# Thread: Another complex analysis problem

1. ## 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?

2. Originally Posted by namelessguy
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).

3. Thanks for your help Opalg. Do you have any suggestion for part b) ?

4. Originally Posted by namelessguy
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 $\text{re}(c)\ne \text{re}(-z_k/(2\pi))\ (1\leqslant k\leqslant n)$.