Question

Find the integral of g(z) around the circle |z-i|=2 in the positive sense when g(z)=1/((z^2)+4)

Answer 1

using cauchy integral formula or residues

Answer 2

using CIF you need to split it up as partial fractions, ohh there is a different way actually

Answer 3

so the contour is disc centred at i , radius of 2

Answer 4

so it extends from -i to +3i

Answer 5

note the two singularities , at +-2i ,only one of these is inside the contour , that is +2i , -2i is outside the contour, so contributes nothing to the integral.

Answer 6

now this is a little trick you can do with these , you can bring factors up the top.

I write something to try and justify this

Answer 7

\[g(z) = \frac{ 1}{(z+2i)(z-2i)} = \frac{ \frac{1}{z+2i} }{z-2i}\]

Answer 8

now , the functions is of the form \[\frac{f(z)}{z-a} dz \]

Answer 9

where \[f(z) = \frac{1}{z+2i} \] and is analytic inside the contour.

Answer 10

now by applying cauchy integral formula all we need to do is sub the singularitiy into f(z) and multiply by the 2pi i

Answer 11

\[\int\limits_{C} g(z) dz = 2 \pi i \times \frac{1}{2i + 2i} = \frac{2 \pi i }{4i} = \frac{\pi}{2}\]

Answer 12

Thank you

Answer 13

how did you find the singularity points?

Answer 14

denominator =0

Answer 15

you can go to quadratic formula if you want, but its pretty easy after a while

Answer 16

so if my denominator was ((z^2)+4)^2...I would just set that equal to zero?

Answer 17

yes you would get the same singluarities, but you would need a slightly different formula

Answer 18

if you hade 1/ (z^2+4)^2

Answer 19

then you would go \[g(z) = \frac{1}{(z+2i)^2 (z-2i)^2} = \frac{ \frac{1}{(z+2i)^2} }{(z-2i)^2} \]

Answer 20

now \[f(z) = \frac{1}{ (z+2i)^2 } \] and its analytic inside the contour

Answer 21

but because you have (z-2i)^2 in the denominator, you need to differentiate the numerator once, then sub the value in

Answer 22

and also divide by 1! , but thats just 1so you ignore it

Answer 23

and then dont forget the 2 pi i

Answer 24

complex analysis is pretty fun in my opinion

Answer 25

I just finished it last semester in university

Answer 26

can you explain the step where I have to differentiate the numerator once then substitute the value in?
I don't understand where you're talking about

Answer 27

\[\int\limits \frac{f(z)}{(z-a)^{n+1} } = 2 \pi i \frac{ f^{(n)} (a) }{n!} \] you remember that?

Answer 28

yes

Answer 29

well its the same thing

Answer 30

you have (z-2i)^2 in the denominator , n=1

Answer 31

ah, the derivative of f(a)
okay

Answer 32

no, its derivative of f(z) evaluated at a , to be technical.

Answer 33

that's what I meant, but yes

Answer 34

you can use those exact same methods to find messy things such as

Answer 35

\[\int\limits_{|z|= 10} \frac{1}{(z-3)^3 ( z+2i)^5 ( z^2 + 1 )^{10} } dz \]

Answer 36

would take a long time, but its possible :)

Answer 37

offcourse that doesnt mean you should forget how to do things like \[\int\limits_{|z|=1} \frac{ 1}{(z-5i)^3(3z+4)^2} dz \]

Answer 38

The poles of g(z), both simple, are 2i and -2i, and only the pole 2i lies on the interior of C. Se let
f(z) = 1/(z + 2i), and let s = 2i. According to the Cauchy Integral Formula,
integral around C of f(z) dz/(z - s) = 2*pi*i*f(s). and f(s) f(2*i) = i/(4*i), so the contour integral evaluates to pi/2 (the equation writer is not functioning at my end).