Sorry for the ambiguity: yes, the complex projective line or Riemannian sphere
like in your link to Wikipedia, where I typed IP^1 for the usual stylized P^1.

I just prefered ramification instead of branching points, so I need not to talk
about 'branch cuts' or have to use chosen inverse function branches.

Edited:

Simplifying my argumentation to the very point it is to say: a polynomial map
in C of degree k (k=2 for the 'inverse' of the square root) has multiplicity k
in infinity, if thought of being extended to P^1.

Find a citation appended (Forster, Riemannsche Flächen)

multiplicity.pdf

Sorry for the ambiguity: yes, the complex projective line or Riemannian sphere
like in your link to Wikipedia, where I typed IP^1 for the usual stylized P^1.

I just prefered ramification instead of branching points, so I need not to talk
about 'branch cuts' or have to use chosen inverse function branches.

Edited:

Simplifying my argumentation to the very point it is to say: a polynomial map
in C of degree k (k=2 for the 'inverse' of the square root) has multiplicity k
in infinity, if thought of being extended to P^1.

Find a citation appended (Forster, Riemannsche Flächen)

multiplicity.pdf

I think it is more easy to look at sqrt as the (partial) invers of squaring,
so branch points are replaced by ramification points (see "Hurwitz" above).

For sqrt(z) that means to extend C ---> C, z |---> z^2 to IP^1 ---> IP^1 by
(x:y) |---> (x^2:y^2) using homogenous coordinates for projective line IP^1.

But then it is 'clear' that the usual 0 in C = (0:1) playes the same role as
infinity = (1:0), thus if one is a ramification point, then the other is it.
And those are the only ones (i.e. the mapping is an unramified covering else,
locally inversible).

Formally: the above is a map of Riemannian surface of degree n = 2 (which is
the generic cardinality of a fibre) and the Riemann-Hurwitz says:

g(X) =  b/2 + n * (g(Y) - 1) + 1, where g = genus, which is 0 for X=Y=P^1
and b = total ramification order = Sum( b , b over all ramifaction points).

So 0 = b/2 + 2 * (0-1) + 1, i.e. b = 2. 

Where b(f,x) = v(f,x) - 1, and v(f,x) = multiplicity for f for x. That notion
fits with polynomial: it is the same as the top degree, Forster 2.3, and in
0 we have multiplicity = 2 and in oo as well: there are 2 points having 
ramifiaction index = 1. Being unramified else they are "0" and "infinity".

Switching to 'square root' we should get: branching exactly in 0 and infinity.


For the actual task I would suggest to study sqrt(z^2 + 1), being more easy.

Then one would have to extend z |---> (z^2-1)^2 to IP^1, which can by done
by 'homogenisation' (z^2-w^2) = w^4+2*w^2*z^2+z^4, having degree 4, and
(x:y) -> (x^4 + x^2*y^2 : y^4 + x^2*y^2)

I have not worked it out.

Hope that is correct. 

And: yes, I know that it is not an answer to Alejendro's critics about csgn(z)
given by 'series' and missed by 'MultiSeries'.

I think it is more easy to look at sqrt as the (partial) invers of squaring,
so branch points are replaced by ramification points (see "Hurwitz" above).

For sqrt(z) that means to extend C ---> C, z |---> z^2 to IP^1 ---> IP^1 by
(x:y) |---> (x^2:y^2) using homogenous coordinates for projective line IP^1.

But then it is 'clear' that the usual 0 in C = (0:1) playes the same role as
infinity = (1:0), thus if one is a ramification point, then the other is it.
And those are the only ones (i.e. the mapping is an unramified covering else,
locally inversible).

Formally: the above is a map of Riemannian surface of degree n = 2 (which is
the generic cardinality of a fibre) and the Riemann-Hurwitz says:

g(X) =  b/2 + n * (g(Y) - 1) + 1, where g = genus, which is 0 for X=Y=P^1
and b = total ramification order = Sum( b , b over all ramifaction points).

So 0 = b/2 + 2 * (0-1) + 1, i.e. b = 2. 

Where b(f,x) = v(f,x) - 1, and v(f,x) = multiplicity for f for x. That notion
fits with polynomial: it is the same as the top degree, Forster 2.3, and in
0 we have multiplicity = 2 and in oo as well: there are 2 points having 
ramifiaction index = 1. Being unramified else they are "0" and "infinity".

