I tried solving this question by taking the derivative of f(z) = u(x,y) +iv(x,y). Then setting the derivative to zero at $z_o$. But doing this did not give any useful insight. I am generally confused by how I can relate f(z) never vanishing and being constant but there being a minimum value among the values of |f(z)|.
2
-
$\begingroup$ math.stackexchange.com/q/3413060/42969 $\endgroup$– Martin RFeb 10 at 19:39
-
$\begingroup$ Without loss of generality you may suppose that $f(z_0)\in\mathopen]0\mathbin;+\infty\mathclose[$. Now, if $f$ isn't constant, there is a smaller $k\geqslant1$ such that $f^{(k)}(z_0)\neq0$, then $f(z_0+h)=f(z_0)+\alpha h^k+o(h^k)$ when $h\to0$ (for some $\alpha\neq0$). For some convenient choice for $h$, one obtains that $|f(z_0+h)|<f(z_0)$; a contradiction. $\endgroup$– jp boucheronFeb 10 at 22:22
Add a comment
|