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Homework Help: Evaluate complex value and domain of analytic function

  1. Aug 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Can anyone help me to check whether the first two are correct or not?
    For the third question, i may need some help on it.

    Evaluate the following if they exist:
    a) sin(3Logi)
    b) (1+i)^(1/3-i) (principal value)

    Determine the largest open set in which the function Log(1-z^n) is analytic. Here n is a positive integer.

    Any help is appreciated!

    2. Relevant equations
    Log(z) = Log(abs(z)) + iArg(z)
    z^a=exp(a*Log(z)) a, z[itex]\inℂ[/itex]

    3. The attempt at a solution
    a) sin(3*Log(i))
    = sin[3(Log(abs(i)) + iArg(i))]
    = sin[3(Log(1)+i(pi/2))]
    = sin[3(0+i*pi/2)]
    = sin[i*3pi/2]
    = -i*sinh[i*(i*3pi/2)]
    = -i*sinh[-3pi/2]
    = i*sinh[3pi/2]

    b)(1+i)^(1/3-i) principal value
    = exp[(1/3-i)*Log(1+i)]
    = exp[(1/3-i)*(Log(abs(1+i))+iArg(1+i))]
    = exp[(1/3-i)*(Log(sqrt(2))+i(pi/4))]

    Third question:
    Let z=x+iy.
    First, I think i need to solve Re{1-z^n}≤0 and Im{1-z^n}=0 for x, y, then the largest domain is the ℂ except the set {z=x+iy| Re{1-z^n}≤0 and Im{1-z^n}=0}.

    → 1-(x+iy)^n =0, then i can't anymore further. I tried to use binomial theorem on (x+iy)^n but just can't do it.
  2. jcsd
  3. Aug 20, 2013 #2


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    Science Advisor

    Okay, since you are asking for the principal value you don't need to do things like "plus a multiple of [itex]i\pi[/itex].

    Yes, that looks correct. Personally, I would have used
    [tex]sin(x)= \frac{e^{ix}- e^{-ix}}{2i}[/tex]
    but, of course, that is the same as your sinh(x).

    I think you can do a little better. first, of course Log(sqrt(2))= (1/2)Log(2).
    Also you can multiply out in the exponential: {(1/6)Log(2)- (pi/4)}+ {(1/3)pi/4- (1/2)Log(2)}i.
    And then use the fact that exp(a+ bi)= exp(a)(cos(b)+ isin(b)).

    You are making a mistake in trying to deal with this in "Cartesian form". [itex]ln(z)= ln(re^{i\theta}= ln(r)+ i\theta[/itex], as you said before, is analytic as long as r> 0. That means this function is analytic as long as |1- z|> 0 which is exactly the same as saying [itex]z\ne 1[/itex].
  4. Aug 21, 2013 #3
    HallsofIvy, so the answer is just z≠1??
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