Complex Analysis

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In the complex derivative,



Can 'h' be anything?

This looks a bit like the directional derivative formula in :



But yet there is a 'unique' ? I would have thought should vary depending on the direction you approach.

(This question of course leads to how the Cauchy-Riemann equations came about)

Thanks

[zzdfa]

Remember the definitions: a function is differentiable only if that limit exists.
If you get 2 different values depending on which direction you approach, then the limit does not exist and hence the function is not differentiable. if it -is- unique, then we say it is differentiable.



for example f(z)=Re(z)+2Im(z)= is not complex differentiable at 0 because the limit doesn't exist.

[/0]

But if you think of the function as a 'contour map' on the x-y plane ()... if every point in a domain has derivatives from every side equal, then surely the function must be constant?

[zzdfa]

It is true that if the directional derivatives of f:R^2->R at each point are the same for every direction, then f is constant.  (because if directional derivative is the same in all directions it implies the gradient is 0 at that point. if the gradient at 0 at every point then f is constant)

However you can't think of a complex function C->C as a contour map, remember the range of the function the complex numbers, so the graph is a surface in C^2, or R^4.


just try a simple example, check that if f(z)=z^2 then  

[/0]

Ah, ok thanks zzdfa =)

Another Q, if you approach a point along the imaginary axis then:



Is the from the in the limit?

Since

It seems that if you set , then it essentially becomes the first limit above:
Since ( is the same as right?)

So is it really ok to take the out?

[zzdfa]

It seems that if you set , then it essentially becomes the first limit above:
Since ( is the same as right?)

Yes the -i is from the 1/i, your argument doesn't work because is actually equal to 

because ik->0 is the same as k->0,
and




I think where you might be getting confused is the association between the function f:C->C and the function f*:R^2->C.

given a function with complex domain, f(z), the associated function f*:R^2->C is defined by f*(x,y)=f(x+yi).

of course in the above we used f to denote both f and f* because you can just look at the number of arguments to figure out which is which.

[/0]

Thanks zzdfa xD


When people talk of a function being differentiable, do they mean differentiable in a neighbourhood, or differentiable everywhere?
Usually people only test differentiability at (0,0), is there any reason why?

The lecture notes give examples or functions which are non-differentiable:
, , , and
Does this mean differentiable nowhere, or does it mean not differentiable at a selection of points?


Also, the tute says that is not analytic at . But

So , , , .

So at , the Cauchy Riemann equations are satisfied, so why isn't the function analytic at z = 0?

Thanks

[zzdfa]

You might see your lecturer use (0,0) alot in examples, when computing derivatives, because it's simplifies the calculations a bit. But of course there is no special significance since you can translate any function by f(z-p).

Generally in complex analysis when we say differentiable we mean differentiable on some open set U.

 You can say that your function f is differentiable at (0,0) with derivative 0, but you can't say it's analytic because analyticity  at p means 'differentiable in an open set containing p'. so to show that f is analytic around p you have to show that the CR equations are satisfied in an open set around p, which is isn't in your case.

[humph]

You might see your lecturer use (0,0) alot in examples, when computing derivatives, because it's simplifies the calculations a bit. But of course there is no special significance since you can translate any function by f(z-p).

Generally in complex analysis when we say differentiable we mean differentiable on some open set U.

 You can say that your function f is differentiable at (0,0) with derivative 0, but you can't say it's analytic because analyticity  at p means 'differentiable in an open set containing p'. so to show that f is analytic around p you have to show that the CR equations are satisfied in an open set around p, which is isn't in your case.

No, when you say differentiable, you just mean differentiable at a point. When it is holomorphic/analytic at a point, then it must be differentiable in a neighbourhood of that point.

[zzdfa]

ah yeah, I even contradicted myself in the next sentence

[humph]

Thanks zzdfa xD


When people talk of a function being differentiable, do they mean differentiable in a neighbourhood, or differentiable everywhere?
Usually people only test differentiability at (0,0), is there any reason why?

The lecture notes give examples or functions which are non-differentiable:
, , , and
Does this mean differentiable nowhere, or does it mean not differentiable at a selection of points?


Also, the tute says that is not analytic at . But

So , , , .

So at , the Cauchy Riemann equations are satisfied, so why isn't the function analytic at z = 0?

Thanks

Who's your tutor? As was mentioned, it's differentiable at zero, but analytic (well I prefer holomorphic) means it has to be differentiable in a neighbourhood of zero, which isn't the case here.

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My tutor's Davidson Ng, but we haven't had a tutorial yet.

I understand what you mean, that it must be differentiable in a neighbourhood... but the theorem that I used is as follows:

Let be given by , for . Assume that all partial derivatives , , , exist, are continuous, and satisfy the Cauchy-Riemann equations at . Then is holomorphic at .


It never says anything about being differentiable in a neighbourhood, only about the continuity of the partial derivatives

[Ahmad]

Where did you see that?

satisfies CR equations at z = 0 and nowhere else, the partials are continuous but f is nowhere analytic.

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I believe it's called the Looman-Menchoff Theorem

The version I quoted is what's in the lecture notes. Is there a difference?

[Ahmad]

Yeah. The result you should remember is that if CR equations are satisfied at a point and the partials are continuous at the point then the function is differentiable at the point. Then being holomorphic/analytic in some open set just means differentiable everywhere in the open set, so CR equations are satisfied + continuous partials everywhere in the open set. Analytic at a single point would mean differentiable in some open set containing the point.