MA425 - Complex Analysis

\(|z_1 z_2| = r_1 r_2\)

\(\langle z_1 z_2 = \theta_1 + \theta_2\)

Proof

18:32 <Evidlo> In my book, I'm finding the maximum of f(z) = |z^{2} + 3z - 1| within the disk |z| < 1. I understand that analytic functions only have extrema on the boundary, but how are you sure f is analytic? Isn't there a cusp at z = 0? 18:33 <Z-module> every polynomial is an entire function: analytic on the whole plane 18:33 <Z-module> (and your "every" should be "every nonconstant") 18:34 <Z-module> Evidlo: z^{2} + 3z - 1 is an analytic complex function. |that| is not analytic (in the complex sense). 18:35 <Z-module> but: if f is a complex function that is: nonconstant, and is analytic in some open disk D, and continuous on the boundary circle of D, then all maxima of the real function |f| occur on that boundary. 18:36 <Z-module> and further, if f is nonzero in the open disk, then all minima of |f| also occur on that boundary. 18:37 <Z-module> (but you do need f nonzero in the open disk for that, as you see by taking f(z) = z on an origin-centered disk) 18:38 <Z-module> Hmm, maybe you're not actually dealing with complex functions and the maximum modulus theorem here? 18:38 <Evidlo> do you really mean nonzero, and not nonconstant? 18:38 <Z-module> yes, nonzero 18:38 <Z-module> look at my f(z) = z example, in the unit disk centered at the origin 18:40 <Z-module> but if f is nonzero and nonconstant in the open disk D, **then** either f is 0 somewhere on the boundary in which case min |f| = 0 is only on the boundary, or else f is nonzero on the whole boundary too in which case the function 1/f is nonconstant and is analytic on the open D too, and we now apply the first theorem. (a max of |1/f| is a min of |f| ) 18:40 <Z-module> (when f is nonzero everywhere in question) 18:41 <Z-module> I should have said: in that last case, 1/f is also continuous on the boundary