Let $f$ be an entire function. Suppose $f(z)$ is real if and only if $z$ is real. Prove that $f$ has at most $1$ zero. How to prove? Totally I have no idea... Please give the solution in detail since I do not understand.


2 Answers 2


The argument principle is useful to solve this problem. There are two ways you can think about the argument principle. One way is you use an integral formula over a closed curve and a number pops out, representing $2\pi i$ times the number of zeros minus the number of poles. The other way is to think about how given a curve $\gamma$, not necessarily closed, the integral $\int_{\gamma}{\frac{f'(z)}{f(z)}dz}$ represents the increase in argument of $f$ along $\gamma$, i.e. how much the image of the path $\gamma$ winds around the origin. And this last number need not be an integer multiple of $2\pi i$.

Of course, these two ways of thinking are really the same, but if you are having trouble evaluating an integral, or you prefer to use a more geometric argument, it can be helpful to think about it the second way. In Gamelin's Complex Analysis p227, there is a nice example illustrating this technique:

We want to find the number of zeros of the polynomial $p(z) = z^6 + 9z^4 + z^3 + 2z + 4$ in the first quadrant. So what we do is to take a curve $\gamma$ that starts at $0$ and goes out to $R$ along the real axis, then moves counter-clockwise along the arc ${|z|=R}$ a quarter of a circle until $iR$, and then moves down the imaginary axis back to $0$. If we determine the increase in argument of $f$ along each of the three pieces of the curve, it will tell us how many zeros there are in that quarter slice. Along the real axis, $p(z) > 0$, so the argument of $f\circ\gamma$ is constant. If we have chosen $R$ large enough, the $z^6$ term will dominate the polynomial around the quarter-circle, so the increase in argument of $f\circ\gamma$ will be approximately $6$ times that of $\gamma$: $6(\frac{\pi}{2}) = 3\pi$. Then we can examine $f\circ\gamma$ for the last segment of our curve and find it contributes an additional $\pi$ to the argument. So we have a total increase of about $4\pi$ in the argument of $f$ along $\gamma$, which means there are $2$ zeros in the first quadrant.

So now to apply this idea to your problem. Let $\gamma$ be a circle of radius $R$ around $0$. The integral $\frac{1}{2\pi i}\int_{\gamma}{\frac{f'(z)}{f(z)}dz}$ will equal the number of zeros contained inside the curve, which will equal the number of zeros of $f$ on the real interval $[-R,R]$, since our hypothesis tells us the zeros can only occur on the real axis. How much can the argument of $f$ increase along $\gamma$? First consider the increase around the upper half of the circle. $\gamma(t) \in \mathbb{R}$ exactly when $t$ equals $0$ or $\pi$, so by our hypothesis, these are the only times $f\circ\gamma$ can cross the real axis. So from $t$ equals $0$ to $\pi$, the argument can only increase by exactly $\pi$ or $-\pi$. Then the same argument applies to the bottom half of the circle $\gamma$. So as $\gamma$ goes counter-clockwise around the origin, $f\circ\gamma$ will have an increase in argument of $2\pi$, $0$, or $-2\pi$. But the last possibility is excluded because our function has no poles.

  • $\begingroup$ In that polynomial calculations, you meant $6(\frac{\pi}{4})$ right? $\endgroup$ Oct 31, 2016 at 20:49
  • $\begingroup$ Also how did you so easily say for last segment $\pi$ arguement increases? $\endgroup$ Oct 31, 2016 at 21:02

Here's a rough sketch. First prove that $f$ cannot have a repeated zero by analyzing the behaviour of $f$ near a zero. Then note that $f+r$ has the same property for any $r \in \mathbb{R}$, so prove that $f$ cannot have two real zeros by using this together with the first part.

  • $\begingroup$ still confused... $\endgroup$
    – Shiquan
    Jan 30, 2014 at 3:41
  • $\begingroup$ If $f(a) = 0$, what do you know about $f$ in a neighbourhood of $a$? And if $a$ is a repeated root, what more can you say about $f$? You can add whatever you tried into your question so that we can know how to help you understand the structure of the problem. $\endgroup$
    – user21820
    Jan 31, 2014 at 9:27
  • $\begingroup$ If $f(a)=0$ we have an $\epsilon$ neighbourhood around $a$ where $f$ is non zero. If $a$ is repeated root, $f'(a)=0$. After that what to do? $\endgroup$ Oct 31, 2016 at 6:55
  • 1
    $\begingroup$ @ArghyaChakraborty: It is better to use the asymptotic expansion. We can assume that $a=0$. If $f$ has a repeated zero at $0$, then $f(z) = a z^2 + o(z^2)$. Intuitively this means that $f$ is like $( z \mapsto z^2 )$ at $0$, which winds twice around the origin. There are two ways to do this rigorously. One is using the argument principle as in the other answer. The other way is by hand. Take $r \in \mathbb{R} \to 0$. Then $f( r \exp(it) ) = a r^2 \exp(2it) + o(r^2)$ for any $t \in \mathbb{R}$, and to get the contradiction all we need is that it crosses the real axis at $t \approx \frac12 π$. $\endgroup$
    – user21820
    Oct 31, 2016 at 11:01
  • $\begingroup$ @ArghyaChakraborty: This crossing can be obtained by the intermediate value theorem easily. Now note that $f$ cannot have two real zeros, if not by Rolle's theorem the restriction to $\mathbb{R}$ has a stationary point and we can add a real constant to $f$ to get a contradiction. $\endgroup$
    – user21820
    Oct 31, 2016 at 11:07

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