Switching to 'square root' we should get: branching exactly in 0 and infinity.


For the actual task I would suggest to study sqrt(z^2 + 1), being more easy.

Then one would have to extend z |---> (z^2-1)^2 to IP^1, which can by done
by 'homogenisation' (z^2-w^2) = w^4+2*w^2*z^2+z^4, having degree 4, and
(x:y) -> (x^4 + x^2*y^2 : y^4 + x^2*y^2)

I have not worked it out.

Hope that is correct. 

And: yes, I know that it is not an answer to Alejendro's critics about csgn(z)
given by 'series' and missed by 'MultiSeries'.

@Alejandro Jakubi 

I was re-reading the Riemann-Hurwitz Theorem (having z^2 in mind) and after
scratching my head (over Forster, Riemannian Surfaces) I would agree. Or not.


However that is not the problem here. As far as I understand Markiyan's task
he finally wants the residue in infinity (given avove through MultiSeries).

For that one also can take a large circle *if the function is analytic beyond
the circle* and compute the following contour integral 

res(f, infinity) = - 1/(2*Pi*I) * Int( f(x), x in circle line)

Example f:= z -> 1/z. Then the residue of f in infinity is -1 (and not 0, as
one might think first having in mind, that f(1/z) is holomorphic in 0). That
either follows by the 'definition' for res in oo.


And I say it seems here one can not use that method, since any large circle
finds points in the branch cut and thus g is not continous on that.

A more easy example would be quad:= z -> z^2 +- 1, g(z)= 1 /sqrt(quad(z)).

Having res in infinity = - 1 (using MultiSeries)

@Alejandro Jakubi 

I was re-reading the Riemann-Hurwitz Theorem (having z^2 in mind) and after
scratching my head (over Forster, Riemannian Surfaces) I would agree. Or not.


However that is not the problem here. As far as I understand Markiyan's task
he finally wants the residue in infinity (given avove through MultiSeries).

For that one also can take a large circle *if the function is analytic beyond
the circle* and compute the following contour integral 

res(f, infinity) = - 1/(2*Pi*I) * Int( f(x), x in circle line)

Example f:= z -> 1/z. Then the residue of f in infinity is -1 (and not 0, as
one might think first having in mind, that f(1/z) is holomorphic in 0). That
either follows by the 'definition' for res in oo.


And I say it seems here one can not use that method, since any large circle
finds points in the branch cut and thus g is not continous on that.

A more easy example would be quad:= z -> z^2 +- 1, g(z)= 1 /sqrt(quad(z)).

Having res in infinity = - 1 (using MultiSeries)

I have doubts, that a large circle avoids the branching in the expression:

We have g := z -> 1/sqrt( quad(z) ) where quad := z -> 4*z^2+4*z+3.

Now we take the circle c:= t -> r^2 * exp(2*Pi*I*t) (r^2, not r !),
a0 := 1/2+1/2*arccos(1/(2*r^2))/Pi, b0 := 1/2-1/2*arccos(1/(2*r^2))/Pi

Then a0 and b0 are between 0 and 1 (reaching 1/4 and 3/4 for large r).

c_a0:= c(a0) = -1/2-1/2*I*r^2*(4-1/(r^4))^(1/2) by "evalc" assuming 1 < r,
c_b0:= c(b0) = -1/2+1/2*I*r^2*(4-1/(r^4))^(1/2)

Both have absolute value = r^2, since they are on the circle. 

Feed them to the quadartic:

  'quad(c_a0)'; 
  evalc(%) assuming 1<=r;  # the same for c_b0

                                  4
                              -4 r  + 3

which is a purely real number and negative, so sqrt of it is not holomorphic
there and will jump.

Hence g is not analytic in c(a0), any given radius 1 <= r.


Especially one can not compute the residue with that kind of contours
by the usual integral formula.

Edited: much more easy take t=1/4.

I have doubts, that a large circle avoids the branching in the expression:

We have g := z -> 1/sqrt( quad(z) ) where quad := z -> 4*z^2+4*z+3.

