1. ## Complex line integral

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

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

2. Hi there!

Let

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

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

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

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.

3. 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 $\displaystyle f(z)=z^{-1+i}$ 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:

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

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

\displaystyle \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

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

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

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