Path dependance of complex conjugate

[ognik]

Hi, an exercise asks to show that $ \int_{0,0}^{1,1} {z}^{*}\,dz $ depends on the path, using the 2 obvious rectangular paths. So I did:

$ \int_{c} {z}^{*}\,dz = \int_{c}(x-iy) \,(dx+idy) = \int_{c}(xdx + ydy) + i\int_{c}(xdy - ydx) = \frac{1}{2}({x}^{2} + {y}^{2}) |_{c} + i(xy - yx)|_{c} $

The real part is an exact differential which is path-independent; also using the 2 double-step paths explicitly $ [{c}_{1} = (o,o) -> (1,0)-> (1,1) $ and $ {c}_{2} = (o,o) -> (0,1)-> (1,1) ] $ confirmed that.

So I expect the imaginary part to show path dependence, but xy-yx evaluates to 0 instead? Where have I gone wrong with this approach?

 

[Euge]

Hi ognik,

You have a right approach but the wrong analysis. The differential $x\, dy - y\, dx$ is not exact (in particular, it's not $d(xy - yx)$), so your reasoning is off point. Recall that a differential $M\, dx + N\, dy$ is exact iff $M_y = N_x$. Since $\frac{\partial}{\partial y}(-y) = -1 \neq 1 = \frac{\partial}{\partial x}(x)$, the differential $x\, dy - y\, dx$ is inexact. Use parametrizations as you would normally do with line integrals to solve this problem.

 

[ognik]

Hi Euge, first may I just check my understanding about the approach I used:
You may have misread, I actually have the real part as exact, IE $ (xdx + ydy); so {M}_{y}=0={N}_{x} $, I think that's right?

Yes, the Imag part$ (xdy−ydx) $ is not exact; so I integrated it to get $ i(xy−yx) $, which cancels to 0 doesn't it?

Now onto parameterization, I am used to param'ing when it makes the eqtn easier (like polar/trig etc), but given that the path C is here 2 straight line segments, I would normally be quite happy to stay with rectilinear cords - so for next time, what should have warned me to parametise?

This next is quite new to me, please review?

There are 2 paths segments, call them C₁ and C₂.
First I'll look at moving 1 to the right, then 1 up:
C₁ is along the real axis, so $ {z}^{*}(t) =t, 0 \le t \le 1, dz = dt $
C₂ is from (1,0) to (1,1), so $ {z}^{*}(t) =1+it, 0 \le t \le 1, dz = idt $
Then $ \int_{0}^{1} t\,dt + \int_{0}^{1} (1+it)\,idt = \frac{1}{2}{t}^{2}|^1_{0} + (i-\frac{1}{2}{t}^{2}) |^1_{0} = \frac{1}{2} + i - \frac{1}{2} - i = 0 $

Similarly for the alternate path, up first, then to the right:
C₁: $ {z}^{*}(t) =it, 0 \le t \le 1, dz = idt $
C₂: $ {z}^{*}(t) =i+t, 0 \le t \le 1, dz = dt $
$ \therefore \int_{0}^{1} it\,idt + \int_{0}^{1} (i+t)\,dt = -\frac{1}{2}{t}^{2}|^1_{0} + (it + \frac{1}{2}{t}^{2}) |^1_{0} = \frac{1}{2} + i - \frac{1}{2} = i $
And as they are not equal, $ {z}^{*} $ is path dependent. YAY!

Can I also conclude the function is not analytic? Because analytic functions should be path independent?
Thanks

 

[Euge]

Hi Euge, first may I just check my understanding about the approach I used:
You may have misread, I actually have the real part as exact, IE $ (xdx + ydy); so {M}_{y}=0={N}_{x} $, I think that's right?

Yes, the Imag part$ (xdy−ydx) $ is not exact; so I integrated it to get $ i(xy−yx) $, which cancels to 0 doesn't it?

I did see that you said the real part is exact, but based off of the presentation of your calculations, it appeared that you used the incorrect assertion that $d(xy - yx) = x\, dy - y\, dx$; otherwise, I cannot see how you obtained $\int_c x\, dy - y\, dx = (xy - yx)\bigg|_{c}$.

Now onto parameterization, I am used to param'ing when it makes the eqtn easier (like polar/trig etc), but given that the path C is here 2 straight line segments, I would normally be quite happy to stay with rectilinear cords - so for next time, what should have warned me to parametise?

Contour integrals are defined using parametrization of curves. That would be a warning. :D

This next is quite new to me, please review?

