How do you solve the posted problem using Rouche's theorem--Complex Analysis

[Steven G]

Let f(z) and g(z) be analytic functions on the bounded domain D that extend continuously to dD and satisfies |f(z) +g(z)| < If(z)|+ Ig(z)| on dD. Show that f(z) and g(z) have the same number of zeros in D, counting multiplicity

I can prove that f(z) and g(z) have no zeros.
At that point I am lost.
Any hints/solutions would be helpful.

 

[fresh_42]

Let f(z) and g(z) be analytic functions on the bounded domain D that extend continuously to dD and satisfies |f(z) +g(z)| < If(z)|+ Ig(z)| on dD. Show that f(z) and g(z) have the same number of zeros in D, counting multiplicity

I can prove that f(z) and g(z) have no zeros.
At that point I am lost.
Any hints/solutions would be helpful.

I'm unsure whether this can help, but this is a summary of complex functions:

An Overview of Complex Differentiation and Integration

I want to shed some light on complex analysis without getting all the technical details in the way which are necessary for the precise treatments that can be found in many excellent standard textbooks.

www.physicsforums.com

It has residue formulas, some tricks, and examples. The residue theorem appears to be called for, maybe applied to $f\pm g.$ But I miss the point where the strict triangle inequality kicks in. Don't we have $|z_1+z_2| \leq |z_1|+|z_2|$ for any complex numbers? So how does strictness change the situation? This is all we have. At least it rules out $f=c\cdot g.$

 

[Steven G]

Yes we do have |z₁+z₂|< |z₁|+|z₂| for complex numbers

 

[fresh_42]

I have found a theorem (Rouché's theorem) that uses the strict inequality and which might be of help here:

Suppose $f$ and $g$ are meromorphic in a neighborhood of $\overline{B}(a;R)$ with no zeros ( $Z$) or poles ( $P$) on the circle $\gamma =\{z\in \mathbb{C}\,|\,|z-a|=R\}$. If $Z_f,Z_g,P_f,P_g$ are the numbers of zeros, resp. poles, of $f$ and $g$ inside $\gamma$ counted according to their multiplicities and if
$$|f(z)+g(z)|<|f(z)|+|g(z)|$$on $\gamma ,$ then
$$Z_f-P_f=Z_g-P_g\;.$$Proof: From the hypothesis
$$\left|\dfrac{f(z)}{g(z)+1}\right|< \left|\dfrac{f(z)}{g(z)}\right|+1$$on $\gamma .$ If $\lambda :=f(z)/g(z)$ and if $\lambda$ is a positive real number then this inequality becomes $\lambda +1<\lambda +1,$ a contradiction. Hence the meromorphic function $f/g$ maps $\gamma$ onto $\Omega:=\mathbb{C}-[0,\infty ).$ If $L$ is a branch of the logarithm on $\Omega$ then $L(f(z)(g(z))$ is a well-defined primitive for $(f/g)\,'(f/g)^{-1}$ in a neighborhood of $\gamma .$ Thus
$$\begin{array}{lll} 0&= \dfrac{1}{2\pi i} \int_\gamma (f/g)\,'(f/g)^{-1} \\[12pt] &= \dfrac{1}{2\pi i} \int_\gamma \left(\dfrac{f\,'}{f}-\dfrac{g\,'}{g}\right)\\[12pt] &= (Z_f-P_f)-(Z_g-P_g). \end{array}$$

 

[blamocur]

Let f(z) and g(z) be analytic functions on the bounded domain D that extend continuously to dD and satisfies |f(z) +g(z)| < If(z)|+ Ig(z)| on dD. Show that f(z) and g(z) have the same number of zeros in D, counting multiplicity

I can prove that f(z) and g(z) have no zeros.
At that point I am lost.
Any hints/solutions would be helpful.

I am very rusty on complex analysis, but is not your problem equivalent to the symmetric version of Rouche's theorem in Wikipedia's page ?

 

[Steven G]

Yes it is! Finally I see the solution. Thanks so much.