{i}^{j}*
*(*j *changes from *p *to 0). the increment *D**w = -C 'ai**p** + O*(*r*) must differ little from -*ih*_{2},
whence

Finally, the function realizing
the conformal mapping of the half-plane Im *z* > 0 onto
the strip with the cut (Fig. 85) has the form

With the supplementary mapping of
our strip witha cut onto the upper half-plane Im *w* >
0 with a cut inclined segment of length 1 (Fig. 86):

where *H = h*_{1}*
+ h*_{2}* *and, using (9), we find the mapping
of the half-plane Im *z* = 0 onto this region:

(we denote *w *again by *w*).
After simple transformations, we find:

The removed segment forms with the
positive axis the angle *p**h*_{1}/*H, *which we denote by *ap *(Fig.
86); introducing the parameter *a*, we obtain finally
the mapping of the upper half-plane Im *z *> 0 onto the
upper half-plane Im *w *> 0 with the removed segment
(0, *e*^{i}^{ap}):

Fig. 86 shows the curves
corresponding for this mapping to the straight lines Im *z*
= const.

For *a *= 1/2, we obtain
the old result (**33.**, Example
2).

3) *Polygon of Fig. *87*
represents a quadrangle with two vertices at infinity. *Having
in mind an application of the symmetry principle, we restrict
consideration to its upper half - the triangle *A*_{1}*A*_{2}*A*_{3}*
*with the angles *a*_{1} = 0, *a*_{2}* *= -*a*, *a*_{3}= 1 + *a*
(**S***a*_{k}
= 1). We call the points of the *x*-axis, corresponding to
the vertices: *a*_{1} = 0, *a*_{2}* *= ¥ , *a*_{3} =
-1; taking into consideration the correspondence of the points *a*_{3}
and *A*_{3}, we have

We employ for the determination of
the constant *C* the fact that for the travel of the point
*z *along the semi-circle *c*_{r}:
*z = re*^{i}^{j }(*j *varies
from *p* to 0) the function, determined by the last integral,
receives the increment

(on the circle *c*_{r},
the function (*z* + 1)^{a} differs little from 1: (*z* + 1)^{a}
= *O*(*r*)). On the other hand, during this
circuit, the corresponding point *w* travels with *A*_{1}*A*_{3}*
*to *A*_{1}*A*_{2},
consequently, *D**w *differs little from -*hi*. Thus, *C*
= *h*/*p*, and the function yielding the conformal mapping of the
upper half-plane onto the upper strip of the polygon of Fig. 87
has the form

Replacing here *z* by *e*^{z},
we obtain the mapping of the strip 0 < *y *< *p *onto the upper half of the
polygon

Since, according to our choice of corresponding
points, there corresponds to the lower edge of the strip the
central line of the polygon, then, by the symmetry principle,
(12) yields the conformal mapping of the strip *p *< *y* < *p *onto the entire polygon.

For rational *a*,
Integral (12) is expressed in terms of elementary function (it
reduces to the integral of the binomial differential). For *n*
= 1, we obtain already the known mapping (cf. Example 5 in **30.**):

For *a *= 1/2,
we have

4) We will find the conformal mapping of the
strip *p *< Im *z*
< *p *onto the plane with
two cut rays(Fig. 88 0 < *a *<
1). Again, we apply the symmetry principle - the upper half of
the region in the *w*-plane represents itself a triangle
with two vertices at infinity and the angles *a*_{1}* *= *a *-1,
*a*_{2}* *= -*a*,
*a*_{3}* *= 2. In order to employ the
Schwarz-Christoffel formula, we map the strip 0<*y<**p* onto the half-plane *z* = *e*^{z}.
Taking into consideration the correspondence of the points, shown
in Fig. 88, we take *a*_{1}=0, *a*_{2
}= -¥, when the point *a*_{3}
falls on the negative semi-axis and we find *a*_{3 }=
-*a*, where *a* is as yet an undetermined positive
number. The Schwarz-Christoffel formula then assumes the form

here *C* is a positive constant, because
the ray *A*_{1}*A*_{2}* *does
not turn for the mapping and, consequently, arg *C* = 0
(cf. the note at the end of **37.**),
*C*_{1}* *is real, because substitution of
positive values of *z *into
(15)* *must yield a real *w*
(cf. Fig. 88). In order to give (15) a simpler form, we set 1/*a** *= - *a*/(*a* - 1),
i.e., *a* = (1 - *a*)/*a* ; thus

