Complex function derivation

**Pinsky** · January 19th 2008, 12:14 PM

I have to write in theory the derivation of a complex variable.
So, when
$\Delta x=0$
$\Delta z=i\Delta y$

i have

$\lim_{\Delta z \rightarrow 0}\frac{f(z+\Delta z)-f(z)}{\Delta z}=\lim_{\Delta y \rightarrow 0} \frac{-i\Delta y}{i\Delta y}=-1$

what do we do to get $\lim_{\Delta y \rightarrow 0} \frac{-i\Delta y}{i\Delta y}$ from $\lim_{\Delta z \rightarrow 0}\frac{f(z+\Delta z)-f(z)}{\Delta z}$

**topsquark** · January 19th 2008, 01:49 PM

Originally Posted by Pinsky

I have to write in theory the derivation of a complex variable.
So, when
$\Delta x=0$
$\Delta z=i\Delta y$

i have

$\lim_{\Delta z \rightarrow 0}\frac{f(z+\Delta z)-f(z)}{\Delta z}=\lim_{\Delta y \rightarrow 0} \frac{-i\Delta y}{i\Delta y}=-1$

what do we do to get $\lim_{\Delta y \rightarrow 0} \frac{-i\Delta y}{i\Delta y}$ from $\lim_{\Delta z \rightarrow 0}\frac{f(z+\Delta z)-f(z)}{\Delta z}$

You don't. You need to know something about your function first.

-Dan

**Pinsky** · January 19th 2008, 02:24 PM

Sorry, forgot to write it.

$f(z)=\bar{z}$

need to find out if its derivable.

**ThePerfectHacker** · January 19th 2008, 04:51 PM

Originally Posted by Pinsky

$f(z)=\bar{z}$

You cannot find the derivate. Just Cauchy-Riemann equations. They do not work.

**topsquark** · January 19th 2008, 04:59 PM

Originally Posted by Pinsky

Sorry, forgot to write it.

$f(z)=\bar{z}$

need to find out if its derivable.

Ahhh!

$f(z) = \bar{z}$

$f^{\prime}(z) = \lim_{h \to 0} \frac{\overline{(z + h)} - \bar{z}}{h}$
where h approaches z from some arbitrary direction to 0.

The trick is that in order to define a derivative it must be the same no matter what route we take to get h to go to 0. For example, let h approach 0 along the real axis:
$f^{\prime}(z) = \lim_{h \to 0} \frac{\overline{(z + h)} - \bar{z}}{h}$

$= \lim_{h \to 0} \frac{\bar{z} + h - \bar{z}}{h}$

$= \lim_{h \to 0} \frac{h}{h}$

$= 1$

But now let h approach 0 along the imaginary axis:
$f^{\prime}(z) = \lim_{h \to 0} \frac{\overline{(z + h)} - \bar{z}}{h}$

$= \lim_{h \to 0} \frac{\bar{z} + \bar{h} - \bar{z}}{h}$

$= \lim_{h \to 0} \frac{\bar{h}}{h}$

In order to evaluate this, let $h = ia$ and the expression becomes:
$= \lim_{a \to 0} \frac{\bar{ia}}{ia}$

$= \lim_{a \to 0} \frac{-ia}{ia}$

$= -1$

Since the two answers are different we know that the derivative of $f(z) = \bar{z}$ cannot be defined.

-Dan

**mr fantastic** · January 19th 2008, 07:43 PM

Originally Posted by ThePerfectHacker

You cannot find the derivate. Just Cauchy-Riemann equations. They do not work.

I'll add a little known (to some, maybe) but useful theorem:

When a function is expressed in the form $f(z, \bar{z})$ , the Cauchy-Riemann relation is $\frac{\partial f}{\partial \bar{z}} = 0$ .

Posting of proof available upon request.

So it's trivial to see that $f(z) = \bar{z}$ is not analytic .....