There are 2 paths segments, call them C₁ and C₂.
First I'll look at moving 1 to the right, then 1 up:
C₁ is along the real axis, so $ {z}^{*}(t) =t, 0 \le t \le 1, dz = dt $
C₂ is from (1,0) to (1,1), so $ {z}^{*}(t) =1+it, 0 \le t \le 1, dz = idt $
Then $ \int_{0}^{1} t\,dt + \int_{0}^{1} (1+it)\,idt = \frac{1}{2}{t}^{2}|^1_{0} + (i-\frac{1}{2}{t}^{2}) |^1_{0} = \frac{1}{2} + i - \frac{1}{2} - i = 0 $

Similarly for the alternate path, up first, then to the right:
C₁: $ {z}^{*}(t) =it, 0 \le t \le 1, dz = idt $
C₂: $ {z}^{*}(t) =i+t, 0 \le t \le 1, dz = dt $
$ \therefore \int_{0}^{1} it\,idt + \int_{0}^{1} (i+t)\,dt = -\frac{1}{2}{t}^{2}|^1_{0} + (it + \frac{1}{2}{t}^{2}) |^1_{0} = \frac{1}{2} + i - \frac{1}{2} = i $
And as they are not equal, $ {z}^{*} $ is path dependent. YAY!

Can I also conclude the function is not analytic? Because analytic functions should be path independent?
Thanks

The parametrization of $C_2$ is incorrect. It should read $z(t) = 1 + it$, $0 \le t \le 1$. Then $z^*(t) = 1 - it$, $0 \le t \le 1$. Also, one does not say that a complex function $f$ is path dependent, but rather a contour integral $\int_{\Gamma} f(z)\, dz$ is path dependent. However, one can say that $f(z) = z^*$ is not analytic since $\int_{(0,0)}^{(1,1)} z^*\, dz$ is path dependent.

 

[ognik]

Thanks Euge - I really value your input in terms of correcting my terminology and notation etc. I recently read an illuminating article on 'mathematical maturity' - I must be pre-teen

My original approach stuff again:

My Im. part: $ \int xdy $: I treat x as a constant w.r.t. y, so $ \int xdy = xy$? Similarly, $ \int y dx = yx $...

"Contour integrals are defined using parametrization of curves" - not something my book tried hard to mention - so please continue to assume I might not know what is otherwise obvious, thanks.
Does this apply also to real line/path integrals?
---------------
I see I mixed up z & z* for both paths, now I get $ \int {z}^{*}dz $ = 2 for the 1st path, and =1 for the 2nd. Therefore the line integral of the function is path dependant (I can be taught :) )

 

[Euge]

My Im. part: $ \int xdy $: I treat x as a constant w.r.t. y, so $ \int xdy = xy$? Similarly, $ \int y dx = yx $...

How do you know $x$ is a constant with respect to $y$? For instance, if the contour of integration is the circle $x^2 + y^2 = 1$, then $x$ depends on $y$; $x = \sqrt{1 - y^2}$ if $x \ge 0$ and $x = -\sqrt{1 - y^2}$ is $x \le 0$.

"Contour integrals are defined using parametrization of curves" - not something my book tried hard to mention - so please continue to assume I might not know what is otherwise obvious, thanks.
Does this apply also to real line/path integrals?

I advise you to review the concept of line integrals in the plane - I recommend Thomas's Calculus. Contour integrals are defined similarly to line integrals. Let $\Omega$ be a domain in $\Bbb C$; let $f : \Omega \to \Bbb C$ be a continuous function. Suppose $\gamma$ be a piecewise-differentiable curve on $\Omega$. The integral $\int_\gamma f(z)\, dz$ is defined as $\int_0^1 f(c(t))\dot{c}(t)\, dt$, where $c(t)$, $0 \le t \le 1$, is a parametrization of $\gamma$.

 

[ognik]

How do you know $x$ is a constant with respect to $y$? For instance, if the contour of integration is the circle $x^2 + y^2 = 1$, then $x$ depends on $y$; $x = \sqrt{1 - y^2}$ if $x \ge 0$ and $x = -\sqrt{1 - y^2}$ is $x \le 0$.

Good point, but with the 2 rectilinear path segments given in this problem, either x or y are constants. EX: Going to the right 1st, we'd have y=0, then x=1 - so aren't x,y independent of each other for that path?

I advise you to review the concept of line integrals in the plane - I recommend Thomas's Calculus. Contour integrals are defined similarly to line integrals. Let $\Omega$ be a domain in $\Bbb C$; let $f : \Omega \to \Bbb C$ be a continuous function. Suppose $\gamma$ be a piecewise-differentiable curve on $\Omega$. The integral $\int_\gamma f(z)\, dz$ is defined as $\int_0^1 f(c(t))\dot{c}(t)\, dt$, where $c(t)$, $0 \le t \le 1$, is a parametrization of $\gamma$.

I'll have a look for that, thanks. I have just browsed through the line integrals section of my book, although it has some examples using param's, it doesn't say you must use params... Would you say it is a must use, or are there situations when staying with x, y is OK?

 

[Euge]

Good point, but with the 2 rectilinear path segments given in this problem, either x or y are constants. EX: Going to the right 1st, we'd have y=0, then x=1 - so aren't x,y independent of each other for that path?

