Complex line integral

**arbolis** · August 29th 2009, 07:10 PM

I must find the value of $\int _{C} z^{-1+i} dz$ , where the integrand is such that $0<\arg z<2\pi$ and $C$ is the positively oriented circle of radius 1.

My attempt : First I parametrize $C$ by $\gamma (t)=\cos t +i \sin t$ .
Second, I try to find $f'(z)$ . The only method I know is to use the definition : $f'(z)=\lim _{h \to 0}\frac{(z+h)^{-1+i}-z^{-1+i}}{h}$ . I don't know how to calculate this limit... I'd like a hint.
Thanks!

**Marine** · August 29th 2009, 10:20 PM

Hi there!

Let

$\gamma : [a,b]\rightarrow C$
$t\rightarrow\gamma(t)$

be a rectifiable, smooth enough curve. Then we have:

$\int_{\gamma}f(z)dz=\int^b_af(\gamma(t))\dot\gamma dt$

You parametrisation is correct, try also applying to it the Euler formula.

Find out about the Cauchy-Riemann differential equations (if you think you'll need a differentiation of the integrand function).

For the integral, check out when a holomorphic function has an antiderivative and most importantly - WHERE.

When you're ready, compare your resuld with the Cauchy's integral formula.

**shawsend** · August 30th 2009, 08:59 AM

Nothing is being said about the multifunction-ness of the integrand. I'm kinda' rusty, but this is how it looks to me:

The function $f(z)=z^{-1+i}$ is a multifunction with an infinite number of overlapping sheets, and the description above does not distinguish which sheet the integration is being performed over. I would write:

$ \begin{aligned} z^{-1+i}&=e^{[(-1+i)\left[\ln|z|+i(arg(z)+2k\pi)\right]} \\ &=e^{[2k\pi i(-1+i)]}z^{-1+i} \end{aligned} $

with $z^{-1+i}=e^{i\, arg(z)(-1+i)},\quad 0<arg(z)<2\pi$ .

For example, take $C$ as a section of the unit circle starting at $z_0=e^{\pi i/4}$ and ending at $z_1=e^{7\pi i/4},$ then the integrand is analytic over any contiguous section of the multifunction over $C,$ so I can use antiderivatives:

$ \begin{aligned} \mathop\int\limits_{C} z^{-1+i}dz&=e^{2k\pi i(-1+i)}\mathop\int\limits_{C} z^{-1+i}dz,\quad 0<arg(z)<2\pi,\quad k\in\mathbb{Z} \\ &=e^{2k\pi i(-1+i)}\left(\frac{1}{i}\;z^{i}\Biggr|_{z_0}^{z_1}\r ight) \\ &=-ie^{2k\pi i(-1+i)}\left(z_1^i-z_0^i\right) \\ &=-ie^{2k\pi i(-1+i)}\left(e^{-arg(z_1)}-e^{-arg(z_0)}\right),\quad 0<arg(z)<2\pi \end{aligned} $

So that

$\mathop\int\limits_{C}z^{-1+i}dz=-i e^{2k\pi i(-1+i)}\left(e^{-7\pi/4}-e^{-\pi/4}\right),\quad k\in\mathbb{Z}$

**Opalg** · August 30th 2009, 10:44 AM

Originally Posted by arbolis

I must find the value of $\int _{C} z^{-1+i} dz$ , where the integrand is such that $0<\arg z<2\pi$ and $C$ is the positively oriented circle of radius 1.

My attempt : First I parametrize $C$ by $\gamma (t)=\cos t +i \sin t$ .

That's a good start, but I would keep to complex exponentials and write the parametrisation as $z=e^{it}$ . Then $dz = ie^{it}dt$ and $z^{-1+i} = \bigl(e^{it}\bigr)^{-1+i} = e^{-t}e^{-it}$ . The integral then becomes $\oint _{C} z^{-1+i} dz = \int_0^{2\pi}\!\!\!e^{-t}e^{-it}ie^{it}dt = i\!\!\int_0^{2\pi}\!\!\!e^{-t}dt$ , which I guess should give you no trouble.

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