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Math Help - Integrations - Complex

  1. #1
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    Integrations - Complex - Comment on my solutions!

    Hi everyone!

    This is my first post in this forum. I've only just picked up math again, after more then 10 years off the bench. Keep that in mind while you're digging through my solving. I'm studying complex analysis and I've just solved, or tried to solve some integration problems. I'd really like someone to look at my solutions, however typing in LaTeX was somewhat gnarly, but I think I got the essence down.

    \[ \oint_{\abs{\left|z\right|} < 1} \frac{1}{z^5 + 3z + 5}\,dz\]

    This one I used Rouche's to find out roots in the denominator. The root inside unit circle can be used to calculate the Residues. However, I couldn't find any zeros within the unit circle, with leads to Res f(z) = 0.


    \[\int_0^{\frac{\pi}{2}} {\frac{cos(2 \theta) } {1+2cos^2 \theta} }\,d\theta\]

    With this one I changed the limits to 0 -> 2\pi Then I used z = e^{i\theta} and 2\cos^2(\theta) = 1+\cos(2\theta). This with Eulers formula, specially Cosines. Gave me:

    \frac{1}{4i}\[ \oint_{\abs{\left|z\right|} < 1} {\frac{z^2 + 1 } {z(z^2+4z+1) }\,dz\]

    I found two simple poles inside the unit circle, z = 0 and z = 2 - 3^0.5 Now I could calculate the residus with Hostpital's

    \[ Res f(z) = \|{\frac{z^2 + 1 } {(3z_k^2+8z_k+1) }\|_ {_{z_k = 0}^{z_k=2-\sqrt{3}}}

    Inserting figures.

    \frac{(2-\sqrt3)^2 + 1 } {(3(2-\sqrt3)^2+8(2-\sqrt3)+1) } +1

    Wish lead to, with a little help for wolframalpha

    \frac{77 + 2 \sqrt3}{61}

    Consequentially

    \frac{2\pi i}{4i} \frac{77 + 2 \sqrt3}{61} = \pi \frac{77 + 2 \sqrt3}{122}

    The numbers feels a bit strange, but this is what I got.

    Please, feel free to comment!
    Last edited by liquidFuzz; September 14th 2010 at 06:03 AM.
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  2. #2
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    I think you have a mistake with your algebra at the beginning.
    I got \displaystyle I = \int_0^{\frac{\pi}{2}} {\frac{cos(2 \theta) } {1+2cos^2 \theta} }\,d\theta = \displaystyle i \int_{|z|=1} \frac{1}{z^2 + 4z + 1} which gives, eventually, I = \frac{ \pi}{2 \sqrt{3}}

    Also, regarding the first question - you can indeed show that the polynomial in your denominator has no zeroes in the unit circle using Rouche's theorem.
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  3. #3
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    OK... Thank you very much!

    I'll redo the rewriting of the integral. Two questions.

    1. \int_0^{\frac{\pi}{2}} f(z) = \frac{1}{4}\int_0^{2\pi} f(z) - Is this a proper way of changing the limits?

    2. About the Rouche stunt I was pulling. Is this a well mannered way of showing possible zeros of the polynom?
    z^5 + 3z + 5; f(z) = z^5, g(z) = 3z + 5
    Now using z = 1 - to show zeros within unit circle
    f(1) = 1, g(1) = 8
    | f(1) | \not> | g(1)|
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  4. #4
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    Quote Originally Posted by liquidFuzz View Post
    OK... Thank you very much!

    I'll redo the rewriting of the integral. Two questions.

    1. \int_0^{\frac{\pi}{2}} f(z) = \frac{1}{4}\int_0^{2\pi} f(z) - Is this a proper way of changing the limits?
    This is not always true. Take, for example, f(z) = sinz and then this is clearly not true.
    Here's what I did:

    \displaystyle I = \int_0^{\frac{\pi}{2}} {\frac{cos(2 \theta) } {1+2cos^2 \theta} }\,d\theta = \displaystyle \frac{1}{2}\int_{- \frac{ \pi}{2}}^{\frac{\pi}{2}} {\frac{cos(2 \theta) } {1+2cos^2 \theta} }\,d\theta

    Where the first equality is correct since the integrand is an even function (verify it).

