complex integral 1/z

**jbpellerin** · September 22nd 2008, 08:15 PM

Ok so I found that first easy integral which was ln((z+1)/z)
but now I don't know how to do the second
someone tried to help me before but he just went straight for the answer and I would like to know what to do not just what the answer is

**choovuck** · September 22nd 2008, 10:03 PM

e.g. for the piece $\int_{[1,i]}d\xi/\xi$ , parametrize the path by $\xi=1-t+it$ where t goes from 0 to 1. Then $d\xi=(-1+i)dt$ , so we get $\int_0^1\frac{i-1}{1-t+it}dt$ , which you can compute if you know the first integral

**jbpellerin** · September 22nd 2008, 10:15 PM

Originally Posted by choovuck

e.g. for the piece $\int_{[1,i]}d\xi/\xi$ , parametrize the path by $\xi=1-t+it$ where t goes from 0 to 1. Then $d\xi=(-1+i)dt$ , so we get $\int_0^1\frac{i-1}{1-t+it}dt$ , which you can compute if you know the first integral

that makes sense
but i'm still not sure how and when to integrate for logs
because sometimes you can use: integral = F(b) - F(a) and sometimes you can't
is this one of the times when I can?
and how do I say that

**choovuck** · September 23rd 2008, 01:40 AM

for any fixed z choose a branch of logarithm so that [z,z+1] is not on the cut (e.g. take the cut $(-\infty,0]$ ) if z is not negative, and take the cut $[0,\infty)$ if it is (i.e if $z<1$ )). for this branch $\log(z+t)$ for $t\in[0,1]$ is a primitive for $\frac{1}{z+t}$ , so $\int_0^1 \frac{1}{z+t}dt=\log(z+1)-\log z$

writing the answer in the form $ln((z+1)/z)$ , as you did, is not correct. you should write as $\log(z+1)-\log z$ where you explain what branch of logarithm you're taking for each particular z

**shawsend** · September 23rd 2008, 07:38 AM

Guess that may be my fault for telling him to write it as $\log\left(\frac{1+z}{z}\right)$ . I'm a little unclear on that too. Sorry about that jb.

Here's your chance to get me back plm if you're reading this.

**jbpellerin** · September 23rd 2008, 09:19 AM

Originally Posted by choovuck

for any fixed z choose a branch of logarithm so that [z,z+1] is not on the cut (e.g. take the cut $(-\infty,0]$ ) if z is not negative, and take the cut $[0,\infty)$ if it is (i.e if $z<1$ )). for this branch $\log(z+t)$ for $t\in[0,1]$ is a primitive for $\frac{1}{z+t}$ , so $\int_0^1 \frac{1}{z+t}dt=\log(z+1)-\log z$

writing the answer in the form $ln((z+1)/z)$ , as you did, is not correct. you should write as $\log(z+1)-\log z$ where you explain what branch of logarithm you're taking for each particular z

I still don't get it
this is really frustrating

I made it all the way up to 4th years honors calculus and now it's like I hit a brick wall

ok so for the parametrization of the first one, I got ln(i-1+1) - ln(1) = ln(i)
is this right?

**shawsend** · September 23rd 2008, 10:08 AM

Alright, I'll risk digging a deeper hole cus' I need to understand it also as I'm planning to take Complex Analysis next semester. Until you can understand totally, the following analysis, you're not going to understand antiderivatives which are multi-functions (like log):

Look first at evaluating the following integral using antiderivatives:

$\mathop\oint\limits_{|z|=1}\frac{1}{z}dz$

At first you'll say, that can't be done since the antiderivative $\log(z)$ is not analytic in and on the contour $|z|=1$ . Remember: $\int_{z_0}^{z_1} f(z)dz=F(z_1)-F(z_0)$ only if the antiderivative $F$ is analytic throughout a domain containing the contour. However, we can split up the unit circle: a top and bottom half-circle each of which can be contained separately in analytic branches of the logarithm function:

$<br /> \mathop\oint\limits_{|z|=1}\frac{1}{z}dz=\mathop\i nt\limits_{\text{Top}}\frac{1}{z}dz=\log(z)\bigg|_ 1^{-1};\quad \log(z)=\ln|r|+i\Theta,\; -\pi/2\leq \Theta\leq 3\pi/2$
$+\mathop\int\limits_{\text{Bottom}}\frac{1}{z}dz =\log(z)\bigg|_{-1}^{1};\quad \log(z)=\ln|r|+i\Theta,\;\pi/2\leq \Theta\leq 5\pi/2$

$=\pi i+(2\pi i-\pi i)=2\pi i$

Note how I segregated the multi-sheet logarithm function (look at the Riemann surface for log(z)) into two separate analytic sheets which ``overlapped'' so as to include the entire contour. Now, it's a little more messy with the antiderivative $\log(z+t)$ but I'm pretty sure the concept is the same.

**jbpellerin** · September 23rd 2008, 10:13 AM

but I'm not integrating over the unit circle.. ?

**shawsend** · September 23rd 2008, 10:29 AM

The contour below is what I think the integral is (I could be wrong). It's still a closed contour around the origin of the function 1/z which has the antiderivative log(z) but you need to write it as that other integral. Look, I did the whole problem after choovuck schooled me above: We have as a parameterization for each leg:

$c_1: z=(1-t)+ti$

$c_2: z=-t+(1-t)i$

$c_3: z=(-1+t)-ti$

$c_4: z=t+(-1+t)i$

Now make those parameterizations into $\int 1/z dz$ , shake, then bake, then express it in the form $\int_0^1 \frac{1}{z+t}dt$ then use the results from that integral to solve the integral over the four legs above.

**jbpellerin** · September 23rd 2008, 10:32 AM

I agree with everything you said in the last post
I just get 4i and not 2pi*i

**shawsend** · September 23rd 2008, 11:33 AM

I'll do $c_1$ :

$\mathop\int\limits_{C_1}\frac{1}{z}dz=\int_0^{1}\f rac{-(1-i)dt}{1-t(1-i)}$
$=\int_0^1 \frac{dt}{-\frac{1}{1-i}+t}dt$

Now use $\int_0^1 \frac{1}{z+t}dt=\log(1+z)+\log(z)$ and so I get:

$\int_0^1 \frac{dt}{-\frac{1}{1-i}+t}dt=\log\left(1-\frac{1}{1-i}\right)-\log\left(\frac{-1}{1-i}\right)$

$=\log(1/2-i/2)-\log(-1/2-i/2)=-\pi i/4+3\pi i/4=\pi i/2$ The other three give the same but after doing $C_2$ the other two are similar and you don't have to do any more calculations.

**choovuck** · September 23rd 2008, 08:04 PM

the result should be $2\pi i$ by Cauchy formula.

shawsend, the integral is indeed over the square you pictured, and it can be easily explained why the integral over it is the same as the integral over the unit circle, and then your computations work and are more straightforward than doing the 4 parametrizations. But the problem states explicitly to do it, so your post #9 is the way to do the problem.

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