maximum modulus principle:

If a function $f$ is analytic and not constant in a given domain $D$, then $|f(z)|$ has no maximum value in $D$. That is, there is no point $z_0$ in the domain such that $|f(z)|≤|f(z_0)|$ for all points $z$ in it.



Suppose that a function $f$ is continuous on a closed bounded region $R$ and that it is analytic and not constant in the interior of $R$. Then the maximum value of $|f(z)|$ in $R$, which is always reached, occurs somewhere on the boundary of $R$ and never in the interior.

Now let $f$ be an analytic function and $z_0\in D$. Let $C_1=\{z: |z|=r_1\}$ be a circle that passes through $z_0$ and assume $|f(z)|\leq|f(z_0)|$ for all $z\in D_1=\{z: |z|\leq r_1\}$.

Let $C_2=\{z: |z|=r_2\}$ be a circle that passes through $z_0$ and assume $|f(z)|\leq|f(z_0)|$ for all $z\in D_2=\{z: |z|\leq r_2\}$.

Constructing these disks which cover the domain $D$, we have that for any $z\in D_1\bigcup\cdots\bigcup D_k$, we have $|f(z)|\leq|f(z_0)|$, so $z_0$ is a point in domain $D$ that the maximum of $|f|$ in $D$ occur there. this is contradiction to maximum modulus principle.

you may see the picture here:


I think somewhere in this constructing is wrong, but I don't know where.

Could anyone help to find the mistake.


  • 2
    $\begingroup$ Can you copy/paste the question in order to see the question when we type comments? $\endgroup$ Dec 14, 2011 at 18:27
  • $\begingroup$ I copied the source over. Feel free to edit, though, asd. $\endgroup$ Dec 14, 2011 at 18:33
  • $\begingroup$ It kind of seems like you're just assuming that $|f(z_0)|$ is as much of a maximum as you need it to be at every step of the construction. Can you do this for a particular $f$ and $D$? Otherwise it's hard to call it a construction. $\endgroup$ Dec 14, 2011 at 18:42
  • $\begingroup$ @Dylan Moreland. thanks for copying. $\endgroup$
    – asd
    Dec 14, 2011 at 19:15

1 Answer 1


There is no reason why your $z_0$ should be the maximum of $f$ along the boundary of any circle you construct. The maximum modulus principle just says the maximum of $f$ on a disc occurs at the boundary. If $z_0$ is a point on the boundary of a disc $B$, there may be $z_1$ on the boundary of $B$ such that $f(z_1) > f(z_0)$. If that is true then, of course, you may and will find $z_2$ in the interior of $B$ with $f(z_2) > f(z_0)$ but with $f(z_1) > f(z_2)$. There is no way to construct an inductive contradiction as you suggest.

  • $\begingroup$ Exactly! My question is why $z_0$ can not be a maximum of $|f|$ in the $D_i$'s. Is there any lemma or theorem that discard this case. $\endgroup$
    – asd
    Dec 14, 2011 at 19:10
  • 1
    $\begingroup$ uhmm -- the maximum modulus principle? Your circles simply cannot constructed the way you are proposing it. $\endgroup$
    – user20266
    Dec 14, 2011 at 19:37

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