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Math Help - complex analysis

  1. #1
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    complex analysis

    1. Evaluate \int_{C(0,2)}\frac{dz}{1+z^2}.

    I know z(t)=2e^{it}, t\in[0,2\pi], dz=i2e^{it}dt. So \int_{C(0,2)}\frac{dz}{1+z^2}=\int_0^{2\pi}\frac{i  2e^{it}}{1+(2e^{it})^2}dt. I use substitution, letting u=2e^{it}, du=i2e^{it}dt, then I get \int_0^{2\pi}\frac{i2e^{it}}{1+(2e^{it})^2}dt=\int  _2^2\frac{du}{1+u^2}=0. Is this okay? I don't know how else to do this...

    2. Let 0<|a|<|b|. Evaluate \int_{C(0,r)}\frac{dz}{(z-a)(z-b)}, where (i) 0<r<|a|, (ii) [tex] |a|<r<|b|, (3) |b|<r.

    Can I get some help with this?
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by dori1123 View Post
    1. Evaluate \int_{C(0,2)}\frac{dz}{1+z^2}.

    I know z(t)=2e^{it}, t\in[0,2\pi], dz=i2e^{it}dt. So \int_{C(0,2)}\frac{dz}{1+z^2}=\int_0^{2\pi}\frac{i  2e^{it}}{1+(2e^{it})^2}dt. I use substitution, letting u=2e^{it}, du=i2e^{it}dt, then I get \int_0^{2\pi}\frac{i2e^{it}}{1+(2e^{it})^2}dt=\int  _2^2\frac{du}{1+u^2}=0. Is this okay? I don't know how else to do this...

    2. Let 0<|a|<|b|. Evaluate \int_{C(0,r)}\frac{dz}{(z-a)(z-b)}, where (i) 0<r<|a|, (ii) [tex] |a|<r<|b|, (3) |b|<r.

    Can I get some help with this?
    Are you familiar with the residue theorem?
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  3. #3
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    Quote Originally Posted by chiph588@ View Post
    Are you familiar with the residue theorem?
    No. Do I have to use that to do my work?
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by dori1123 View Post
    No. Do I have to use that to do my work?
    Residue theorem - Wikipedia, the free encyclopedia

    You don't need it for this problem, but it's an extremely useful theorem.

    I'm pretty sure your answer is correct.
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