# Proving that an integral is a pure imaginary

1. Feb 2, 2013

### JPaquim

1. The problem statement, all variables and given/known data

Show that $\int_{\gamma}\ f^*(z)\ f'(z)\ dz$ is a pure imaginary for any piecewise smooth closed curve $\gamma$ and any $C^1$ function $f$ whose domain contains an open set containing the image of $\gamma$

2. The attempt at a solution

I have tried to approach it from some different angles. I've tried to Taylor expand everything, to see if some interesting things would happen, but I found none. I've also tried to expand it as such:

$\int_{\gamma}\ f^*(z)\ f'(z)\ dz = \int_{\gamma}\ [(u - iv)(\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x})](dx + idy) =$

$\int_{\gamma}\ (u\frac{\partial u}{\partial x}+v\frac{\partial v}{\partial x})dx+(u\frac{\partial u}{\partial y} + v\frac{\partial v}{\partial y})dy + i\int_{\gamma}\ (u\frac{\partial v}{\partial x}-v\frac{\partial u}{\partial x})dx+(u\frac{\partial v}{\partial y}-v\frac{\partial u}{\partial y})dy$

What can I do with this? Is this the way to go?

Cheers

2. Feb 2, 2013

### Dick

It's a lot simpler than that. A number is pure imaginary if its complex conjugate is equal to its negative. Hint: use integration by parts. You are starting from the integral of f*df.

Last edited: Feb 2, 2013
3. Feb 2, 2013

### JPaquim

I thought of that, but integration by parts produces $-\int_\gamma\ f\cdot(f^*)'dz$, but I have no guarantee that $f^*$ is differentiable... And even if it were, I'd still have to show that $-\int_\gamma\ f\cdot(f^*)'dz = -\int_\gamma\ f\cdot(f')^*dz$, which doesn't seem trivial...

4. Feb 2, 2013

### Dick

Your are making this seem a LOT harder than it is. If f is differentiable then f* is differentiable. That's almost trivial. As is (f*)'=(f')*. We are just talking C^1 differentiabilty here, the existence of continuous partial derviatives, not complex differentiability. If your curve is z=γ(t) then you can express all of the derivatives as d/dt.

Last edited: Feb 2, 2013
5. Feb 3, 2013

### JPaquim

Oh, that makes a lot of sense! So one gets:

$\int_{\gamma}\ f^*(z)\ f'(z)\ dz = \int_a^b\ f(\gamma(t))^*\cdot f'(\gamma(t))\cdot \gamma'(t)\ dt = \int_a^b\ ((f\circ\gamma)(t))^*\cdot \frac{d}{dt}(f\circ\gamma)(t)\ dt$
$= [\ |\ f(\gamma(t))\ |^2\ ]_a^b - \int_a^b\ \frac{d}{dt}((f\circ\gamma)^*)(t) \cdot (f\circ\gamma)(t)\ dt = -\int_a^b\ (f'(\gamma(t)))^* \cdot \gamma'(t) \cdot f(\gamma(t))\ dt$
$= -\int_{\gamma}\ f(z)\ (f'(z))^*\ dz$

Last edited: Feb 3, 2013
6. Feb 3, 2013

### Dick

Right! It's actually (ff*)=|f|^2 in the difference term for the integration by parts, but that doesn't make any difference.

7. Feb 3, 2013

### JPaquim

You're totally right, my mistake. I've already corrected it.

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