The correspondence of the points *z *= -*a *and *w* = *ie*^{i}^{ap}* *yields

whence, taking into consideration that *C *and
*C*_{1} are real, we find

Substituting into (16)* **z **= e*^{z},
we find the required mapping

Fig. 88 likewise shows the correspondence of the lines for this mapping.

5) The *polygon *in Fig. 89 (0 £ *a *£
3/2 ) represents itself a quadrangle with two vertices at ¥ and and the angles *a*_{1
}= *a*_{ }- 2, *a*_{2 }= 2, *a*_{3 }= - *a*, *a*_{4
}= -*b*. The Schwarz-Christoffel integral has the
form

(*a *¹ 0, *a *¹ 1). We employ for the
determination of the constants *C*and* b*:1) that
the ray *A*_{2}*A*_{3} becomes the
positive semi-axis, whence arg *C* = 0, i.e., *C*
is a positive constant; 2) the correspondence of the points *z*
= - *b *and *w* = *ai*. Separating in (18),
after substituting *z* = *be*^{i}^{p}, *w = al*, the
real and imaginary parts, we obtain the two equations

]

which also permit to find (although approximately) the unknown constants.

In particular, for *a
*= 1/2, we find *b* = 3, *C = *3Ö3*a*/32 and the mapping function
becomes

For *a **= *1,
we have instead of (18)

the constants are given by

6) *The polygon in Fig. 90 represents a
pentagon. *We consider its right half - the quadrangle with
the angles *a*_{1}=1/2,
*a*_{2}=*a*_{4}=0,
*a*_{3}=3/2; (**S***a*_{k}=2). With *a*_{1} = 0, *a*_{2 }= 1, *a*_{3 }= *a*², *a*_{4 }= ¥, the conformal mapping of the upper *z-*half-plane onto this quadrangle
is yielded by the function

In order to determine the constants *a*
and *C*, we consider the change of *w* as *z*
moves along the semi-circles *C*_{R} and *c*_{r
}with the centre* **z **=
*1 and infinitely large and small radius, respectively. There
corresponds to the first circuit the move with the ray A_{1}A_{4}
to the ray *A*_{3}*A*_{4 }*, whence **D**w = H + O*(1/*R*);
on the other hand, for large |*z*|,
the root in the integral is approximately equal to 1, whence

comparing these expressions, we find *C = Hi/**p**. *There corresponds to
the second circuit the move of the ray *A*_{1}*A*_{2} to *A*_{3}*A*_{2},
whence *D**w = ih + O*(*r*)
and the Schwarz-Christoffel integral yields

comparison of these two expressions yields

Substituting (z - *a*²)/z = 1/*w *²
reduces the integral to the form

when the integral is readily evaluated.
Substituting the found value of *a* and *C *and
integrating, we obtain

There corresponds to the auxiliary cut along the
imaginary axis in the *w*-plane the cut along the negative
semi-axis in the *z*-plane,
whence, by the symmetry principle, the function obtained yields
the mapping of the *z**-*plane
with removed positive semi-axis onto the entire given pentagon.

Setting *z *= *z*²,
we obtain the final mapping of the upper *z*-half-plane
onto all of the pentagon:

7) The *angles of the polygon *in
Fig. 91 are *a*_{1 }= *a*_{3 }=*
a*_{5} = 0, *a*_{2
}= *a*_{4} = 3/2.
Let the points corresponding to the vertices be: *a*_{1}*
= -a*, *a*_{2 }*= *-1, *a*_{3
}= -*b, a*_{4}*=*0, *a*_{5}
= ¥, when the function mapping the
upper half-plane onto this polygon has the form

Integrating along the infinitely large
semi-circle with centre at the origin, we find *C*(-*i**p*) = -*ih*_{3},
whence* C* = *h*_{3}/*p*
. Integrating along the infinitely small semi-circles with centre
at *z* = *-a *and* z *= -*b, *we find
*

whence

The last two equations allow to find *a*
and* b*; Integral (24) may be expressed in terms of
elementary functions.

