Algebraic functions

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Find the singular points of the equation y3−3y+2x=0y^{3}-3y+2x = 0.

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Remark.
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Algebraic functions satisfy an equation f(x,y)=0f(x,y) = 0, where f(x,y)f(x,y) is a polynomial in the complex variables xx and yy. If both of these variables remain real, then the equation written represents an algebraic curve. If the equation has degree nn in yy, then it defines nn branches y1,y2,…,yny_{1}, y_{2}, \ldots , y_{n} of a multivalued function of xx. They represent regular functions of xx in a domain where their values are finite and unequal. If x0x_{0} is a point lying in such a domain, then each of the functions yνy_{\nu } expands in integer powers of x−x0x-x_{0}. If, however, at the point x0x_{0} several of the yνy_{\nu } coincide, then in a neighborhood of this point the functions yνy_{\nu } expand in fractional powers of x−x0x-x_{0} by the formula

y=∑μ=μ0∞aν(x−x0)μm, y = \sum _{\mu =\mu _{0}}^{\infty } a_{\nu }(x-x_{0})^{\frac{\mu }{m}},

where the number μ0\mu_{0} may be negative, and mm does not exceed nn. The values aνa_{\nu } can be found by substituting these series into the equation. The singular points themselves can be found from the equations:

f(x,y)=0,∂f∂y=0. f(x,y) = 0, \quad \frac{\partial f}{\partial y} = 0.

Find the singular points of the equation y3−3y2+x6=0y^{3}-3y^{2}+x^{6} = 0.

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Find the singular points of the equation y3−3y+2x2(2−x2)=0y^{3}-3y+2x^{2}(2-x^{2}) = 0.

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Find the singular points of the equation y3−3y+2x3(2−x3)=0y^{3}-3y+2x^{3}(2-x^{3}) = 0.

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Let the function w=φ(z)w=\varphi (z) be regular (possibly many-valued) in the annular region 0<∣z∣<ρ0 < \left|z\right| < \rho, ρ>0\rho > 0, and satisfy there identically an equation of the form

F0(z)wl+F1(z)wl−1+⋯+Fl−1(z)w+Fl(z)=0, F_{0}(z) w^{l} + F_{1}(z) w^{l-1} + \cdots + F_{l-1}(z) w + F_{l}(z) = 0,

where F0(z),F1(z),…,Fl−1(z),Fl(z)F_{0}(z), F_{1}(z), \ldots , F_{l-1}(z), F_{l}(z) are regular in a neighbourhood of the point z=0z=0. If there exists a power series c0+c1z+c2z2+⋯c_{0}+c_{1}z+c_{2}z^{2}+\cdots such that the function (φ(z)−c0−c1z−c2z2−⋯−cn−1zn−1)z−n(\varphi (z)-c_{0}-c_{1}z-c_{2}z^{2}-\cdots -c_{n-1}z^{n-1})z^{-n} remains bounded in the neighbourhood of z=0z=0 for infinitely many values of nn, then φ(z)\varphi (z) is regular in the neighbourhood of z=0z=0, and we have φ(z)=c0+c1z+c2z2+⋯+cnzn+⋯\varphi (z)=c_{0}+c_{1}z+c_{2}z^{2}+\cdots +c_{n}z^{n}+\cdots.

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Let f(z)f(z) be an analytic function, z=x+iyz=x+iy, and let the square of the absolute value of f(z)f(z)

∣f(z)∣2=φ(x,y) \left|f(z)\right|^{2} = \varphi (x,y)

be an algebraic function of the real variables xx and yy. Then the function f(z)f(z) is itself an algebraic function of zz.

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Remark.
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H. A. Schwarz, oral communication.

Suppose (f,D)(f, D) and (g,D)(g, D) are function elements, PP is a polynomial in two variables, and P(f,g)=0P(f, g) = 0 in DD. Suppose ff and gg can be analytically continued along a curve γ\gamma, to f1f_{1} and g1g_{1}. Prove that P(f1,g1)=0P\left(f_{1}, g_{1}\right) = 0. Extend this to more than two functions. Is there such a theorem for some class of functions PP which is larger than the polynomials?

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References

[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.
[PóSz98] Pólya G., Szegő G. Problems and Theorems in Analysis. Springer, 1998.
[Rud87] Rudin W. Real and Complex Analysis. 3rd ed., McGraw-Hill, 1987.