Log(z) gives an anti-derivative of 1/z

[squenshl]

How do I calculate
$$\int_C$$ dz/1-z when C is a line segment from 1/2 to $$\sqrt{3i}$$.
I know that in the complement of the negative real axis, log(z) gives an anti-derivative of 1/z.

 

[ice109]

you need to use calculus of residues

 

[elibj123]

you need to use calculus of residues

I don't see a convinient way to use residues since this is not an integral of a close curve.

What you need to do, is to do a parametrization z(t) of the line segement.
Then you substitute z with a corresponding expression of t, and dz with a corresponding expression of dt, and then the integral comes down to integration over a real variable.

 

[HallsofIvy]

How do I calculate
$$\int_C$$ dz/1-z when C is a line segment from 1/2 to $$\sqrt{3i}$$.
I know that in the complement of the negative real axis, log(z) gives an anti-derivative of 1/z.

I hope that last was a typo. The derivative of log(z) is 1/z. The anti-derivative is zlog(z)- z. Also, was that supposed to be [tex]i\sqrt{3}[/itex] rather than $$\sqrt{3i}$$? That is, is the i supposed to be inside the square root or not?<br /> <br /> Looks like tedious but straightforward computation from me. Assuming you mean [tex]i\sqrt{3}[/itex], C can be written $z= (1/2)(1+ t)+i\sqrt{3} t$ so that $dx= (-1/2+ i\sqrt{3})dt$ and $1- z= (1/2)(1+ t)- i\sqrt{3}t$ with t going from 0 to 1.<br /> <br /> Put those into the integral and you should be able to reduce it to two integrals, one of the form $1/z^2$ and the other $1/(z^2+ 1)$.<br /> <br /> If you <b>really</b> meant [tex]\sqrt{3i}[/itex], then <br /> [tex]\sqrt{3i}= \sqrt{\frac{3}{2}}+ i\sqrt{\frac{3}{2}}[/itex][/tex][/tex][/tex][/tex]

 

[squenshl]

It was a typo. Oopsy daisys.
dz not dx

 

[squenshl]

Can we not use the Cauchy integral formula because it is not a closed curve.

 

[squenshl]

I didn't do it that way but i got -0.944 + 0.117i

 

[jackmell]

You can only use Cauchy's Integral Formula if it's a closed contour. Of course you could add two more legs and make it closed but that's not necessary. It helps if you write it clearly as an integral:

$\mathop\int\limits_{1/2}^{\sqrt{\frac{3}{2}}+i\sqrt{\frac{3}{2}}} \frac{1}{z}dz$

Keep in mind it's logarithmic antiderivative can be made analytic along a straight-line contour between the end points so you can just evaluate the antiderivative at those endpoints as long as you use an analytic extension of log in a region containing the contour, say the branch that has a cut along the negative real axis. Can you then use that branch and compute:

$\log(z)\biggr|_{1/2}^{\sqrt{\frac{3}{2}}+i\sqrt{\frac{3}{2}}$