* It is not difficult to verify that one must take here the negative value of the root.

8) In
conclusion, we present the example of the mapping of the circle |*z*|
< 1 onto the inside of the polygon - the five-point star of
Fig. 92. This domain represents a decagon, five angles of which
are *a *= 5/7 and five *b
*= 1/5.We will use (6) of **38.**; in order to find the points of the circles,
corresponding to the vertices of the star; we consider a tenth of
it - the triangle *A*_{1}*B*_{1}*O
*(Fig. 92). This triangle may be mapped onto the sector 0
< arg *z* < *p*/5, |*z*| < 1 so that the point *A*_{1
}lies at 1 and *B*_{1}at *e*^{i}^{p}^{/5}.
By the symmetry principle, our mapping continues into the entire
star, where the points *A*_{k }become *a*_{k
}= *e*^{i2}^{p}^{(k-1)/5}.
(the fifth roots of 1) and *b*_{k }= *e*^{ik}^{p}^{/5}.(the
fifth roots of -1), *k* = 1, 2, ··· , 5. By the
strength of the uniqueness, its mapping may be found from (6) of **38.**
which consequently has the form

(we have employed the obvious
identities **P**(*z - a*_{k}) = *z*^{5}
- 1, **P**(*z - b*_{k}) = *z*^{5}
+ 1 .

The constant *C *can be
taken to be real, since it determines the dimension of *OB*_{k}*
= R *of the star; since the point *z* = -1 becomes an
apex of the star and *z* = 0 its centre, then

after setting *t* = {(1 + *x*^{5})/(1
- *x*^{5})}², this integral becomes one
expressible in term of the Euler *G-*function:

(cf. **90.**) Thus,

**40.
Rounding of corners****. **In
many practical problems, one has to consider that the actual
angles of polygons under consideration are always rounded off. Wel
present here approximate methods for making allowance for such
rounding.

1) *Rounding
of an angle less than **p**. *We will first find a function which realizes
mapping of the upper *z*-half-plane onto the upper *z*-half-plane
from which a small area has been removed, bounded by the segment
(-1, 1) and the arc of a curve resting on this segment and
touching the the real axis at its ends * (Fig. 93). We consider
for this purpose the mapping

of the upper *z*-half-plane onto the upper *z*_{1}-half-plane
with half the unit circle removed and in the capacity of our
curve we choose the half of an ellipse with semi-axes 1 and 1 + *h*,
near the the semi-circle (Fig. 93). There remains now to find the
mapping of the upper *z*_{1}-half-plane with the
removed half ellipse onto the *z*-half-plane. The last
problem is solved in an elementary manner. By the similarity
mapping *z*_{2} = *z*_{1}/*c, *where

we move the foci of the ellipse to
the points ±*i*; we then apply the mapping *z*_{2}
= (*z*_{3} - 1/*z*_{3})/2 and
obtain in the *z*_{3}-plane instead of the ellipse
the circle with radius

finally, the mapping *z* =
(*z*_{3}/*r* + *r*/*z*_{3})/2
yields the upper half-plane. We have now

or, taking the expressions for *r*
and *c* into account,

* The function of 2) in **34.**** **is not satisfactory, because its arc
does not touch the axis.

With the aid of the supplementary linear transformations

we obtain the more general result: The function**

where *b*_{1}* = b - a, b*_{2}*
= b + a*, yields the mapping of the upper half-plane Im* z
*> 0 onto the upper half-plane Im *z
**> *0 with removal of the small area,
bounded by the segment (*b - a*, *b + a*) and the
arc lying on this segment and touching it at its ends; the
quantity *h*, proportional to he maximum
ordinate value of the curve, is assumed to be small of higher
order with respect to *a* (Fig. 94).