    Now let  \phi = 2 \theta \Rightarrow d \theta = \frac{d \phi}{2} and you get, using 1 + 2cos^2( \theta) = 2 + cos(2 \theta) = 2 +  cos (\phi), that:
    \displaystyle I = \frac{1}{4} \int_{- \pi}^{\pi} {\frac{cos( \phi) } {2 + cos( \phi)} }\,d\phi \displaystyle = \frac{1}{4} \int_{- \pi}^{\pi} 1 - {\frac{2 } {2 + cos \phi} }\,d\phi

    And now I see that I have an algebra mistake too I missed a  2 \pi.

    Now you change variables and solve the integral over |z| =1 like you did before.

    2. About the Rouche stunt I was pulling. Is this a well mannered way of showing possible zeros of the polynom?
    z^5 + 3z + 5; f(z) = z^5, g(z) = 3z + 5
    Now using z = 1 - to show zeros within unit circle
    f(1) = 1, g(1) = 8
    | f(1) | \not> | g(1)|
    Not exactly. To use Rouche's theorem you have to have that |g| > |f| for all z such that |z|=1
    To show this:
    |f(z)| = |z^5| = |z|^5 = 1^5 = 1
    |g(z)| = |3z+5| \ge 2 (for |z|=1)
    And so |g(z)| > |f(z)| \forall z \in \mathbb{S}^1
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  5. #5
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    Thanks again!

    I know that cosine is an even function... How would I go about verifying that more than just mentioning it when I rewrite the boundaries? I missed the angle = double angle, but I used it to simplify the integral. Well, I changed the boundaries but I missed out on rewriting the angles. Your rewriting really help me, the count feel down in the slot and made a clear cling.

    Now I feel unsure of the Rouche thing... For instance, why do you show that |g(z)| = |3z+5| \ge 2 for |z|=1 why bigger or equally to 2 in particular?
    Last edited by liquidFuzz; September 14th 2010 at 10:55 AM.
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  6. #6
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    Quote Originally Posted by liquidFuzz View Post
    Thanks again!

    I know that cosine is an even function... How would I go about verifying that more than just mentioning it when I rewrite the boundaries? I missed the angle = double angle, but I used it to simplify the integral. Well, I changed the boundaries but I missed out on rewriting the angles. Your rewriting really help me, the count feel down in the slot and made a clear cling.
    Since cos \theta = cos( - \theta) \ \forall \theta \in \mathbb{R}, you get
    integrand = f( \theta) = \frac{cos(2 \theta)}{1 + cos^2 \theta} = \frac{cos(-2 \theta)}{1 + cos^2 ( - \theta)} = f( - \theta) \ \forall \theta \in \mathbb{R}, so it is even.

    Now I feel unsure of the Rouche thing... For instance, why do you show that |g(z)| = |3z+5| \ge 2 for |z|=1 why bigger or equally to 2 in particular?
    For the Rouche proof, I used the reverse triangle inequality - |x-y| \ge |x| - |y|:
    |5 + 3z| = |5 - (-3z)| \ge |5| - |-3z| = 5 -3|z| = 5 -3 = 2
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  7. #7
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    Once again, thank you!

    Just one thing. Your way of showing that cosine is even, cos -z = cos z, seems to be the same thing as saying, Cosine is an even function. If I really had to show that it's even maybe I could use the Taylor series for cosine to show that it's even, or Maclaurine since it's obviously centered at zero in this case. That gives me the terms of the series and they are clearly even.
    Last edited by liquidFuzz; September 14th 2010 at 12:10 PM.
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  8. #8
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    Quote Originally Posted by liquidFuzz View Post
    Once again, thank you!

