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Copmlex Analysis

  1. Mar 18, 2009 #1
    (i) can someone explain why sin z isn't bounded if z is complex

    (ii) also can someone explain how branch cuts work and what their use is? i've been sitting in tutorials wondering what's going on for the last few weeks!

    (iii) does [itex]e^z[/itex] have a branch cut?
  2. jcsd
  3. Mar 18, 2009 #2


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    For the first one, work out what sin(x+iy) is in terms of real functions of x and y. Branch cuts are ways of restricting the selecting part of the complex plane so that a 'function' that would be multiple valued (like sqrt(z), if z^2=y then also (-z)^2=y) can be restricted to a single value (a branch). Is e^z one of those cases where there are multiple values of y such that y=e^z?
  4. Mar 18, 2009 #3
    (i) [itex]\sin(x+iy)=i[\sin{x} \cosh{y} + \cos{x} \sinh{y}][/itex] which diverges due to the hyperbolic trig functions and hence can't be bounded-is that ok?

    (ii)so if we have some f(z) that is multivalued i.e. there is more than one z that produces the same value of f(z) then is it true that these different values of z are the different "branches"?
    and so we restrict the copmlex plane in such a way so that we only have one value of z that corresponds to that f(z)-this is essentially cutting out the branches we don't want -are these true - leaving us with the "principal" branch?

    (iii)in my notes, when it starts talking about this it first discusses how we get a discontinuity in arg(z) on the negative real axis - what is the connection between this and the branches?

    (iv)for [itex]z=|z|e^{i \theta}[/itex], when we choose the principal argument is that essentialy a branch cut?

    (v)for the example you give,
    [itex]f(z)=|z|^{\frac{1}{2}}e^{i \frac{\theta}{2}}[/itex] and so for [itex]\theta \rightarrow \theta + 4k \pi , k \in \mathbb{Z}[/itex], we get different branches - we make a branch cut by restricting the argument to a value in the interval [itex] \left[-\pi,\pi \right] [/itex] - is that true?

    (vi)[itex]e^z=e^{x+iy}=e^x(\cos{y}=i \sin{y})[/itex] - is it true that [itex]y \rightarrow y + 2 \pi[/itex] would give the same value and so this is a multivalued function and therefore has multiple branches?
  5. Mar 18, 2009 #4


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    i) your expansion sin(z) is a bit off. cos(iy)=cosh(y), not i*cosh(y), but your conclusion is correct. And yes, a branch cut lets you define a single valued function from a multiple values 'function', by selecting a uniform choice of one of the values. For the rest of your questions, you seem to be confusing exp and log. It's true that exp(z)=exp(z+2*pi*i). But that doesn't mean exp(z) is multivalued, it means that its inverse, log, is multivalued. log needs a branch cut to be well defined, not exp(z).
  6. Mar 18, 2009 #5
    why does it need a branch cut to be well defined?

    also can u explain point (iii) in post 3?
  7. Mar 18, 2009 #6


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    Because I could define log(1) to be either 0 or 2*pi*i among other choices. Defining log(|z|*exp(i*arg(z))=log(|z|)+i*arg(z). The first case corresponds to picking arg(1)=0 and the second to arg(1)=2*pi. They are both equally valid choices. If I extend the arg continuously in the first case, when I approach the negative real axis by circling ABOVE the origin I wind up near arg(z)=pi. When I circle BELOW I wind up near arg(z)=-pi. They don't match. So I cut out the negative real axis from the domain and don't worry about it. Starting with 2*pi gives me another branch of the log.
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