Homework Help: Complex analysis proof- HELP

1. Apr 29, 2010

g1990

1. The problem statement, all variables and given/known data
f(z) is a complex function (not necessarily analytic) on a domain D in C. The directional derivative is Dwf(z0)=lim(t->0) (f(z0+tw)-f(z0))/t, where w is a unit directional vector in C. There are three parts to the question:
a. Give an example of a function that is not differentiable at any point but that has directional derivatives is very direction w at every point z0. Very that your example satisfies the required properties
b. If f is differentiable at z0, show that there is a constant c such that Dwf(z0)=cw for every w
c.Assume that the real and imaginary parts of f has continuous partials. Show that if there exists a constant c such that Dwf(z0) exists and equals cw for every w, then f is differentiable at zo.

2. Relevant equations
Cauchy Riemann: du/dx=dv/dy and du/dy=-dv/dx

3. The attempt at a solution
for a, I know that my function can't satisfy the CR eqns, b/c it can't be differentiable, but I don't know how to plug a specific function into my directional derivative definition

2. Apr 29, 2010

Dick

Ok, here's a practice function. Suppose f(x+iy)=x. What's the directional derivative of f at z0=0 in the direction w=1+0i? How about in the direction w=0+1i? Just plug z0 and w into your definition of directional derivative.

3. Apr 29, 2010

g1990

the first one would be 1 and the second one zero.

So in general, for this function, Dwf(z0)=f(z0+t(u+iv))-f(z0)/t=(Re(z)+ut-Re(z))/t=u
where u+iv is the unit directional vector. So that would be an example of a function that is not differentiable anywhere but which has directional derivatives everywhere, right?

4. Apr 29, 2010

Dick

Sure. f(x+iy)=x doesn't satisfy Cauchy-Riemann, even though it has directional derivatives everywhere.

5. Apr 29, 2010

g1990

wow thanks! for part b, what I was thinking is that if f is differentiable at some z0, any two given directional derivatives must be equal. I'm not sure how to spin that into there exists a c such that every directional derivative equals cw though. One thing is that c should be the gradient at z0, but we haven't officially learned that yet, so I shouldn't have to use it.

6. Apr 29, 2010

Dick

Well, no, any two directional derivatives won't be equal. If Dwf=cw then it depends on w. This isn't the complex derivative. Try expressing the directional derivative in terms of the partial derivatives of f=u+iv. That way you can use Cauchy-Riemann.

7. Apr 29, 2010

g1990

Okay, so if f(x,y)=u(x,y)+iv(x,y) and w=w1+iw2 is the unit directional number, then
the directional derivative is
lim(t->0)[u(x+tw1,y+tw2)-u(x,y)]/t+[v(x+tw1,y+tw2)-v(x,y)]/t
I'm not sure how to spit that up even more to extract the partials, or to get w out of the equation.

8. Apr 29, 2010

Dick

du(x,y)/dt=du(x,y)/dx*dx/dt+du(x,y)/dy*dy/dt. That's basically the gradient formula. Some of those 'd's should be partial. And don't forget the 'i' in front of v(x,y).

9. Apr 29, 2010

g1990

okay so, let me get this straight. I have that the directional derivative equals
du(x,y)/dt+idv(x,y)/dt=du(x,y)/dx*dx/dt+du(x,y)/dy*dy/dt+i(dv(x,y)/dx*dx/dt+dv(x,y)/dy*dy/dt). Now, I know that du(x,y)/dx is just du/dx, one of the terms of the CR eqns, and likewise for the other three. I suspect that dx/dt=w1 and so on, but I'm not sure why. Also, should I be putting another i in front of all of the dy/dt's?

10. Apr 29, 2010

g1990

11. Apr 30, 2010

Dick

You are giving up too quickly. You are almost there. Sure dx/dt=w1, x(t)=x0+tw1, dx(t)/dt=w1. No, you don't need any more i's. Why would you? Now just use Cauchy-Riemann. Put c1=du/dx=dv/dy and c2=du/dy=(-dv/dx). Can you factor (w1+iw2) out?

12. Apr 30, 2010

g1990

okay so- I have that dx/dt=w1 and dy/dt=w2.
Then,
du(x,y)/dt+dv(x,y)/dt=du(x,y)/dx*dx/dt+du(x,y)/dy*dy/dt+i(dv(x,y)/dx*dx/dt+dv(x,y)/dy*dy/dt)=du/dx*w1+du/dy*w2+idv/dx*w1+idv/dy*w2=(du/dx+idv/dx)w1+(-idu/dy+dv/dy)iw2=(c1+ic2)w1+(c1+ic2)iw2=(c1+ic2)(w1+iw2)

so my constant is just c1+ic2=du/dx+idv/dx. That seems right to me- is that what you meant?

13. Apr 30, 2010

Dick

Yes, the constant c1+ic2 is the complex derivative.