Cauchy-Riemann equations

**jezzyjez** · December 21st 2009, 04:44 AM

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

**mr fantastic** · December 21st 2009, 04:49 AM

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 $f = \frac{x + 1}{x^2 + 2x + 1 + y^2} - i \frac{y}{x^2 + 2x + 1 + y^2}$ .

**jezzyjez** · December 21st 2009, 05:13 AM

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??

**tonio** · December 21st 2009, 05:23 AM

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

**mr fantastic** · December 21st 2009, 08:13 PM

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).

**Bruno J.** · December 21st 2009, 09:14 PM

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.

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