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Let: $f, x, u, y, v : \mathbb{C} \to \mathbb{R}$ be functions in the complex variable $s$.

I made this claim

If $f(s)≠0 \,\pmod{2\pi}$,

then $x(s)=0,u(s)=0, y(s)=0,v(s)=0$, (this first implication is true from an earlier analysis)

then $x(s)=u(s)=0, y(s)=-v(s)=0$,

then $x(s)-u(s)=0, y(s)+v(s)=0$,

then $(x(s)-u(s))+i (y(s)+v(s))=0$

which give an equation of the form $h(s)=0$.

I suspect that I made a kind of hidden errors or a kind of a wrong logical step. My question is about the correcteness of these steps.

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1 Answer 1

up vote 1 down vote accepted

Your question seems to boil down to the following:

If $x(s) = u(s) = y(s) = v(s) = 0$, is $(x(s) - u(s)) + i(y(s) + v(s)) = 0$?

If this is indeed your question, the answer is yes.

$$(x(s) - u(s)) + i(y(s) + v(s)) = (0 - 0) + i(0 - 0) = 0 + 0i = 0$$

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@RH1: Let me know if I have misunderstood your question. –  Michael Albanese Jan 6 '13 at 8:52
    
@ Michael Albanese: Yes, this is my question. Thank you. However, I feel that there are other roots of the last equation in addition to those verifying the first one. –  ZE1 Jan 6 '13 at 9:25
1  
There are. Any $s$ for which $x(s) = u(s)$ and $y(s) = -v(s)$ will also give the same result. –  Michael Albanese Jan 6 '13 at 9:27
    
@ Michael Albanese: So, the implication is not true. –  ZE1 Jan 6 '13 at 9:31
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No, the implication is true. What you wrote is $f(s) \neq 0 \,\pmod{2\pi} \implies h(s) = 0$, where $h(s)$ is the complex number you form from the functions $x, y, u,$ and $v$. However the converse of what you wrote, that is $h(s) = 0 \implies f(s) \neq 0 \, \pmod{2\pi}$, is not necessarily true as my previous comment shows. –  Michael Albanese Jan 6 '13 at 9:34

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