*****Loop integral 2009*****?

1) J=∫[L] dz/(2009 + z^2009) =

2πi∑[i=1 to 2009] Res(Zi), where Zi are the roots of

2009+z^2009=0 ==> Zi^2009=-2009,

(all Zi are inside the unit circle |z| = 1)

and

Res(Zi)=

lim[z->Zi] (z-Zi)/(2009 + z^2009)=[Lopital's rule]=

lim[z->Zi] 1/(2009*z^2008)=1/(2009*Zi^2008)=

Zi/(2009*Zi^2009)=-Zi/2009² ==>

∑[i=1 to 2009] Res(Zi) =

(-1/2009²)∑[i=1 to 2009] Zi =

(-1/2009²)*0 = 0 ==> J=0

Edit:

Mistake due to hurrying.

2009^(1/2009)>1

|Zi|=2009^(1/2009)=1.00379...

1) L: |z| = 1 ==> All poles Zi are OUT of the unit circle |z|=1 (but very close to it) ==> J=0

2) L: | 2z - √7| + | 2z + √7| = 8

| z - √7/2| + | z + √7/2| = 4

√[(x-√7/2)²+y²]+√[(x+√7/2)²+y²]=4

√[(x+√7/2)²+y²]=4-√[(x-√7/2)²+y²]

(x+√7/2)²=

16-8√[(x-√7/2)²+y²]+(x-√7/2)²

2√7x=16-8√[(x-√7/2)²+y²]

√7x=8-4√[(x-√7/2)²+y²]

4√[(x-√7/2)²+y²]=8-√7x

16(x²-x√7+7/4+y²)=64-16x√7+7x²

9x²+16y²=36

L: x²/2²+y²/1.5²=1 ==> All Zi are inside the ellipse ==> J=0.

I think the ellipse is correct now.

Edit2:

At the beginning of solving (2), I was thinking to use the inequality

|a|+|b|≥|a+b|, which, used for L2: |2z - √7|+|2z + √7|=8 would lead to

L3: |z|≤2, but this can't be used, since L2 is inside L3,

as the equation of the ellipse showed later.

Very nice question!

Tabula,

first off great question, did the exponent have to be so large!!

I will following the same train of thought as Roveta, however for convenience, I solve

int ( dz/ ( n + z^n) dz ) ; n being a natural number

for both (1), (2)

we have,

L = int ( ( 1/ (n + z^n ) dz)

L = sum ( Res(z = zi) i = 1 to n )

Where zi are the solutions to

n + z^n = 0 --> z^n = -n (*)

that for (1) lie within |z| = 1

to solve (*) I will employ De Moivre's formula,

i.e,

if z^n = w ; w being another complex number

then the solution z are,

z = r^(1/n) * ( exp( ( (t + 2*k*pi)/n )*i) k = 0 to n - 1

which clearly are all distinct

r = magnitude of w

t = principle of w

here r = n , t = (3/2)pi ( as n is a natural number)

Hence the solutions z are

z = n^(1/n) * exp( ( ( (3/2)pi + 2*k*pi )/n )* i)

z = n^(1/n) * exp( ( ( (3/2) + 2*k )pi/n )* i) k = 0 to n - 1

we observe for all residues to lie within |z| = 1 we have the condition

|n^(1/n)| <= 1 --> n <= 1

and as such no solutions exist as n is a natural number

Hence for (1)

(1) = 0

For (2).... this obviously requires a bit more work, I will keep you posted (pardon the pun)

David