Now we take the circle c:= t -> r^2 * exp(2*Pi*I*t) (r^2, not r !),
a0 := 1/2+1/2*arccos(1/(2*r^2))/Pi, b0 := 1/2-1/2*arccos(1/(2*r^2))/Pi

Then a0 and b0 are between 0 and 1 (reaching 1/4 and 3/4 for large r).

c_a0:= c(a0) = -1/2-1/2*I*r^2*(4-1/(r^4))^(1/2) by "evalc" assuming 1 < r,
c_b0:= c(b0) = -1/2+1/2*I*r^2*(4-1/(r^4))^(1/2)

Both have absolute value = r^2, since they are on the circle. 

Feed them to the quadartic:

  'quad(c_a0)'; 
  evalc(%) assuming 1<=r;  # the same for c_b0

                                  4
                              -4 r  + 3

which is a purely real number and negative, so sqrt of it is not holomorphic
there and will jump.

Hence g is not analytic in c(a0), any given radius 1 <= r.


Especially one can not compute the residue with that kind of contours
by the usual integral formula.

Edited: much more easy take t=1/4.

Yes, I looked it up in Cartan Chap III.5.1, having doubts in Wikipedia. Then:

-1/z^2 * g(1/z); MultiSeries:-series(%,z=0,2); simplify(%);

                          -1              2      3
                   - 1/2 z   + 1/4 - 1/8 z  + O(z )


For the integral I have doubts now as well: usually it is assumed that the
function is continous along the path (Rudin 10.8, avoiding diff forms, or
Freitag). Have to ponder that more carefully.

Yes, I looked it up in Cartan Chap III.5.1, having doubts in Wikipedia. Then:

-1/z^2 * g(1/z); MultiSeries:-series(%,z=0,2); simplify(%);

                          -1              2      3
                   - 1/2 z   + 1/4 - 1/8 z  + O(z )


For the integral I have doubts now as well: usually it is assumed that the
function is continous along the path (Rudin 10.8, avoiding diff forms, or
Freitag). Have to ponder that more carefully.

series(g(1/z), z); # puzzling by csgn(1/z)

MultiSeries:-series(g(1/z), z, 2);

                        1/2      1/2
                       4        4     2      3
                       ---- z - ---- z  + O(z )
                        4        8

so the residue is zero, I think.


PS: http://en.wikipedia.org/wiki/Residue_%28complex_analysis%29#Residue_at_infinity
My impression is that Maple seems to use only the 2nd case (using showstat(residue))
and its final formula (and yours is the 1st case).

series(g(1/z), z); # puzzling by csgn(1/z)

MultiSeries:-series(g(1/z), z, 2);

                        1/2      1/2
                       4        4     2      3
                       ---- z - ---- z  + O(z )
                        4        8

so the residue is zero, I think.


PS: http://en.wikipedia.org/wiki/Residue_%28complex_analysis%29#Residue_at_infinity
My impression is that Maple seems to use only the 2nd case (using showstat(residue))
and its final formula (and yours is the 1st case).

I lost much feeling for complex analysis :-(

4*z^2+4*z+3;
subs(z=x+I*y, %);
u,v:=evalc(Re(%)), evalc(Im(%));

{0=v, u <= 0}; # that determines the branch cut
solve(%, [x,y]);

                               1/2               1/2
                              2                 2
           [[x = -1/2, y <= - ----], [x = -1/2, ---- <= y]]
                               2                 2

Which describes the (closed) interval between the solutions for 4*z^2+4*z+3,
as you said. So 1/sqrt(...) is holomorphic. In the plane except that range.

But why a residue exists in complex infinity? Must it be not meromorphic there
(and the sqrt may prevent such) ?

I lost much feeling for complex analysis :-(

4*z^2+4*z+3;
subs(z=x+I*y, %);
u,v:=evalc(Re(%)), evalc(Im(%));

{0=v, u <= 0}; # that determines the branch cut
solve(%, [x,y]);

                               1/2               1/2
                              2                 2
           [[x = -1/2, y <= - ----], [x = -1/2, ---- <= y]]
                               2                 2

Which describes the (closed) interval between the solutions for 4*z^2+4*z+3,
as you said. So 1/sqrt(...) is holomorphic. In the plane except that range.

But why a residue exists in complex infinity? Must it be not meromorphic there
(and the sqrt may prevent such) ?

Here is my worksheet for the simple example 1 / sqrt(z)

MP_contour_bySqr.mws

MP_contour_bySqr.pdf