# Proving that an integral is a pure imaginary

• JPaquim
In summary, the homework statement is that the integral of f*df is a pure imaginary for any piecewise smooth closed curve and any C^1 function f. The attempt at a solution has been tried from different angles, but found to be unsuccessful. One possible solution is to use integration by parts, but this is easier said than done as f may not be differentiable.
JPaquim

## Homework Statement

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

JPaquim said:

## Homework Statement

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

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.

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Dick said:
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.

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...

JPaquim said:
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...

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:
Dick said:
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.

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

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

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

## 1. What does it mean for an integral to be a pure imaginary?

For an integral to be a pure imaginary, it means that the result of the integral is a complex number with no real component. In other words, the integral evaluates to a number that is a multiple of the imaginary unit, i, without any real numbers involved.

## 2. How can I prove that an integral is a pure imaginary?

To prove that an integral is a pure imaginary, you can use the fundamental theorem of calculus, which states that the integral of a function can be evaluated by finding the antiderivative of the function and evaluating it at the upper and lower limits of integration. If the result is a multiple of i, then the integral is a pure imaginary.

## 3. Can all integrals be proven to be pure imaginary?

No, not all integrals can be proven to be pure imaginary. Only integrals of functions that have an imaginary component can be proven to be pure imaginary. If a function has no imaginary component, then its integral will also have no imaginary component.

## 4. What is the significance of proving that an integral is a pure imaginary?

The significance of proving that an integral is a pure imaginary is that it can help in simplifying complex calculations and solving certain types of integrals. It can also provide insight into the behavior and properties of the function being integrated.

## 5. Are there any practical applications of proving that an integral is a pure imaginary?

Yes, there are practical applications of proving that an integral is a pure imaginary. For example, in physics, integrals of certain functions that represent physical quantities can be proven to be pure imaginary, providing valuable information about the behavior of the physical system being studied. Additionally, in engineering and mathematics, the proof of a pure imaginary integral can be used to simplify calculations and solve problems.

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