# Does $f(s)\not= 0 \Rightarrow g(s)=0$ imply that $g(s)=0\Rightarrow f(s)\not= 0$?

I have the following implication: $$f(s)\not= 0\Rightarrow g(s)=0$$ Then we can deduce that its converse is also true. $$g(s)\not= 0\Rightarrow f(s)=0$$ where $f:\mathbb{C}→\mathbb{R},\,\,g:\mathbb{C}→\mathbb{C}$ and $g$ is analytic, but $f$ is not.

My question is as follows: can I deduce that $$g(s)=0⇒f(s)≠0\,\,\,?$$

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What is the relation between $f$ and $g$? Are they connected is some ways? – Babak S. Jan 5 '13 at 9:01
I have $g(s)=ϕ(s)g(2-s)$ and $f=arg(ϕ(s))$ the complex argument of $ϕ(s)$. – ZE1 Jan 5 '13 at 9:03
That is a contrapostive not converse (difference, contrapostive is logically equivalent to the intial statement; knowing this saves you work.) – PyRulez Jan 6 '13 at 3:02
What does analytic mean? – PyRulez Jan 6 '13 at 3:21

No. What if both functions are the zero function?

Also, you mean "contrapositive" and not "converse" in your second line.

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No! Put $P$ for $f(s) = 0$, $Q$ for $g(s) = 0$.

Then you are given $\neg P \Rightarrow Q$.

As you say, you can contrapose to get $\neg Q \Rightarrow P$.

But it would be a fallacy to infer $Q \Rightarrow \neg P$.

You can't normally reverse conditionals, after all! You can have $A \Rightarrow B$ true while the transposed $B \Rightarrow A$ is false.

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