2
$\begingroup$

By the link Proof that a function is holomorphic it is told

Elementary operations or compositions of holomorphic functions give holomorphic functions on the maximal domain where the functions are defined. This is a consequence of the rules of derivation for product, ratio and compositions of functions.

But per my understand I have counterexample $ f(z) = f(x + \imath y) = \frac{x - \imath y}{x + \imath y}. $ I calculate by dividing complex that $ f(x + \imath y) = 1 - 2\imath\frac{xy}{x^2 + y^2} = u(x, y) + \imath v(x, y). $ Then I verify Cauchy-Riemann criteria $ \frac{\partial u}{\partial x} = 0, $ and while this $ \frac{\partial v}{\partial y} = -2x\frac{x^2 - y^2}{(x^2 + y^2)^2}, $ which means that $f$ is not holomorphic.

Did I made mistake in calculations or this means that cited statement is not correct?

  • 6
    Is your numerator holomorphic?2011-07-13

2 Answers 2

5

Take the representation $z=x+y i$. Then the function $f:\mathbb{C}\rightarrow\mathbb{C}$ defined by $f(z)=\overline{z}=x-y i$ is not a holomorphic function. It is antiholomorphic. See http://en.wikipedia.org/wiki/Complex_conjugate

1

$x-iy$ is not a holomorphic function. The Cauchy-Riemann equations are not satisfied: $\partial x/\partial x = 1$ while $\partial (-y)/\partial y = -1$. So the equation $\partial u/\partial x = \partial v / \partial y$ does not hold at any point $(x,y)$. So this function is not holomorphic, and so your example is not a counterexample.