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Complex, holomorphic and integration

  1. Oct 25, 2007 #1
    1.Represent the following rational functions as sums of elementary fractions and find the primitive functions (indefinite integrals);
    (a) f(z)= z-2/z^2 +1


    2.Now, I am pretty sure that I have to split up this equation into real and imaginary parts in order to show that it is holomorphic. Once I have show that, I can use that fact to find a g'(z)=f(z)


    3. The attempt at a solution:

    So I tried f(z)=f(x+iy)= x+iy-2/(x+iy)^2 +1
    u= x-2/(x+iy)^2 +1 and v= y/(x+iy)^2 +1

    but by Cauchy-Reimann I am not getting du/dx=dv/dy, even though I know I should.
    du/dx= -x^2 -y^2 +4x+1+4iy and
    dv/dy= x^2 + y^2 +1

    so I got frustrated, and attempted to do the integral anyways, first by substitution, then by parts;
    This is my best attempt so far:

    Int[z/z^2+1] + Int[-2/z^2+1]
    Int[z/z^2+1]=1/2ln|z^2+1|
    Int[-2/z^2+1]= Int[-2/(z+i)(z-i)]
    I dont feel like listing everything, but by partial fractions:
    Int[-2/z^2+1]= Int[i/z-i] + Int[-i/z-i]= iln(z-i) - iln(z+i)

    Final answer: 1/2ln|z^2+1|+ iln(z-i) - iln(z+i) +c

    I would like to know if my answer makes even remote sense, and if you guys can figure out why my cauchy-reimann equation isnt working, that would be awesome too.
     
  2. jcsd
  3. Oct 25, 2007 #2

    Dick

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    Your integral looks reasonable. The reason Cauchy-Riemann isn't working is that if you are writing f(x+iy)=u(x,y)+v(x,y)*i, u and v should be real functions. You haven't gotten rid of all of the i's in them.
     
  4. Oct 25, 2007 #3
    i'm not sure what constitutes a elementary function (i always thought rational functions were elementary!).

    But the reason your not getting the cauchy reimann equations satisfied, well look at how you split up u(x,y) and v(x,y)

    u= x-2/(x+iy)^2 +1 and v= y/(x+iy)^2 +1

    Notice something wrong? Remember that the C-R equations work when u and v are of the form f(x,y)=u(x,y)+iv(x,y), where u and v are REAL functions of two variables. Your u and v above have complex numbers in them!

    Where is the function:

    g(z) = 1/2ln|z^2+1|+ iln(z-i) - iln(z+i) +c

    differentiable? We know that wherever it IS differentiable, it's derivative is your original function - but the problem is asking for the subset of the plane under which this is true.
     
  5. Oct 25, 2007 #4
    Thanks Dick, thats such a morale boost!

    So I'm trying to remove the i's:
    I multiplied out x+iy-2/x^2 - y^2 + 2ixy +1
    I'm pretty sure I cant do u=x-2/x^2 - y^2 +1 and v=y/2xy
    I am honestly out of ideas :(
     
  6. Oct 25, 2007 #5

    Dick

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    If you have a fraction like c/(a+bi), to make the denominator real, multiply top and bottom by (a-bi). That will put all of the i's in the numerator where you can collect them.
     
  7. Oct 25, 2007 #6
    Just for fun, Maple gave me 2/z+ln(z)+z for that integral.
    How it got there, I have no idea, but it looks much cleaner :)
     
  8. Oct 25, 2007 #7
    omg, I cant believe I forgot the conjugate... thanks again
     
  9. Oct 25, 2007 #8
    Ok, so that took forever, but it pretty much worked, thanks for the help guys!
     
  10. Oct 25, 2007 #9

    Dick

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    If you differentiate that, do you get what you started with? I don't think so. I think you typed something wrong into Maple.
     
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