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# Differentiability in the complex plane. watch

1. Given that for some constants a and b, show that v is the imaginary part of some differentiable function f only if a = 6 and b = -3.

I'm not sure how to approach this. I think it's linked to Cauchy-Riemann equations somehow but can't figure out how.
2. Remember that if v is continuously differentiable then

That + C-R equations should be enough I think.
3. (Original post by DFranklin)
Remember that if v is continuously differentiable then

That + C-R equations should be enough I think.
I've figured out and I get for eachso the equation holds but I still don't understand how to deduce that a can only be 6 and b can only be -3.
4. Subscribing.
5. (Original post by canine101)
I've figured out and I get for eachso the equation holds but I still don't understand how to deduce that a can only be 6 and b can only be -3.
You just need to use the Cauchy-Riemann equations:

You can find from your expression for .

By the Cauchy-Riemann equations, , so we also have an expression for .

We can then integrate the expression for with respect to x to get an expression for (which will include an unknown function of y).

We can then differentiate this expression for with respect to y to obtain an expression for .

And using the Cauchy-Riemann equations once more, you get the result you need. Can you see where to go from here?
6. What mark said. First find , then integrate to give a value of in terms of your constants and . Now you can partially differentiate wrt and can compare this result with the by the uniqueness of the partial derivative.

On a similar note, I'm not sure it would be sufficient to simply integrate your and to find two values of then compare, as we would not be using the uniqueness of the partial derivative and thus wouldn't be fulfilling the 'only if' part of the question... although I could be wrong.
7. To spell out my hint a little stronger:

Use the Cauchy-Riemann equations and then work out what must be.

I would have thought this is the standard approach, but it does assume that if f is complex differentiable on an open set it must be C^2 on that set - which you may not have had proved yet.

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Updated: April 7, 2011
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