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Homework Help: Understanding Analyticity and Continuity in Complex Analysis

  1. Mar 25, 2012 #1
    1. The problem statement, all variables and given/known data

    Determine where the function f(x + iy) = 2sin(x) + iy^2 + 4(ix - y) is differentiable and where it is analytic.

    3. The attempt at a solution

    f(x + iy) = 2sin(x) -4y + i(y^2 +4x)

    Through C-R equations:

    du/dx = 2 cos x
    dv/dy = 2y

    du/dy = -4
    dv/dx = 4

    So the C-R equations hold only if y = cos (x)

    1) Supposedly these partial derivatives are continuous everywhere, I do not understand what this exactly means, what is an example of a partial derivative that is not continuous everywhere?

    Also it says f is differentiable only on the curve y = cos (x), but it's not analytic since it's not differentiable in the e-neighbourhood at a point on the curve.

    2) I do know that for a function to be analytic it must be differentiable at a given point, and also in the e-neighbourhood of it, but how is this exactly determined, and what does it mean to be differentiable in an e-neighbourhood of a point in this curve specifically?

    If I draw out y = cos (x), pick out a point and look at the e-neighbourhood around this point, all you get is the point on that line, a portion of the line going through, and nothing else around it.
  2. jcsd
  3. Mar 25, 2012 #2
    You need to look at [itex]f[/itex] as a function of the complex variable [itex]z=x+iy[/itex]. We then write [itex]f(z)=f(x+iy)=u(x,y)+iv(x,y)[/itex]. Now, a necessary and sufficient condition for [itex]f'(z)[/itex] to exist is that u and v satisfy the CR equations and that the CR partials are continuous. You know a function is continuous at a point x if f(x) exist, the limit there exist and the limit is equal to f(x). Surely those CR partials you computed, they too are just functions right, meet those requirements. They are continuous everywhere in fact unlike the partial with respect to x of say for example, [itex]f(x)=\sqrt{x}[/itex] which is not continuous at the origin. And the partials you computed satisfy the CR equations when [itex]y=\cos(x)[/itex]. Again, you need to consider all of this in terms of a complex function over the z-plane with


    Each u and v are surfaces of course and so from what you computed above, f(z) is differentiable only along the line [itex]y=\cos(x)[/itex] in the complex plane. You could if you wanted draw u(x,y) and v(x,y) over the x-y- plane, the line [itex]y=\cos(x)[/itex] in the x-y plane, and even traces over u and v directly above the line [itex]y=\cos(x)[/itex]. Make them yellow. Now those yellow traces over u and v represent the only spots on those two surfaces, and therefore the only z=x+iy locations of f(z), where the function f=u+iv is differentialbe. However by definition, a function is analytic only if it is differentialbe in a [itex]\delta[/itex]-neighborhood of a point and a [itex]\delta[/itex]-neighborhood is a circle around a point. Well, since f is only differentialbe along a line, it of course isn't differentiable everywhere in some circular neighborhood, however small the circle, around any of those points along the line so it's not analytic anywhere.
    Last edited: Mar 25, 2012
  4. Mar 25, 2012 #3
    Thanks for your detailed response, I'm starting to understand it now.

    You said:
    What do you mean by the limit there exists and the limit is equal to f(x)?

    Suppose we take y = cos (x), O.K, at a given point x; f(x) exists. But I don't get what you mean by the limit existing, and also that the limit is equal to f(x)?

    I plotted the surfaces and the line y = cos (x), it's a bit tricky to see what's happening, but do you mean it's differentiable where there is an intersection between the two surfaces and the curve y = cos (x)?

    How would one do this without visualising? This question has a fairly easy surface and line to visualise, but some other questions are well beyond me attempting to visualise them in an exam, mathematically how would one determine this?
  5. Mar 26, 2012 #4
    That's the definition of continuity. Take x=1. Then [itex]\lim_{x\to 1} \cos(x)=\cos(1)[/itex]. So, the function is defined at x=1, it's limit there exists and that limit is equal to the function at x=1. Ergo, it's continuous there.

    Ok, maybe that wasn't the best thing to tell you. Forget about drawing the real and imaginary parts of the function for now. Rather, just remember if the partials of the function satisfy the CR equations and those partials are continuous where they satisfy the CR equations, then the function is differentiable where the partials satisfy the CR equations. In your case, the partials satisfy the CR equations whenever [itex]z(t)=t+i\cos(t)[/itex]. That's the line in the complex z-plane equiv. to y=cos(x) for z=x+iy. Therefore, the function is differentiable on that line. However since analyticity is defined for a region, a circular region about a point, there exists no circular region about any of the points along the line, no matter how small the region, where your function is differentiable throughout the region. Therefore, the function is not analytic anywhere.
  6. Mar 26, 2012 #5
    Yes, it makes sense now, hopefully this will stick. Thanks so much!
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