- #1
ognik
- 626
- 2
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.
---------------------
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?
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.
---------------------
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?
