Yes, but I thought you posted a 'false' question, which is not the case.

One can (more or less) reduce to the following problem

given X^2+Y^2 = exp(arctan(Y/X)*A) over the Reals - solve for Y.

Or shorter: X^2+Y^2 = (alpha)^(arctan(Y/X)), 0 < alpha.

@Andreas_93 

I understood your question the wrong way and thought you want to switch between polar and cartesian coordinates.

@Andreas_93 

I understood your question the wrong way and thought you want to switch between polar and cartesian coordinates.

`2F1`(a,b+lambda,c+lambda,z)=         # last line in (2.5)
    (1-z)^a*(1 + a*(c-b)/(c+lambda)*zeta + O(1/lambda^2));
subs(a=1, %):
subs(b=0,c=1, lambda=n, zeta = z/(z-1),%);
subs(z=2, %);

                                           2        1
              2F1(1, n, 1 + n, 2) = -1 - ----- - O(----)
                                         1 + n       2
                                                    n

`2F1`(a,b+lambda,c+lambda,z)=         # last line in (2.5)
    (1-z)^a*(1 + a*(c-b)/(c+lambda)*zeta + O(1/lambda^2));
subs(a=1, %):
subs(b=0,c=1, lambda=n, zeta = z/(z-1),%);
subs(z=2, %);

                                           2        1
              2F1(1, n, 1 + n, 2) = -1 - ----- - O(----)
                                         1 + n       2
                                                    n

The Sum can be written in terms of hypergeom([1, n],[n+1],2)

  Sum(1/binomial(n, k), k = 1 .. n); 
  value(%): evalc(Re(%)): # obviously it is real
  simplify(%) assuming 0< n: simplify(%, size):
  convert(%, hypergeom);

      -1/2*(n+1)*(hypergeom([1, n],[n+1],2)-1)/n

Temme, http://arxiv.org/PS_cache/math/pdf/0205/0205065v1.pdf, Eq (2.5) 
using a=1,b=0,c=1 gives `2F1`(1,n,n+1,2) ~ -1-2/(n+1) - O(1/n^2), so

  subs(hypergeom([1, n],[n+1],2) = (-1-2/(n+1)), %);
  simplify(%): expand(%);

      1+2/n

For a numerical cross check use a non-integer n = 0.5 + 10^8 (to avoid
long computational times) and at least 15 Digits.

The Sum can be written in terms of hypergeom([1, n],[n+1],2)

  Sum(1/binomial(n, k), k = 1 .. n); 
  value(%): evalc(Re(%)): # obviously it is real
  simplify(%) assuming 0< n: simplify(%, size):
  convert(%, hypergeom);

      -1/2*(n+1)*(hypergeom([1, n],[n+1],2)-1)/n

Temme, http://arxiv.org/PS_cache/math/pdf/0205/0205065v1.pdf, Eq (2.5) 
using a=1,b=0,c=1 gives `2F1`(1,n,n+1,2) ~ -1-2/(n+1) - O(1/n^2), so

  subs(hypergeom([1, n],[n+1],2) = (-1-2/(n+1)), %);
  simplify(%): expand(%);

      1+2/n

For a numerical cross check use a non-integer n = 0.5 + 10^8 (to avoid
long computational times) and at least 15 Digits.

+1, sorry ... and limit(hypergeom([1, n],[n+1],2), n=infinity) = -1

+1, sorry ... and limit(hypergeom([1, n],[n+1],2), n=infinity) = -1

More detailed I did it as shown in the attached sheet.

MP_Limit_Hirnyk.mws

More detailed I did it as shown in the attached sheet.

MP_Limit_Hirnyk.mws

interface(version);

  Classic Worksheet Interface, Maple 9.50, Windows, Dec 8 2004 Build ID 174210

alpha * x - conjugate(alpha) * x = 1;
solve(%, x);

  1/(alpha-conjugate(alpha))


No answer in M12.02 or M 15.01 ...

@Markiyan Hirnyk 

In your example I still would do t=arccos(x) to get an algebraic equation,
with cases to be considered, and I guess Maple does such and combines
them into an answer (using algebraic identies for trigonometrics).

Mine is by 'electric paper and pencil' only.

Generally agreed and the solution is fine.

But more generally one will find limitations and may be it is worth to enrich
the given proc in case of more than 1 parameter in the solution given by
Maple, for example having t = 2*sin(z)^2 there will be 3 (two booleans
and an integer one).

Where it would be worth to expalin the notations in the help at 'RootOf'
and the above used command.

BTW it seems, that solve ignores the condition on the variable t, no?

@Markiyan Hirnyk 

In your example I still would do t=arccos(x) to get an algebraic equation,
with cases to be considered, and I guess Maple does such and combines
them into an answer (using algebraic identies for trigonometrics).

Mine is by 'electric paper and pencil' only.

Generally agreed and the solution is fine.

But more generally one will find limitations and may be it is worth to enrich
the given proc in case of more than 1 parameter in the solution given by
Maple, for example having t = 2*sin(z)^2 there will be 3 (two booleans
and an integer one).

Where it would be worth to expalin the notations in the help at 'RootOf'
and the above used command.

BTW it seems, that solve ignores the condition on the variable t, no?

http://www.experimentalmath.info/
http://www.google.com/search?q=experimental%2Bmathematics

because the other link needs to login to Google (and more)