MHB Do Cauchy-Riemann Conditions Guarantee Analyticity?

[ognik]

Hi - just started complex analysis for the 1st time. I have been a little confused as to the chicken and egg-ness of Cauchy-Riemann conditions...

1) Wiki says:
"Then f = u + iv is complex-differentiable at that point if and only if the partial derivatives of u and v satisfy the Cauchy–Riemann equations (1a) and (1b) at that point"; that seems clear enough to me.

My book however says:
"Cauchy–Riemann conditions are necessary for the existence of a derivative of f (z); that is, if df/dz exists, the Cauchy–Riemann conditions must hold"
To me the highlighted part implies that if you can differentiate f(z), then the C-R conditions will hold, which kind of contradicts the "if and only if" of the wiki definition? I'd appreciate is clarity on that...

Also I have read that the pd's must ALSO be continuous for the existence of a derivative of f (z)?

2) I understand that if f(z) is differentiable as above - at & near some point $ {z}_{0} $, then f(z) is 'complex analytic' at that point. This apparently means that if complex analytic, the Real and Imag parts each always satisfy the Laplace equation?

So - we test a complex equation to see if the C-R conditions are satisfied, if they are then the eqtn is analytic (at that point, everywhere makes it an entire function) AND we know the Laplace eqtns hold ?

3) To confuse me further, in search of a clearer explanation I found stuff on the web which says: "All complex functions f(z) are infinitely differentiable and, in fact, analytic where defined"?

I am a tad confused between the above 3 points and would appreciate something like a bullet list of what I really need to understand.
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4) Soldering on, an identity exercise I am stuck on is:
The functions u(x, y) and v(x, y) are the real and imaginary parts, respectively, of an analytic function w(z).
Show that $ \pd{u}{x}\pd{u}{y}+\pd{v}{x}\pd{v}{y} = 0 $ and give a geometric interpretation

Sounds simple, Cauchy-Riemann applies for analytic function, so $ {u}_{x} = {v}_{y} $ and $ {u}_{y} = -{v}_{x} $
$ \therefore {u}_{x} - {v}_{y} = 0 = {u}_{y} + {v}_{x}$, pd both sides w.r.t. y gives
$ {u}_{xy} - {v}_{yy} = {u}_{yy} + {v}_{xy}, \therefore {u}_{xy} - {v}_{xy} ={u}_{yy} + {v}_{yy} $ ... not useful.

I also tried $ {u}_{x} = {v}_{y} \therefore {u}_{xy} = {v}_{yy} $ and
$ {u}_{y} = -{v}_{x} \therefore {v}_{xy} = -{u}_{yy} $
$ \therefore {u}_{xy} + {v}_{xy} = {v}_{yy} -{u}_{yy} $ also not useful

I also fiddled with the laplace eqtns to no avail ($ \nabla^2{u} = 0 = \nabla^2{v} $ )
A hint please?

 

[ognik]

I think I should break this down a bit :-)

Wiki has a definition: "A function is analytic if and only if its Taylor series about x0 converges to the function in some neighborhood for every x0 in its domain"

Isn't the point about all Taylor series that they will converge to the function?

And am I right to say that, to be able to build a Taylor series, the function has to be infinitely differentiable anyway?
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Another from Wiki: "The sole existence of partial derivatives satisfying the Cauchy–Riemann equations is not enough to ensure complex differentiability at that point. It is necessary that u and v be real differentiable, which is a stronger condition than the existence of the partial derivatives"

What is the difference between 'real differentiable' and 'the existence of partial derivatives'?