** Instead of we write again *z*
and *z**..*

Now, let the function *w = f*(*z*)
yield the conformal mapping of the upper half-plane Im *z *> 0
onto some polygon *D*. where the point *b* corresponds to the vertex *B
*of the angle of the polygon less than *p*. Joining the
supplementary mapping *z* = *g*_{b}(*z*) with the
aid of function (2), we find the conformal mapping

of the upper *z-*half-plane onto the
domain which is obtained from *D *by
rounding the angle *B* in a sufficiently small
neighbourhood of the width of this angle (Fig. 94). By applying
the method of this example, one can round off all corners of *D *less than *p*.

2) *Rounding of corners larger than **p*. Without limiting the generality,
on may assume that the vertex *A*_{1} of the
multi-polygon *D*, the corner
of which we round, lies at the point *w* = 0, the side *A*_{1}*A*_{2}
goes along the positive semi-axis and *a*_{1} = 0,
but the prototypes -*a*_{1}, -*a*_{2}
, ····, -*a*_{n} of the remaining
vertices of *D *are negative
(this can always be assured by supplementary bi-linear
transformations of the plane). Under these assumptions, one may
write the function, yielding the conformal mapping of the upper *z-*half-plane
onto the multi-polygon *D*,
with the aid of the Schwarz-Christoffel integral in the form

where

and *C* is a positive constant (arg *C*
= 0 due to our choice of the segment *A*_{1}*A*_{2}).

In order to round the angle at the vertex *A*_{1},
we consider instead of (4) the function

where *b *and *g *are constants subject to
definition; we will assume to *b *be
a small positive number (in each case *b *<
*a*_{n}). According to 5) in **38.**, the
function

yields the mapping of the half-plane Im *z *>
0 onto a polygon with sides, parallel to the sides of *D*, where the point *z* = -*b* corresponds to the vertex *B",
*lying on the negative axis *u* (the angle for it
equals *a*_{1}*p*) and the remaining vertices *A*_{2}*",*···,*A*_{n}*"
*correspond to the points *a*_{1}*, a*_{2}*,
*··· ,* a*_{n}* *(this
polygon is shown by broken lines in Fig. 95).

We consider yet the function

which maps the polygon of Im *z* > 0
onto the polygon with the vertices *A*_{1}*', A*_{2}*',
*··· ,* A*_{n}*'* (it is
shown in Fig. 95 by solid lines). For every fixed *z*, the
vector *w* determined by (5) is obtained by addition of
the vectors *f*_{1}(*z*) and *f*_{2}(*z*).
By executing this addition, we verify the fact that when *z*
describes the real axis, the point *w* will describe the
closed path *A*_{1}*,A*_{2}*,*···,*A*_{n}*BA*_{1}*,
*which apart from *BA*_{1 }consists of
segments, parallel to the corresponding sides of the given
polygon (solid lines in Fig. 95.

In order to obtain for this the parametric
equation of the segment *BA*_{1}, we introduce the
positive parameter *t* = -*z *(0 < *t*
< *b*). Equation (5) then
yields

whence the tangent (tan) of the angle of
inclination of the tangent to *BA*_{1}* *with
the *u-*axis will be

We see from this expression that in the case of an angle
larger than *p *(i.e., for 1
< *a*_{1 }< 2),
tan *j *will be equal to 0 at *t
= *0, corresponding to *A*_{1} and equal tan *a*_{1}*p*
at the point *t = **b, *corresponding
to *B*. Thus, with the angle larger than *p*, the arc *BA*_{1 }really
surrounds the vertex *A*_{1}**.*

** For **a*_{1 }<
1, we have *dv/du|*_{t = 0 }*= *tan
*a*_{1}*p *, *dv/du|*_{t = }_{b}_{
}*= *0 and the arc *A*_{1}*B
*does not round the angle. You may achieve
rounding in this case by taking instead of 5) the function

where *b *> 0; however,
such a method is less convenient than the one given at the start
of this section.

According to the boundary
correspondence principle, the function (5) yields the conformal
mapping of the half-plane Im *z *>* *0 onto the
domain bounded by the contours *A*_{1}*,A*_{2}*,*···,*A*_{n}*BA*_{1}*.
*By varying the constants *C*, *b
*and *g *, we may
obtain that the domain *D* will
differ conveniently little from the given multi-angle domain D.

