complex analysis

**journyman** · May 24th 2010, 06:03 AM

hello. I would like some direction in how to approach these problems

1.Calculate the line integral
$<br /> \int\limits_{-ipi}^{ipi}({e^z-z^2}) dz<br />$ calculated along circle at origin & radius r=pi

2.Find the sum of the fourier series of f(x)= 1 -x on [-pi, pi]

the 1st i tried substituting x+ iy but it looks strange and the parametrization of the line, should I change to polar coordinates? Im trying to use z(t)= a.t + b

the 2nd i set f(x)= 1-sinx which makes it an odd function. but Im not sure how to solve it. like how to get values of n in sin(nt)dt.

**chisigma** · May 24th 2010, 10:48 AM

The function $f(z)= e^{z} - z^{2}$ is analyitic in the whole complex plane, so that is...

$\int_{\gamma} f(z)\cdot dz =0$ (1)

... for any closed path $\gamma$ . That means that the integral of f(*) from $-i \pi$ to $+ i \pi$ doesn't depend from the path connecting the limits of integration, so that we can choose the direct path along the $i \omega$ axis and is...

$\int_{- i \pi}^ {+ i \pi} (e^{z} - z^{2})\cdot dz = i \int_{-\pi}^{+ \pi} (e^{i \omega} + \omega^{2}) \cdot d\omega = i \int_{-\pi}^{+ \pi} (cos \omega + \omega^{2}) \cdot d\omega = \frac{2}{3} i \pi^{3}$ (2)

Kind regards

$\chi$ $\sigma$

**chisigma** · May 24th 2010, 11:42 AM

Second question: is...

$f(x)= \frac{a_{0}}{2} + \sum_{n=1}^{\infty} a_{n} \cos n x + b_{n} \sin n x$ (1)

... where...

$a_{0} = \frac{1}{\pi}\cdot \int_{-\pi} ^{ + \pi} (1-x)\cdot dx = 1$

$a_{n} = \frac{1}{\pi}\cdot \int_{-\pi} ^{ + \pi} (1-x)\cdot \cos nx \cdot dx= 0$

$b_{n} = \frac{1}{\pi}\cdot \int_{-\pi} ^{ + \pi} (1-x)\cdot \sin nx \cdot dx= 2\cdot \frac {(-1)^{n}}{n}$ (2)

... so that is...

$f(x) = 1 - 2 \sin x + \sin 2x - \frac{2}{3} \sin 3x + \dots$ (3)

Kind regards

$\chi$ $\sigma$

**journyman** · May 25th 2010, 03:29 AM

Thanks for your help. Could you give me an idea of what steps you used or did you just plug in formulas?

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