Question

Help please! Complex analysis.

Answer 1

Evaluate the contour integral via residue theorem.

cos (z)/1-2sin(z)

Answer 2

z0=pi/6

Answer 3

Over what contour?

Answer 4

circle of R=3/2

Answer 5

Centered at the origin?

Answer 6

yes

Answer 7

found the residue at z0=pi/6

r(pi/6)=-sqrt3/2

Answer 8

Alright, so by the residue theorem the integral should evaluate to \(2\pi i\) times the residue, right?

Answer 9

right but the book says the answer is -(pi)(i)

Answer 10

which would mean my residue would have to -1/2

Answer 11

In that case you must have made a mistake with the residue computation. What approach did you use to determine it?

Answer 12

simple pole method

Answer 13

Here's an approach involving the series expansions:
\[(z-\pi/6)\frac{\cos z}{1-2\sin z}=\frac{\color{red}1(z-\pi/6)-\frac{(z-\pi/6)^3}{2!}+\frac{(z-\pi/6)^4}{4!}-\cdots}{1\color{red}{-2}\left((z-\pi/6)-\frac{(z-\pi/6)^3}{3!}+\frac{(z-\pi/6)^5}{5!}-\cdots\right)}\to -\frac{1}{2}\]

Answer 14

but can I also do \[\lim_{z \rightarrow \pi/6} (z-z_0)f(z)\]

Answer 15

with out expansion

Answer 16

Perhaps it's a matter of semantics, but around \(z_0=\dfrac{\pi}{6}\), the series expansions IS the same as the function itself.

Answer 17

sorry i'm terrible with that equation maker

Answer 18

No worries, it comes with practice :)

Evaluating the limit
\[\lim_{z\to\pi/6}\left(z-\dfrac{\pi}{6}\right)\frac{\cos z}{1-2\sin z}\]
can be a bit unwieldy without considering the expansions...

Answer 19

(sinz-1/2) (cosz/-2(sinz-1/2)) =cosz at pi/6

Answer 20

wouldn't your (z-z0)= sinz-1/2

Answer 21

No, the definition of residue for a function \(f(z)\) at a (simple) pole \(z_0\) is given as
\[\newcommand{res}{\text{Res}}
\res\{f(z),z_0\}=\lim_{z\to z_0}(z-z_0)f(z)\]

Answer 22

You seem to be replacing \(z-z_0\) with \(\sin z-\sin(z_0)\).

Answer 23

oh you're right!

Answer 24

so in general then, with sins and cos and e^iz you would have to expand then

Answer 25

right?

Answer 26

I wouldn't say that's the case in general, but it's certainly a good way to think about approaching the computation.

For example, if you're given the function \(f(z)=\dfrac{e^z}{z}\), you can tell right away that there's a simple pole at \(z=0\), and so
\[\res\left\{\frac{e^z}{z},0\right\}=\lim_{z\to0}z\frac{e^z}{z}=\lim_{z\to0}e^z=1\]
with no expansion needed. Of course, if you had elected to use the expansion, yo'd get
\[\res\left\{\frac{e^z}{z},0\right\}=\lim_{z\to0}\frac{z+z^2+\frac{z^3}{2!}+\cdots}{z}=1\]

Answer 27

I see, but what about with sin and cos in the denominator

Answer 28

you would at that point have to expand right

Answer 29

I'm only going to answer yes because that's how I would do it :)

Answer 30

I'm sure there are other methods available for that sort of thing.

Answer 31

lol is there any other way to do it?

Answer 32

I know r(z0)=g(z0)/h'(z0)

Answer 33

There is another way that involves Laurent series expansions (a sort of extension of Taylor series). These sorts of series have the form
\[\sum_{k=-\infty}^\infty a_k(z-c)^k\]
and it turns out that the \(k=-1\) term gives the residue.

If I recall correctly, you can also use Cauchy's integral formula. Both these methods are closely related.

Answer 34

oh yeah, now I remember. This was the first way we learned it. Okay, thanks so much for your help!

Answer 35

Think you can answer one more quick question?

Answer 36

Possibly. I have to admit it's been a while since I've taken complex analysis but I'll do my best.

Answer 37

okay, why is it that b^-1 is the residue. More clearly, why is the coefficient in front of
1/(z-z0) the residue? does that mean left over?

Answer 38

Consider the definition given here: http://en.wikipedia.org/wiki/Laurent_series

You find that
\[a_{-1}=\frac{1}{2\pi i}\oint f(z)\,dz\]
I remember learning that evaluating this integral is the same as the residue, but I forget the theory behind it.

Answer 39

the version I know is f(a) =1/2ipi integrate f(z) dz

Answer 40

I know the theory behind that but not sure that translate to a coefficient.
Anyways , I really appreciate the help man!

Answer 41

You're welcome!

Answer 42

Take care