Zeros of Holomorphic Functions

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Suppose Ω\Omega is a region, φ∈H(Ω),φ′\varphi \in H(\Omega ), \varphi^{\prime } has no zero in Ω,f∈H(φ(Ω)),g=f∘φ,z0∈Ω\Omega , f \in H(\varphi (\Omega )), g = f \circ \varphi , z_{0} \in \Omega, and w0=φ(z0)w_{0} = \varphi \left(z_{0}\right). Prove that if ff has a zero of order mm at w0w_{0}, then gg also has a zero of order mm at z0z_{0}. How is this modified if φ′\varphi^{\prime } has a zero of order kk at z0z_{0}?

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Determine the order mm of the zero z=az = a of the function f(z)f(z), if:

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(1)

f(z)=(cos⁡3z−cos⁡5z)2(1−cos⁡2z)3,a=0f(z) = (\cos 3 z-\cos 5 z)^{2}(1-\cos 2 z)^{3}, \quad a = 0;

(2)

f(z)=(z2−π2)3sin⁡3z,a=πf(z) = \left(z^{2}-\pi^{2}\right)^{3} \sin^{3} z, \quad a = \pi;

(3)

f(z)=(z2+π2)2(e2z−1)4,a=πif(z) = \left(z^{2}+\pi^{2}\right)^{2}\left(e^{2 z}-1\right)^{4}, \quad a = \pi i;

(4)

f(z)=(z4+2z3−2z−1)2(eiπz+1)3,a=−1f(z) = \left(z^{4}+2 z^{3}-2 z-1\right)^{2}\left(e^{i \pi z}+1\right)^{3}, a = -1.

Let the function f(z)f(z) be regular in the closure Gˉ\bar{G} of a domain GG, and f(z)≢f(z) \not\equiv const. Prove that only a finite number of solutions of the equation f(z)=af(z) = a (for an arbitrary fixed value aa) lie in the domain GG.

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Let the functions f(z)f(z) and g(z)g(z) be regular at the point z=az = a, with f(a)=g(a)=0f(a) = g(a) = 0. Prove that:

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(1)

lim⁡z→af(z)g(z)=lim⁡z→af′(z)g′(z)\lim_{z \rightarrow a} \frac{f(z)}{g(z)} = \lim_{z \rightarrow a} \frac{f^{\prime }(z)}{g^{\prime }(z)};

(2)

lim⁡z→af(z)g(z)=lim⁡z→a{g′(z)f2(z)f′(z)g2(z)}\lim_{z \rightarrow a} \frac{f(z)}{g(z)} = \lim_{z \rightarrow a}\left\{ \frac{g^{\prime }(z) f^{2}(z)}{f^{\prime }(z) g^{2}(z)}\right\}.

References

[Rud87] Rudin W. Real and Complex Analysis. 3rd ed., McGraw-Hill, 1987.
[ШПК15] Шабунин М.И., Половинкин Е.С., Карлов М.И. Сборник задач по теории функций комплексного переменного. 4-е изд., БИНОМ. Лаборатория знаний, 2015.