2
$\begingroup$

How do I evaluate $ \ \frac{1}{2 \pi i} \oint_{|z|=3} \frac{f' (z)}{f(z)}\,dz ,~~\mbox{ where }~ f(z) = \frac{z^2 (z-i )^3 e^z}{3 (z+2 )^4 (3z - 18 )^5} \ $ ?

The singularities are z = -2 and z = 6. But the $ \frac{f' (z)}{f(z)}\ $ part does not make much sense, it looks like L'Hospital rule but it's really not. It'd be too tedious if I just calculate $ \frac{f' (z)}{f(z)}\ $. In fact I'm not even sure if I can... It also looks like I might want to use residue theorem, but why $ \frac{f' (z)}{f(z)}\ $?

I am confused about both why this question is asked this way and how to solve it. In the mean time it'll be great if somebody can just tell me how to solve it...

Thanks.

Edit: It might be way way easier to just use Argument principle by Cauchy. But I'm not sure about how to count Poles and Zeros. I mean, poles I know, but WHAT ARE ZEROS? Could somebody solve the problem using this theorem? Thanks!

  • 2
    Do you know the argument principle?2012-12-05
  • 0
    I'm intrigued to notice, per `int(vdu) = uv - int(udv)`, that `udv` looks a lot like `1/(udv)` . Luckily I'm out of school and don't have to solve integrals :-)2012-12-05
  • 0
    @Sanchez: interesting, I am reading that right now. Seems to be what the question tends to ask.2012-12-05
  • 0
    Zeros means $a$ such that $f(a) = 0$. In this case they are 0 and $i$, but you will have to count the multiplicities too.2012-12-05
  • 0
    Oh wow got it! It's this simple.2012-12-05

3 Answers 3

0

I can't tell you why, but you are correct to use the residue theorem here.

The singularities of $\frac{f'(z)}{f(z)}$ are the points where $f(z)=0$ therefore $z=0$ and $z=i$ and where the denominator of $f'(z)=0$ therefore $z=-2$ and $z=6$.

Now we need to calculate the residua on all the points but $z=6$ because it is outside of the circle of radius $3$.

Since your quuestion is tagged as homework I'll not calculate the residues unless you ask.

  • 0
    Haha thanks! It's a practice problem but I just figured if it'll get more attention from somebody who are doing similar hws. So do I calculate residue of f(z) or f'(z)/f(z) ? How do I do it if it's the latter? Does it mean I have to get f'(z) first anyway and then use partial fraction?2012-12-05
  • 0
    Its, the latter because that is the function you are integrating. And you were correct, I just studied this this semester. To calculate the residue there you have a formula. When I was studying this I made a question here about how to prove that formula, the link is here: http://math.stackexchange.com/questions/232165/residues-and-laurent-expansion Be carefull with the order of the poles in this function.2012-12-05
1

Zeros in $\,\{z\in\Bbb C\;;\;|z|<3\}\,$:

$$z^2(z-i)^3e^z=0\Longleftrightarrow z^2(z-i)^3=0$$

and thus we have 5 zeros within the wanted domain, and since we have 4 poles there then...

Note: Zeros and poles are counted with their multiplicity, of course.

  • 0
    Why do you think that, @Sanchez ?2012-12-05
  • 0
    Oh, I see what you meant...hehe. Gracias2012-12-05
  • 0
    Um, since according to the argument principle integral of f'(z)/f(z) = 2pi*i(N-P) where N and P denote respectively the number of zeros and poles of f(z) inside the contour C, with each zero and pole counted as many times as its multiplicity, respectively order, indicates. The answer to this question is simply N - P which equals to 5 - 4 = 1 ? Wow.2012-12-05
1

Hint: Note that $\dfrac{f'(z)}{f(z)}=\dfrac{\mathrm{d}}{\mathrm{d}z}\log(f(z))$. $$ \frac{\mathrm{d}}{\mathrm{d}z}\log\left(\frac{z^2(z-i)^3e^z}{3(z+2)^4(3z-18)^5}\right) =\frac2z+\frac3{z-i}+1-\frac4{z+2}-\frac5{z-6} $$

  • 0
    This works too! Answer = 2 + 3 - 4 = 1 by reading off the residues. Correct?2012-12-05
  • 0
    Indeed :-) It becomes obvious why the function has this form.2012-12-05