Proving Integral of Log(z+5) is 0 Around Contour z=1

[soulsearching]

Can anyone help me with this pls?

How can you prove that the integral of f(z) around the contour z= 1 is 0

where f(z) is Log(z+5)

Thx





I know Log(z) is ln r + i (theta). But i don't know how that applies to this situation.

Also, do I solve it as a normal integral or use the Cauchy Goursat theorem to prove that its integral is zero?

 

[tiny-tim]

Welcome to PF!

How can you prove that the integral of f(z) around the contour z= 1 is 0
…
Also, do I solve it as a normal integral or use the Cauchy Goursat theorem to prove that its integral is zero?

Hi soulsearching! Welcome to PF!

Do you mean |z| = 1?

If so, then z + 5 is just the circle of radius 1 and centre at 5. It doesn't include the origin (z + 5 = 0), so yes … always do it the easy way … use Cauchy Goursat!

 

[soulsearching]

Hi soulsearching! Welcome to PF!

Do you mean |z| = 1?

If so, then z = 5 is just the circle of radius 1 and centre at 5. It doesn't include the origin (z + 5 = 0), so yes … always do it the easy way … use Cauchy Goursat!

So the point z = 5 is outside the contour, so the integral vanishes right? Thank you Tim.



When trying to show the same for this equation here around the same contour

f(Z) = (4z^2 -4z +5)^-1

I found the singularities here to be (1/2 + i) and (1/2-i). Is the integral zero because both singularities are outside the circle of radius 1? Thank you.

I have a major exam tmr and my professor is not being of much help, but I really appreciate your help. :-)

 

[ircdan]

So the point z = 5 is outside the contour, so the integral vanishes right? Thank you Tim.



When trying to show the same for this equation here around the same contour

f(Z) = (4z^2 -4z +5)^-1

I found the singularities here to be (1/2 + i) and (1/2-i). Is the integral zero because both singularities are outside the circle of radius 1? Thank you.

I have a major exam tmr and my professor is not being of much help, but I really appreciate your help. :-)

Yes.

And in this case your integrand is analytic inside and on the circle |z| = 1, so again it's 0 by cauchy's.

 

[soulsearching]

Thanks Dan.

 

[tiny-tim]

When trying to show the same for this equation here around the same contour

f(Z) = (4z^2 -4z +5)^-1

I found the singularities here to be (1/2 + i) and (1/2-i). Is the integral zero because both singularities are outside the circle of radius 1?

Looks good to me! ​

 

[soulsearching]

Also when trying to find the integral of (1/8z^3 -1) around the contour c=1.

I found the singularities to be 1/2, 1/2exp(2pi/3), and 1/2exp(4pi/3)

What is the next step here. Do I just assume the integral is 6pi(i) after using partial fractions to find the numerators of the 3 fractions.

The answer is actually 0, but I don't understand how it was solved. I know it might have something to do with the rule that says, the sum of the integral of a C1, C2 and C3 is equal to the integral of the outside contour, but how do i know the orientation of the three contours inside the big contour C.

Thanks, hope I am not asking too many questions