# Cauchy-Riemann equations

• Dec 21st 2009, 03:44 AM
jezzyjez
Cauchy-Riemann equations
I understand how the equations work but i can't seem to be able to determine whether "1 / (1 + z)" satisfies the cauchy-riemann equations and hence determine whether f ' (z) exists.

Any help would be greatly apprichiated
• Dec 21st 2009, 03:49 AM
mr fantastic
Quote:

Originally Posted by jezzyjez
I understand how the equations work but i can't seem to be able to determine whether "1 / (1 + z)" satisfies the cauchy-riemann equations and hence determine whether f ' (z) exists.

Any help would be greatly apprichiated

If you substitute $z = x + iy$ and multiply numerator and denominator by the complex conjugate of the denominator and simplify, you will get $f = \frac{x + 1}{x^2 + 2x + 1 + y^2} - i \frac{y}{x^2 + 2x + 1 + y^2}$.
• Dec 21st 2009, 04:13 AM
jezzyjez
Yeh cool i get that now thanks, am i missing a trick with the last part of the question

"determine whether f ' (x) exists"

or does it literally exist when the equations satisfy the C-R equations??
• Dec 21st 2009, 04:23 AM
tonio
Quote:

Originally Posted by jezzyjez
I understand how the equations work but i can't seem to be able to determine whether "1 / (1 + z)" satisfies the cauchy-riemann equations and hence determine whether f ' (z) exists.

Any help would be greatly apprichiated

You can evaluate the partial derivatives with the function as Mr. Fantastic showed you, or else realize (most books have it) that the CR-equations

are equivalent to $\frac{\partial f}{\partial\,\overline{z}}=0$ , where we have that $\frac{\partial}{\partial\,\overline{z}}=\frac{1}{2 }\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)$ , so putting $z=x+iy$, we get:

$f(z)=f(x,y)=\frac{1}{1+x+iy}\Longrightarrow \frac{\partial f}{\partial x}=-\,\frac{1}{(1+x+iy)^2}\,,\,\,\frac{\partial f}{\partial y}=-\,\frac{i}{(1+x+iy)^2}$.

Check the above and conclude your function is analytic in $\mathbb{C}\setminus \{1\}$

Tonio
• Dec 21st 2009, 07:13 PM
mr fantastic
Quote:

Originally Posted by tonio
You can evaluate the partial derivatives with the function as Mr. Fantastic showed you, or else realize (most books have it) that the CR-equations

are equivalent to $\frac{\partial f}{\partial\,\overline{z}}=0$ , where we have that $\frac{\partial}{\partial\,\overline{z}}=\frac{1}{2 }\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)$ , so putting $z=x+iy$, we get:

$f(z)=f(x,y)=\frac{1}{1+x+iy}\Longrightarrow \frac{\partial f}{\partial x}=-\,\frac{1}{(1+x+iy)^2}\,,\,\,\frac{\partial f}{\partial y}=-\,\frac{i}{(1+x+iy)^2}$.

Check the above and conclude your function is analytic in $\mathbb{C}\setminus \{{\color{red}-1}\}$

Tonio

A thread of related interest: http://www.mathhelpforum.com/math-he...erivation.html

(And a correction of a minor typo in red).
• Dec 21st 2009, 08:14 PM
Bruno J.
Quote:

Originally Posted by jezzyjez
or does it literally exist when the equations satisfy the C-R equations??

The function is analytic at a point if and only if the C-R equations are satisfied in some neighbourhood of the point.