Question

ok, complex analysis problem....How do you show that the integral of z dz (z =x+iy) statisfies the Cauchy-Reimann equatioins?

Answer 1

The cauchy riemann equations says given z=u(x,y)+iv(x,y).
\[\frac{\delta u}{\delta x}=\frac{\delta v}{\delta y}\]
And:
\[\frac{\delta u}{\delta y}=-\frac{\delta v}{\delta x}\]
Tells you if the function is analytic. As far as using:
\[\int\limits_C z dz\]
I'm not sure.

Answer 2

yeh, I know the formulas, but wondered how I could prove they satisfied them...

Answer 3

cruffo, any idea?

Answer 4

I had to pull out the ol' book for this one. give me a sec to refresh :)

Answer 5

I've never taken complex analysis. This is just what I know.

Answer 6

thanks anyway :-)

Answer 7

this may help:
\[\int z dz = \int(x+iy)(dx+idy)\]
breaking up into real and imaginary parts
\[= \int(xdx-ydy) + i\int (ydx + xdy)\]

Answer 8

ok, I have that first eqn. in my notes....ok, the second step is some of what I needed....

Answer 9

But with that, what would du/dx be and dv/dx be?

Answer 10

right....
f(z) = u(z) +iv(z).
does it make since if
\[u(z) = \int (xdx-ydy)\]
Then use something similar to the fund. Thm. of Calc.?

Answer 11

whoa...now where did you get eqn 2 from?? Did you distribute the ( ) and then separate them?

Answer 12

Yah, just distribute (x+iy)(dx +idy) then seperate real from imaginary parts.

Answer 13

tes, I see how you sub in u(z) = the \[\int\limits_{?}^{?}\] (x dx -y dy)

Answer 14

I watched my friend derive this for me once :D

Answer 15

but for the C-R eqns check, I need du/dx and dv/dy wherer do I get them from?

Answer 16

Can't you rewrite the integral using green's theorem and set the real and imaginary parts equal? O.O

Answer 17

oh, wait...I have z = e^it .............I have solved the integral to =0, but the solution says " it is 0 by the Caucy-Gourset theorm (contour is closed and integrand function f(z)=?? can't read it, maybe = z? and is analytica aty all points interior to C."

Answer 18

I have the derivation in my notes....just like you showed (finally found it) but ....with tha above written as the solution, maybe I don't NEED tp show how the C_R eqns are met??

Answer 19

I see Green's Theorem in here, and it connects to ...oh. oh, see what he did...he DID use Green's Th. to show that the integral meets the C-R eqns.. :-)

Answer 20

So using the derivation and going the long way and applying G's Th to it shows them meeting the C-R's

Answer 21

and what IS the Cauchy-Goursat the solution refers to??

Answer 22

French mathematician E. Goursat (ca 1900) gave a proof of the Integral Theorem, which requires only that f'(z) exists, but not analyticity.

Answer 23

so why does the solution say "the integral of z dz is 0 by the Cauchy-Gouset Theroem"? I found that the integral was = 0 by using the z = e^it he gave us and integrating between 0 and 2pi. (using the polare representation near the end.....how is that using the C-G Th??

Answer 24

I just remember Green's theorem. I can't wait to take complex analysis :DDDD

Answer 25

it's loads of fun! :-(

Answer 26

does satisfying the C-R Th MEAN it is analytic??

Answer 27

Many authors do not require continuity of partial derivatives, but then they can't use Green's Thm in proving the Cauchy Integral Thm. Instead they use a version of the Cauchy-Gouset Thm. That may be what the solution is referring to, but it doesn't mater in this case as the partials are continuous.

Answer 28

If f is analytic then it satisfies the C-R equations and conversely.
So yah, C-R iff analytic

Answer 29

Have soemthing else in my notes....f(z) = integral from 0 to 2pi [u dx/dt - v dy/dt] + i[u dy/dt + v dx/dt] dt. Is this similar to what you wrote originally? the integral of (x + iy) (dx + dy) etc???