We will show by a simple example
that it does so. Consider the polygon in Fig. 96, a particular
case of the triangle of Example 3 of **39.****
**We will assume that the points *A*_{1}*,A*_{2}*,*
and *A*_{3}* *correspond to the points 0, 1
and ¥ of the real axis; then the
Schwarz-Christoffel integral is written in the form

.

where *C* is a positive constant (on the
segment (0,1), *w* must have positive values). In
correspondence with what has been said above, we set instead

This function transfers the segment (0,1) to the
positive semi-axis, and we require that for the passage through
the point z = 1 it undergo the increment *ih*; whence, as
in **39., **we obtain

Moreover, we require the point *z = -**b *correspond* *to the point *B
= **r **- i**r** *so that for small *b *the arc *BA*_{1}*
*will be close to the arc of a circle of radius *r*. After setting *z* = -*t*,
this becomes

Separating in it real and imaginary parts and integrating, we arrive at the relations

The three relations (8) and (9) allow to find *r*, *C* and *g** *as functions of the
parameter *b*. We have for
small *b*:

The mapping function then assumes the form (with
accuracy to small higher order in *b*)

We have become familiar in this chapter with
several problems of the theory of conformal mapping, relating to
a range of the problems formulated at the beginning of **28.**
The reader will find in the following two chapters more examples
of such problems. In Chapter III, conformal mappings will be
encountered in connection with the solution of several boundary
value problems of the theory of plane vector fields, closely
related to applications. Chapter IV is devoted to the variational
principles in the theory of conformal mappings; we consider there
the behaviour of conformal mappings for changes of the boundaries
of the mapped domains (Problem 3) as well as certain approximate
formulae.

We will not touch the methods of approximate
computations of conformal mappings (except the method of grids in
**45.**, which may be used for this purpose). The
reader may become familiar with the analytical methods of such
computations through Chapter V of the book by L.B.Kantorovich and
V.M. Krylov [9]. For many practical purposes,
the preferable methods of computations, using physical analogues
- the methods of such calculations with application of not
complicated specialized devices and wiring patterns are described
in the book by P.F. Filchchakov and V.I. Panchishina [1].

The reader will find practical methods in the book by Koppenfels and Schtalman [13].

**Literature of Chapter II**

[1] V.I.Smirnov, Course of higher Mathematics, Vol. III,
Chapter II, Gostechizdat 1954.

[2]
A.I.Markychevich, Theory of analytic functions, Gostechizdat 1950

[3]
M.A. Lavrentjev, Conformal mapping with applications to certain
problems of mechanics, Gostechizdat 1947

[4]
K. Caratheodori, Conformal mapping, translated from english,
OHNTI, 1934

[5]
R. Courant, Dirichlet Principle, conformal mapping and minimal
surfaces, translated from English, IL, 1953

[6]
G.M.Golyzin, Geometrical theory of functions of a complex
variable, Gostechizdat 1952

[7]
M.V.Keldysh, Conformal mapping of multi-connected domains onto
canonical domains, Uspekh mat. nauk, vyp. VI, 1939,pp. 90 - 119

[8]
G.M.Gluzin, L.V.Kantorovich et al., Conformal mapping of
simply-connected and multi-connected domains, OHTI, 1937

[9] L.V.Kantorovich,
B.I.Keldysh, Approximate methods of higher analysis, Fizmatgiz,
1962

[10] P.F.Filchchakov, Theory of filtration under
hydrotechnical constructions, Vol. I, Izd-vo AN USSR, Kiev, 1959

[11] P.F.Filchchakov, V.I.Paichishin, Integrators. Modelling
of potential fields by electronic circuits. Izd vo AN USSR, Kiev
1961

[12] G.N.Polozhii, Effective solution of poblems of
approximate conformal mapping of singly- and multi-connected
domains and determination of Christoffel-Schwarz coefficients
witht the iad of electronic analogues. Ukr. Matem. Journal, 7.
No. 4, (1956). 423 - 432.

[13] V. Koppenfels, Tz. Schtalman, Practice of conformal
mappings, translated from German, IL, 1963.