Homework Help: Use C-R eqns to determine points whose fn's are analytic

1. May 3, 2012

bugatti79

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

Use the Cauchy Riemann equations to those points whose functions are analytic

$f(z)=x^2-y^2-x+iy(2x+1)$

2. Relevant equations

C-R eqn's

$u_x=v_y, u_y=-v_x, z(x,y)=x+i y$

3. The attempt at a solution

$u(x,y)=x^2-y^2-x$
$v(x,y)=y(2x+1)$

$u_x=2x-1$
$u_y=-2y=-v_x$
$v_y=2x+1 \therefore u_x \ne v_y \implies$

the function f(z) is not analytic...Not sure what the question is asking regarding 'determining those points'

Can some shed some light if I need to do more?

Thanks

2. May 3, 2012

micromass

Surely there are some x and y such that

$$y(2x+1)=2x+1$$

These are the points in which the function is analytic. (don't forget to check the second Cauchy-Riemann equation)...

3. May 3, 2012

bugatti79

If y=1 then the equation holds.

Not sure I understand you at all. If the function doesn't satisfy the C-R equations then is it not analytic? How can one proceed?

Although I have noticed that $u_{xx}+u_{yy}=0$ if this is anything to do with it..?

4. May 3, 2012

jackmell

You need to distinguish analyticity and differentiability. A function is analytic at a point z0 if it's differentiable in a neighborhood of that point and it's differentiable at z0 if it satisfies the CR equation and the partials are continuous. Yours don't so it's not differentiable anywhere. Also, can check if it's a function of $\overline{z}$ and if
$$\frac{df}{d\overline{z}}\neq 0$$
then it's not differentiable. Note your function equals $z^2-\overline{z}$.

5. May 4, 2012

bugatti79

So we have established that this function is not analytic because it is not complex differentiable since it does not satisfy the C-R equations, hence we cannot determine those points? Right?

Is this interpretation correct?

$\displaystyle \frac{df}{dz}=\frac{d}{dz}(z^2-\overline{z})= 2z$ where $\frac{d}{d\overline{z} }(-\overline{z})$ does not exist..right?

What about this final one $h(z)=\cos(2x)\cosh(2y)-i \sin(2x)\sinh(2y)$

This is not analytic either since it does not satisfy the C-R equations, ie

$u_y=-\frac{1}{2} \cos(2x)\sinh(2y) \ne -v_x$...?

6. May 4, 2012

jackmell

I probably should not have brought up the conjugate thing. That just makes is more confussing. It's also confusing to me too. If the derivative exists, then you can use the ordinary rules of differentiation and conclude:
$$\frac{d}{dz} z^2=2z$$

but the derivative of $z^2-\overline{z}$ does not exists so you can't just differentiate it with respect to z and conclude that is the complex derivative. Probably best for now to just focus on f=u+iv and work through the CR equations without considering if f can be represented in terms of z and it's conjugate.

I think this one satisfies the CR equations. Ain't that just cos(2z)?

7. May 4, 2012

bugatti79

Sorry, I had a brain cramp assuming the derivative of cosh is -sinh.

Ok, this function satisfies the C-R hence it is complex differentiable and hence it is analytic.
Not sure what to do next though regarding determining the 'points'....?

Thanks