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Let $C$ be a simple closed curve in the complex plane, and let $f$ be holomorphic on an open set containing $C$. Is there a condition on the signs of Im $f$ and Re $f$ on $C$ that guarantees the existence of a zero of $f$ in the interior of $C$? (Shih's theorem is not quite what I want; there the condition is $\text{Re }\overline{z} \cdot f(z) > 0$.) (See M.-H. Shih, "Bolzano's Theorem for Functions of a complex variable, Am. Math. Monthly v. 89 (1982), 210-211.)

If the number of zeros of $f$ inside $C =$ the number of zeros of the identity in $f(C)$ then it should be sufficient that $f(C)$ passes through each quadrant in cyclic order (which can be rewritten as conditions on the imaginary and real parts as originally requested.) When is this equality the case?

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"The number of zeros of identity" is the strange expression: this number is always $0$ or $1$. Assuming that $f$ is analytic in an open set containing $C$ and its interior region, and has no zeros on $C$, the number of zeros inside $C$ is always equal to the index of the curve $f(C)$ about $0$, $$\frac{1}{2\pi i}\int_{f(C)}\frac{dw}{w}=\frac{1}{2\pi i}\int_C\frac{f'(z)}{f(z)}dz.$$ All other counting formulas for zeros follow from this, as Robert Israel wrote.

This can be counted in terms of the numbers of crossing of the coordinate axes by $u=\Re f$ and $v=\Im f$. For example, consider those points on $C$ where $u(z)>0$ and $v$ changes sign at $z$ (as $z$ moves on $C$ in positive direction). Then the index equals to the number of changes from $-$ to $+$ minus the number of changes from $+$ to $-$ at these points.

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  • $\begingroup$ Of course, this only works if $f$ is defined on all the 'interior' of the curve. $\endgroup$
    – Simon Henry
    Commented Sep 7, 2015 at 13:14
  • $\begingroup$ I used the expression "number of zeros of the identity in $f(C)$" to denote the winding number of $f(C)$ around zero. Then (it seems to me) the equality I mentioned holds for holomorphic functions, and then one version of the condition I asked for would be (as I said) that $f(C)$ passes through all four quadrants in cyclic order--this is just a condition on the real and imaginary parts of $f(z), z \in C$. Is this indeed a sufficient condition to conclude that the open set bounded by $C$ contains a zero of $f$? $\endgroup$
    – student
    Commented Sep 7, 2015 at 15:00
  • $\begingroup$ Your sufficient condition is not stated clearly in this remark, but perhaps if you state it clearly it will become correct. $\endgroup$
    – Alexandre Eremenko
    Commented Sep 7, 2015 at 18:12
  • $\begingroup$ @Simon Henry: you are right. I thought that this was stated in the question but now I see it was not:-) $\endgroup$
    – Alexandre Eremenko
    Commented Sep 7, 2015 at 18:14
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I don't know what kind of condition you're looking for, but whatever condition it is must essentially work through the Argument Principle: as $z$ goes around the (positively oriented) curve $C$, it must force $f(z)$ to have winding number $\ge 1$ around $0$.

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  • $\begingroup$ My edit (adding the second paragraph) and your reply just overlapped. Are you saying that the answer to my question "when is this equality the case" is "always"? $\endgroup$
    – student
    Commented Sep 7, 2015 at 5:46

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