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Math Help - Another complex analysis problem

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    Another complex analysis problem

    a) Let \widehat{f}(z)=\int_{-\infty}^{\infty} f(t) e^{-2 \pi i zt} dt
    Assume that f has bounded support and smooth.
    Show, by integration by parts that \mid \widehat{f}(x+iy) \mid \leq \frac{A}{1+x^2} if \mid y \mid \geq 0
    b) Write P(z)=a_n(2\pi iz)^n + a_{n-1}(2\pi iz)^{n-1} +...+a_0 where a_i are complex constants.
    Find a real number c such that P(z) does not vanish on the line L= \{z:z=x+ic, x\in R \}

    For a) I let u=f(t) and dv=e^{-2 \pi i zt}dt. Then integrate by parts. Is this the right approach?
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    Quote Originally Posted by namelessguy View Post
    For a) I let u=f(t) and dv=e^{-2 \pi i zt}dt. Then integrate by parts. Is this the right approach?
    Yes. You'll need to integrate by parts twice, and then use the fact that f''(t) is bounded (because f is smooth).
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    Thanks for your help Opalg. Do you have any suggestion for part b) ?
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    Quote Originally Posted by namelessguy View Post
    b) Write P(z)=a_n(2\pi iz)^n + a_{n-1}(2\pi iz)^{n-1} +...+a_0 where a_i are complex constants.
    Find a real number c such that P(z) does not vanish on the line L= \{z:z=x+ic, x\in R \}
    P(z) is a polynomial of degree n, so it has n roots z_k\ (1\leqslant k\leqslant n). You want c to satisfy P(2\pi ix-2\pi c)\ne0 for all real x. This is equivalent to 2\pi ix-2\pi c \ne z_k, or c\ne ix - (z_k/(2\pi)), for all real x and for 1\leqslant k\leqslant n. The condition for this is \text{re}(c)\ne \text{re}(-z_k/(2\pi))\ (1\leqslant k\leqslant n).
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