Wait, going to the right 1st, we'd have $y = 0$ and $x = t$, $0 \le t \le 1$, right? Anyway, I assumed you meant $\int x\, dy = xy$ and $\int y\, dx = yx$ for all curves $C$, which wouldn't make sense.

I'll have a look for that, thanks. I have just browsed through the line integrals section of my book, although it has some examples using param's, it doesn't say you must use params... Would you say it is a must use, or are there situations when staying with x, y is OK?

If you have a definition of the line integral in the plane which does not use parametrization, please post it here, and I'll make some comments. However, based on the way the question is posed, it seems like you're picking up line integration from the examples without checking the original formulation.

 

[ognik]

Hi - I am happy with your judgement - kindof made up my mind to use params. in future, but still curious as to whether x,y cords are usable?

I've attached an extract at the beginning of the section on line integrals...
Thanks again

View attachment 4608

 

[Euge]

Hi - I am happy with your judgement - kindof made up my mind to use params. in future, but still curious as to whether x,y cords are usable?

I've attached an extract at the beginning of the section on line integrals...
Thanks again

Notice that in the extract, parametrizations are used to compute the integrals in the examples. It sounds to me that you're suggesting there is an alternative to define line integrals without parametrizations, even implicitly; I was just curious to know, from your knowledge, how you would do that.

 

[ognik]

As I read it, example 1.9.2 uses polar params for a circular path, which I was always happy with. But then they use the same function with a rectangular path - and stay with x,y?

Also, example 1.9.1 doesn't use params?

At the start, the book shows: $ W = \int F · dr = \int F_x(x, y, z) dx+\int F_y(x, y, z) dy+\int F_z(x, y, z) dz $, which - using components, is how I would do a line integral in rectangular cords...

 

[Euge]

As I read it, example 1.9.2 uses polar params for a circular path, which I was always happy with. But then they use the same function with a rectangular path - and stay with x,y?

They stay with $x$ and $y$, but it's a parametrization. Consider the straight line segment from $(0,0)$ to $(1,0)$. Every point on the segment has $y = 0$ and $0 \le x \le 1$. So you may use $x$ as a parameter variable.

Also, example 1.9.1 doesn't use params?

Read where is says "The first integral cannot be evaluated until..." and where it says "If we select the path...". These are important, for it indicates the use of parametrization.

At the start, the book shows: $ W = \int F · dr = \int F_x(x, y, z) dx+\int F_y(x, y, z) dy+\int F_z(x, y, z) dz $, which - using components, is how I would do a line integral in rectangular cords...

Then how would you compute e.g. $\int F_x(x,y,z)\, dx$ if your curve is implicitly defined and neither $y$ nor $z$ can be written explicitly in terms of $x$?

 

[ognik]

They stay with $x$ and $y$, but it's a parametrization. Consider the straight line segment from $(0,0)$ to $(1,0)$. Every point on the segment has $y = 0$ and $0 \le x \le 1$. So you may use $x$ as a parameter variable.

Gotcha. I agree that when I used t, it was easier to see.

Then how would you compute e.g. $\int F_x(x,y,z)\, dx$ if your curve is implicitly defined and neither $y$ nor $z$ can be written explicitly in terms of $x$?

I would parameterize ;)
I get that I should parameterize, just trying to put to rest whatever was in my head about staying with x,y; even if not advisable, if I had an equation Y(x)=blah, for a rectangular path, it is possible to get the right answer without parameterizing ...although x is also a parameter here isn't it?

 

[Euge]

it is possible to get the right answer without parameterizing ...although x is also a parameter here isn't it?

That sounds confusing (getting the answer without parametrizing but having $x$ as a parameter). Here's the thing. If a curve $C$ is given by the graph of a continuous function $f(x)$ over an interval $[a,b]$, then you may "stay with $x$ and $y$" by letting $x$ be the parameter and writing $y = f(x)$, $a\le x \le b$. Then e.g. $\int_C y\, dx = \int_a^b f(x)\, dx$ and (if $f$ is differentiable) $\int_C x\, dy = \int_a^b xf'(x)\, dx$.

 

[ognik]

Thanks Euge, the 'x is also a parameter' was an afterthought, should have made that clear.

Is this right - Line integrals are always w.r.t. a path, and that path will always result in some form of parameterization?

Except if the function is conservative?

 

[Euge]

Maybe I've made this unclear, so let me try again. Line integrals are defined using parametrizations. In this way, you convert a line integral to a Riemann integral, the one you're used to.

You can avoid resorting to explicit use of parametrizations when dealing with line integrals of conservative vector fields, but that's because of the fundamental theorem of line integrals, whose proof uses parametrizations (to follow the definition of a line integral).

 

[ognik]

I guess the root cause is that I struggle with definitions when I don't understand what makes them true or necessary; not to mention the gaps in my 'maths for engineers' background. I think this is one I will just accept - before I drive us both nuts :-)