**ThePerfectHacker** · January 19th 2008, 07:50 PM

Originally Posted by mr fantastic

When a function is expressed in the form $f(z, \bar{z})$ , the Cauchy-Riemann relation is $\frac{\partial f}{\partial \bar{z}} = 0$ .

What do you mean "... of the form $f(z,\bar z)$ ... " and what do you mean by $\partial_{\bar z} f$ ?

**mr fantastic** · January 19th 2008, 09:09 PM

Originally Posted by ThePerfectHacker

What do you mean "... of the form $f(z,\bar z)$ ... " and what do you mean by $\partial_{\bar z} f$ ?

As you probably surmised but wanted to confirm:

".... of the form $f(z, \bar{z})$ ...." means a function depending explicitly on $z$ and $\bar{z}$ .

Egs. $f(z, \bar{z}) = \bar{z}$ , $f(z, \bar{z}) = z^2$ , $f(z, \bar{z}) = z \bar{z}^2$ .

$\partial_{\bar z} f$ means take the partial derivative of $f(z, \bar{z})$ wrt $\bar{z}$ .

Egs. If $f(z, \bar{z}) = \bar{z}$ then $\partial_{\bar z} f = 1$ .

If $f(z, \bar{z}) = z^2$ then $\partial_{\bar z} f = 0$ .

If $f(z, \bar{z}) = z \bar{z}^2$ then $\partial_{\bar z} f = 2 z \bar{z}$ .

And of course, all complex functions of the form $f(x, y) = u(x, y) + i v(x, y)$ can be expressed in the form $f(z, \bar{z})$ , in which case the more common formulation of C-R would probably be more easily applied ..... The formulation of Cauchy-Riemann I posted is rarely taught, in my experience,

(And please be kind enough not to retort with "And that's because it's

."

)

and can make life easier or even trivial on occassion. A case in point is the question that created this thread.

**ThePerfectHacker** · January 20th 2008, 09:12 AM

Originally Posted by mr fantastic

As you probably surmised but wanted to confirm:

".... of the form $f(z, \bar{z})$ ...." means a function depending explicitly on $z$ and $\bar{z}$ .

Egs. $f(z, \bar{z}) = \bar{z}$ , $f(z, \bar{z}) = z^2$ , $f(z, \bar{z}) = z \bar{z}^2$ .

$\partial_{\bar z} f$ means take the partial derivative of $f(z, \bar{z})$ wrt $\bar{z}$ .

As you might guess I am going to have a problem with this. I am thinking how to formally define " $f(z,\bar z)$ "? And even worse a partial derivative to $\bar z$ . What you wrote above, I am sure is correct, but it seems like a physics-type theorem. I am thinking how to state it in mathematical terms. Maybe what we are saying $f(z,\bar z)$ can be written as some composition of complex-valued function having a certain property?

The formulation of Cauchy-Riemann I posted is rarely taught, in my experience,

I never took a course in complex analysis, so I really have no idea. But I can see why you would say that. I would assume that the majority of a basic course in complex analysis consists of complex-integration (which happens to be the fundamental concept).

**mr fantastic** · January 20th 2008, 10:53 AM

Originally Posted by ThePerfectHacker

As you might guess I am going to have a problem with this. I am thinking how to formally define " $f(z,\bar z)$ "? And even worse a partial derivative to $\bar z$ . What you wrote above, I am sure is correct, but it seems like a physics-type theorem. I am thinking how to state it in mathematical terms. Maybe what we are saying $f(z,\bar z)$ can be written as some composition of complex-valued function having a certain property?

I never took a course in complex analysis, so I really have no idea. But I can see why you would say that. I would assume that the majority of a basic course in complex analysis consists of complex-integration (which happens to be the fundamental concept).

Yep, I did kinda guess that one

I guess I better not post my proof

.

I have seen the result a couple of times over the years in complex analysis textbooks - given as a question for the student to prove.