Answer 30

so to know it's analytic (as mentioned in the solution, then I have to show it satisfies the C-R Th.....right?

Answer 31

yes, just assuming a more general f(z) = u(z) + iv(z) instead of just f(z) = z

Answer 32

oh, that's why they are different.....so If I want to sub in Greens' Th I should use this last I typed...

Answer 33

from what would the " integral from 0 to 2pi [u dx/dt - v dy/dt] + i[u dy/dt + v dx/dt] dt." come from? In othere words, what would be the original statement?

Answer 34

Yah, you may want to check out the proof at http://en.wikipedia.org/wiki/Cauchy%27s_integral_theorem

You're question seems to be a specific example of that proof.

Answer 35

the integral of [u + iv] [dx/dt i (dy/dt)] dt ?

Answer 36

and this from oh, I see it, just the integral of f(z) dz :-)
Thanks, I'll check out that proof.

Answer 37

seems like the "dt" part is just specifying that x and y may be parametric equations of the variable t. Should just "cancel out" to give
(u +iv)(udy +ivdx). Not sure, the book I'm looking at doesn't use that notation.

Answer 38

ok, and now I have distributed that

Answer 39

and it looks like what I have to sub in green's Th. for....

Answer 40

Ok, I found that
\[\int_\Gamma (udx +vdy) = \int_a^b \left(u \frac{dx}{dt} + v \frac{dy}{dt} \right)dt\]
where Gamma is any piecewise-smooth curve in R2 of the form (x,y) = (x(t),y(t)), and t varies between a and b.

That may be what you have referenced in your notes.

Answer 41

yes, have soemthing like that! :-) thanks!

Answer 42

you will do very well when you take this course!!!! It's just been 30+ years since I have integrated and worked with partial integrations.....

Answer 43

I took complex variables in summer 2006, so it's been about 5 years! I should probably audit it - I had no idea what I was doing at the time!

Answer 44

and it seems I need these much higher math classes than I took for my secondary math educ. major in the '70's just so I can teach the equivalent of high school geometry in college to new education majors....(I've been doing it for 2 yrs now and algebra before that for 6), but the university has decided ANY education major needs MORE higher math to be legit.....go figure after teaching for 30+ years of teaching the high school algebra and geomtry !!!!

Answer 45

my SON will be taking some of these classes soon for his math minor.....geesh!!

Answer 46

Yep, the community colleges in my area requires masters in something (doesn't matter what) and 18 hours of grad math courses. The local 4-year university requires the equivalent of a masters in math to teach any college level math courses.

Answer 47

yeh, our college got snagged by the Southeren acredation folks and are forcing the issue...AND the 9 hrs of grad math for my masters aren't even going to count!!! I had 30 hrs of math just for my BA degree!

Answer 48

ohhh.... that's sad! Are you going for a masters in math ed?

Answer 49

I was thinking that this approach may be cheating but I'm sure you've noticed that f(z) = z is the derivative of F(z) = (1/2) z^2. Then F is analytic on all C it is certainly analytic on any path (so should satisfy C-R equations). But to go the long way, if we write
\[F(x,y) = (x+iy)^2 = (x^2 - y^2) + i(2xy) = u + iv.\]
Then
\[u_x = 2x = v_y \quad \text{ and} \quad u_y = -2y = -v_x\]
Showing that
\[\int_\Gamma z \;dz\]
satisfies the C-R equations.

But like I said, that may not be what the problem want you to do.

Answer 50

Of course, I forgot the 1/2 out in front :)

Answer 51

I already HAVE my masters in math Education, but that's not good enough to teach the EXACT same thing in college! Another lady is in the same predicament...She taught AP stats in High school but has to take these extra classess also in order to teach regular stats in college!

Answer 52

Weird. I work in Texas, and there is a big push for people to receive MS in Math Education. Many people with such a degree are teaching college math classes (including classes geared toward secondary math teachers). I even know people with a MA in Education, with a concentration in mathematics who are teaching at the community college level without restriction. I wonder if there's gonna be some changes made soon.