In mathematics, the maximum modulus principle in complex analysis states that if f is a holomorphic function, then the modulus |f | cannot exhibit a strict local maximum that is properly within the domain of f. In other words, either f is locally a constant function, or, for any point z0 inside the domain of f there exist other points arbitrarily close to z0 at which |f | takes larger values. Formal statement Let f be a holomorphic function on some connected open subset D of the complex plane ℂ and taking complex values. If z0 is a point in D such that for all z in some neighborhood of z0, then f is constant on D. This statement can be viewed as a special case of the open mapping theorem, which states that a nonconstant holomorphic function maps open sets to open sets: If |f| attains a local maximum at z, then the image of a sufficiently small open neighborhood of z cannot be open, so f is constant. Related statement Suppose that is a bounded nonempty open subset of . Let be the closure of . Suppose that is a continuous function that is holomorphic on . Then attains a maximum at some point of the boundary of . This follows from the first version as follows. Since is compact and nonempty, the continuous function attains a maximum at some point of . If is not on the boundary, then the maximum modulus principle implies that is constant, so also attains the same maximum at any point of the boundary. Minimum modulus principle For a holomorphic function f on a connected open set D of , if z0 is a point in D such that for all z in some neighborhood of z0, then f is constant on D. Proof: Apply the maximum modulus principle to . Sketches of proofs Using the maximum principle for harmonic functions One can use the equality for complex natural logarithms to deduce that is a harmonic function. Since z0 is a local maximum for this function also, it follows from the maximum principle that is constant. Then, using the Cauchy–Riemann equations we show that = 0, and thus that is constant as well. Similar reasoning shows that can only have a local minimum (which necessarily has value 0) at an isolated zero of . Using Gauss's mean value theorem Another proof works