And don't be shy in facing the truth, TPH - progress in mathematics has always been due to physicists

**ThePerfectHacker** · January 20th 2008, 12:37 PM

Originally Posted by mr fantastic

And don't be shy in facing the truth, TPH - progress in mathematics has always been due to physicists

Remove the word always.

It depends on what period. In ancient Greece, through the Medieval Ages, there was no physics and it was done only for math reasons. Later (at what I call) the Modern Age of math, Fermat and Descrates will build the fondation that will inspire Newton, in the 1600's. In the late 1600's and 1700's Newton began to create ideas of mathematics which were based on physics. This lead to studying differencial equations of mechanics. But soon the Bernoulli's, Lagrange, Fourier, and Euler (along with other mathematicians) added to what math used in physics. They developed more advanced methods in calculus, differencial equations, even began (Fourier) solving partial differencial equations, .... We are now at the 1800's. Close to 150 years the majority of the mathematics that is taught to engineering and physics students was devoloped during this time. Yes during this time much of mathematics we use today come from physics.

Fermat did not just no create a foundation of the mathematics that was mentioned above. There was another road in mathematics that he devoloped, a road is is not as commonly known as the one above. It was number theory (in fact he is called the father of modern number theory). Number theory is far far removed from any physical laws. It was exactly how math was done in ancient Greece, just for the sake of it. Fermat's problems inspired other mathematicians so solve them, and in effect create newer mathematics. People such as Gauss, Euler, Lagrange devoloped newer mathematics which was entirely removed from physics. This is an example, of a large area of math for which physics was not responsible for.

We should not forget the beginning of the 1800's as well. It was during this time when Galois, Abel, and others, were developing basic ideas for group and field theory. This area of math was again done for pure algebraic reasons with no connection to physics at all. It still contines to this date. And the amazing thing about this subject of math is that it even inspired physics (according to topsquark, concepts of field theory are used in quantum mechanics). This is again an area of math not influenced by physics at all.

There are more things to say, but I am not going to go there. For example, a lot mathematics of the 20th century and beyond is so far removed from the real world that it is not inspired by anything else.

It seems to me that without physics we would have had still much mathematics because by what said above. However, a big portion of mathematics would not be in existence today. Namely, analysis. After Calculus was formed mathematians where bothered by its informalities which lead them to a theory. Otherwise we might never had the same analysis as we have today.

**ThePerfectHacker** · January 21st 2008, 01:38 PM

Originally Posted by mr fantastic

I'll add a little known (to some, maybe) but useful theorem:

When a function is expressed in the form $f(z, \bar{z})$ , the Cauchy-Riemann relation is $\frac{\partial f}{\partial \bar{z}} = 0$ .

Posting of proof available upon request.

So it's trivial to see that $f(z) = \bar{z}$ is not analytic .....

Maybe this is what it means mathematically.

Let $f:\mathbb{C}\mapsto \mathbb{C}$ be an analytic function. Suppose there exists $\bold{g}: \mathbb{C}\mapsto \mathbb{C}^2$ defined as $\bold{g}(z) = (z,\bar z)$ and $F: \mathbb{C}^2\mapsto \mathbb{C}$ such that $F(\bold{g}(z)) = f(z)$ then $\partial_2 F = 0$ .

Example: Consider $f(z) = |z|^2$ then if we let $F(z_1,z_2) = z_1z_2$ then $F(\bold{g}(z)) = z\bar z = |z| = f(z)$ but $\partial_1 F = z_2 \not = 0$ .

**mr fantastic** · January 21st 2008, 08:02 PM

Originally Posted by ThePerfectHacker

[snip]
Example: Consider $f(z) = |z|^2$ then if we let $F(z_1,z_2) = z_1z_2$ then $F(\bold{g}(z)) = z\bar z = |z| = f(z)$ but $\partial_1 F = z_2 \not = 0$ .

I'd just blunder on in and say $f = |z|^2 = z \bar{z} \therefore \frac{\partial f}{\partial \bar{z}} = z \neq 0 \therefore \text{not analytic}$ ......

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