    Just one thing. Your way of showing that cosine is even, cos -z = cos z, seems to be the same thing as saying, Cosine is an even function. If I really had to show that it's even maybe I could use the Taylor series for cosine to show that it's even, or Maclaurine since it's obviously centered at zero in this case. That gives me the terms of the series and they are clearly even.
    Oh, sorry about that - I misunderstood your question. Yes, you can definitely use the cosine taylor expansion to prove that it's an even function.
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  9. #9
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    I did the rewriting of the integral. I don't get the same result as you do... Maybe you could look for errors in this?

    \int {\frac {cos (\theta)}{2+cos (\theta)}\,d\theta} - Using z = e^{i\theta} we get d\theta = \frac {dz}{iz} This with Eulers formulas, cosine.

    Now the algebra:
    \frac{\frac{z+z^{-1}}{2}}{z(2 + \frac{z+z^{-1}}{2})} = \frac{\frac{z+z^{-1}}{2}}{(\frac{4z}{2} + \frac{z^2+1}{2})} = \frac{z+z^{-1}}{z^2+4z+1} = \frac{z^{2}+1}{z(z^2+4z+1)} - This is where I start to calculate the residues.
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  10. #10
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    Oh, the +2-2 stunt you're pulling is nice. I'll redo the integral using it. First I didn't understand what you meant by missing a 2\pi
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  11. #11
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    I redid the integral and tried to break it down, so errors would be evident. Please point out any errors!

    \displaystyle\int^{\frac{\pi}{2}}_0 \frac{cos(2\theta)}{1+cos^2(\theta)}\,d\theta = \displaystyle\int^{\frac{\pi}{2}}_0 \frac{cos(2\theta)}{2+cos(2\theta)}\,d\theta = \frac {1}{2} \displaystyle\int^{\pi}_0 \frac{cos(\phi)}{2+cos(\phi)}\,d\phi = \frac {1}{2}\displaystyle\int^{\pi}_0 \frac{cos(\phi) +2 - 2}{2+cos(\phi)}\,d\phi = \frac {1}{2}\displaystyle\int^{\pi}_0 1 - \frac{2}{2+cos(\phi)}\,d\phi = even integer = \frac {1}{2} \frac {1}{2}\displaystyle\int^{\pi}_{-\pi} 1 - \frac{2}{2+cos(\phi)}\,d\phi = linearity = \frac{2\pi}{4}  - \frac{1}{4}\displaystyle\int^{\pi}_{-\pi}  \frac{2}{2+cos(\phi)}\,{d\phi}

    Now I'm using z=e^{i\phi} and Eulers formulas, cosine. This gives: \frac{\pi}{2}  - \frac{1}{4i} \displaystyle\int_{|z|=1}  \frac{2}{2+\frac{z+z^{-1}}{2}}\,\frac{dz}{iz} = \frac{\pi}{2}  - \frac{1}{4i} \displaystyle\int_{|z|=1}  \frac{2}{\frac{4z+z^2+1}{2}}\,{dz} = \frac{\pi}{2}  - \frac{1}{4i} \displaystyle\int_{|z|=1}  \frac{4}{4z+z^2+1}\,{dz} = \frac{\pi}{2}  - \frac{4}{4i} \displaystyle\int_{|z|=1}  \frac{1}{4z+z^2+1}\,{dz}

    Finding the divisor's roots.

    z^2 + 4 z + 1 = 0 \Rightarrow z = \sqrt{3}-2, z = -\sqrt{3}-2 where z = \sqrt{3}-2 is in the unit circle.

    Residue for a simple pole

    Res I = \left[ \frac{1}{2z_k + 4} \right]_{z_k = \sqrt{3} - 2} = \frac{1}{2(\sqrt{3} - 2) + 4} = \frac{1}{2\sqrt{3}}

    Now I use \displaystyle\int f(z)\,{dz} = 2{\pi}i ~ Res f(z)

    - \frac{1}{i} ~ \frac{1}{2\sqrt{3}} ~2{\pi}i = - \frac{\pi}{\sqrt{3}}

    Result
    \frac{\pi}{2}- \frac{\pi}{\sqrt{3}}
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  12. #12
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    Looks pretty fine to me!
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