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zzdfa:

--- Quote from: /0 on July 25, 2010, 02:12:34 am ---
It seems that if you set , then it essentially becomes the first limit above:
Since ( is the same as right?)


--- End quote ---
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:

--- Quote from: zzdfa on July 29, 2010, 11:51:49 pm ---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.


--- End quote ---
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 :-\

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