In deriving the necessary conditions for a function to be differentiable at a point , why do we get ? Would not it be ?

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- Jan 2nd 2009, 08:33 PMmanjohn12Cauchy Riemann Equations
In deriving the necessary conditions for a function to be differentiable at a point , why do we get ? Would not it be ?

- Jan 2nd 2009, 09:02 PMAryth
Because the Cauchy Riemann equations are:

where is the complex conjugate of z = x + iy.

Where u and v are the real and imaginary parts of:

- Jan 3rd 2009, 01:55 AMMoo
- Jan 3rd 2009, 07:34 AMAryth
You're right... I edited it.

- Jan 3rd 2009, 08:00 AMHallsofIvy
I'm surprised that advertising like this would be allowed.

- Jan 3rd 2009, 08:33 AMHallsofIvy
Fundamentally, the answer is the "chain rule". Since z= z+ iy, and

Crucial point: 1/i= -i.

Another way of looking at the Cauchy-Riemann equations is this: With f(z)= f(x+ iy)= u(x,y)+ iv(x,y), the derivative, at , is the given by the limit just as for real numbers. And, just like limits for real numbers, in order that the limit exist, we must get the same result approaching along any path. In particular, if we approach along a line parallel to the real axis, h is real so we have

[tex]= \frac{\partial u}{\partial x}+ i\frac{\partial v}{\partial u}

Approaching instead along a line parallel to the imaginary axis, h is imaginary and we can use "ih", with i real, instead. Now we have

and the important difference is that "i" in the denominator. It will cancel "i" in the v part and remember that 1/i= -i so

In order that the limit exist, those two must be the same. Equating real and imaginary parts,

and

- Jan 3rd 2009, 09:53 AMJester