MHB Contour Integration: Calculating the Integral of $1/z$ with Residue Formula

[Fermat1]

Calculate the integral of $1/z$ around $C$, where $C$ is any contour going from $-\sqrt{3}+i$ to $-\sqrt{3}-i$ and is contained in the set of complex numbers whose real part is negative.

My answer: Let $f=1/z$ Then $f$ has a simple pole at $z=0$ with residue 1. How do I calculate the winding number so I can use the residue formula?

Thanks

 

[Prove It]

Calculate the integral of $1/z$ around $C$, where $C$ is any contour going from $-\sqrt{3}+i$ to $-\sqrt{3}-i$ and is contained in the set of complex numbers whose real part is negative.

My answer: Let $f=1/z$ Then $f$ has a simple pole at $z=0$ with residue 1. How do I calculate the winding number so I can use the residue formula?

Thanks

Choose any contour that doesn't include z = 0. Maybe the semicircle to the left of these points centred at $\displaystyle \begin{align*} \left( -\sqrt{3} , 0 \right) \end{align*}$.

 

[Fermat1]

Choose any contour that doesn't include z = 0. Maybe the semicircle to the left of these points centred at $\displaystyle \begin{align*} \left( -\sqrt{3} , 0 \right) \end{align*}$.

So the value depends on whether the contour encloses zero?

 

[Prove It]

So the value depends on whether the contour encloses zero?

The contour CAN'T enclose z = 0 + 0i, surely you can see its real part is NOT negative.

So yes, the value of the integral will change if your contour contains this point. But you are told specifically to NOT include it.

 

[Fermat1]

The contour CAN'T enclose z = 0 + 0i, surely you can see its real part is NOT negative.

So yes, the value of the integral will change if your contour contains this point. But you are told specifically to NOT include it.

ok sorry, I was a bit thrown by you saying to choose a contour. So in fact the integral is zero?

 

[chisigma]

Calculate the integral of $1/z$ around $C$, where $C$ is any contour going from $-\sqrt{3}+i$ to $-\sqrt{3}-i$ and is contained in the set of complex numbers whose real part is negative.

My answer: Let $f=1/z$ Then $f$ has a simple pole at $z=0$ with residue 1. How do I calculate the winding number so I can use the residue formula?

Thanks

Any path that leaves to the left point z = 0 is valid, so that you can choose a straight path. Setting $\displaystyle z= -\sqrt{3} + i\ t$ You have...

$\displaystyle \int_{C} \frac{d z}{z} = \int_{1}^{-1} \frac{d t}{- \sqrt{3} + i\ t} = i \ln \frac{\sqrt{3}- i}{\sqrt{3} + i}\ (1)$

Kind regards

$\chi$ $\sigma$

 

[Fermat1]

Any path that leaves to the left point z = 0 is valid, so that you can choose a straight path. Setting $\displaystyle z= -\sqrt{3} + i\ t$ You have...

$\displaystyle \int_{C} \frac{d z}{z} = \int_{1}^{-1} \frac{d t}{- \sqrt{3} + i\ t} = i \ln \frac{\sqrt{3}- i}{\sqrt{3} + i}\ (1)$

Kind regards

$\chi$ $\sigma$

I wanted to use cauchy's residue formula.

 

[Prove It]

I wanted to use cauchy's residue formula.

How can a contour that doesn't have any singular points in it possibly give any residue?

 

[chisigma]

I wanted to use cauchy's residue formula.

The basic concept is well understood observing the figure...View attachment 2771

It is required to compute the integral $\displaystyle \int_{A B} \frac{d z}{z}$ along a path located entirely in the left half-plane. Two possible paths are AB and ACB. But f(z) is analytic inside the triangle ABC so that is $\displaystyle \int_{A B C A} \frac{d z}{z} = 0$ and that means that the integral computed along the paths AB and ACB are the same, i.e. the value of the integral doesn't depend from the path if it located entirely in the left half plane. The most comfortable path is of course the straigh line AB...

Kind regards

$\chi$ $\sigma$

 

[Fermat1]

Any path that leaves to the left point z = 0 is valid, so that you can choose a straight path. Setting $\displaystyle z= -\sqrt{3} + i\ t$ You have...

$\displaystyle \int_{C} \frac{d z}{z} = \int_{1}^{-1} \frac{d t}{- \sqrt{3} + i\ t} = i \ln \frac{\sqrt{3}- i}{\sqrt{3} + i}\ (1)$

Kind regards

$\chi$ $\sigma$

you should have factored in $z'=i$. So I get $ln(-{\sqrt{3}}-i)-ln(-{\sqrt{3}}+i)$. Can I use log laws even though its complex log? Cheers

 

[chisigma]

you should have factored in $z'=i$. So I get $ln(-{\sqrt{3}}-i)-ln(-{\sqrt{3}}+i)$. Can I use log laws even though its complex log? Cheers

Yes, You can! (Happy) ...

Kind regards

$\chi$ $\sigma$

 

[Fermat1]

Yes, You can! (Happy) ...

Kind regards

$\chi$ $\sigma$

So you admit you made a mistake? Regarding the subtraction law, Wikipedia says they may differ by a multiple of 2pi

 

[MarkFL]

So you admit you made a mistake?...

Perhaps a more diplomatic approach would be:

"Was I correct in needing to factor in $i$?"

 

[chisigma]

So you admit you made a mistake? Regarding the subtraction law, Wikipedia says they may differ by a multiple of 2pi

View attachment 2772

If Wikipedia says so ... what can we do as humans? (Tmi)...

Kind regards

$\chi$ $\sigma$