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Math Help - Complex analysis -pth roots

  1. #1
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    Complex analysis -pth roots

    Let f be a quadratic polynomial function of z with two different roots z_1 and z_2. Given that a branch of  z of the square root of  f exists in a domain D, demonstrate that neither z_1 nor z_2 can belong to D. If D had a double root, would the analogous statement be true?
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  2. #2
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    Re: Complex analysis -pth roots

    I know that f(z)=a_0+a_1(z)+a_2(z^2) and I know that g(z)^2=z. So if z has two different roots, then how could g(z)^2 = z which implies that \sqrt{z} and z_1 and z_2 are all in D. Is that where I need to go?
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