Vtyjkyuk

I was given the a contour integral problem which I am rather unsure about.

First, I was asked to classify the singularities and required branch cuts for

$$f(z)=frac1z^3log[(1-z)]^2.$$

I know that $z=0$ is a pole of order 3, and $z=1$ is a branch point, and thus for the keyhole contour, I need to perform a branch cut of $[1,infty)$. I also know that the residue at $z=0$ is $1$, which I obtained using a Laurent expansion.

What confuses me is the required phases $e^iphi$ necessary to obtain the integral

$$int_1^infty fracln(x-1)x^3 textdx$$

from the contour integral

$$oint_Gamma f(z)text dz,$$

where $Gamma$ is the keyhole contour.

I know that for normal keyhole contour integrals with positive axis branch cut, for the straight portion of the contour approaching from $x=infty$ to $x=1$, I need to include a phase of $e^2ipi$. However, I also noticed that the function inside the logarithm is $x-1$ instead of $1-x$, so I suspect I need to include a phase $e^ipi$ representing the minus one somewhere. I am confused as to how to account for them, the signs of the phase for the straight portions of the contour, and where I should even be placing them in the first place. All help is appreciated, thank you!

You'll have 3 integrals for your keyhole contour (I've gone CCW around the point $z=1$),
$int_infty^1 f + int_z-1 f + int_1^infty f$.

The natural logarithm has its usual branch cut along the $-Re$ axis. The argument (angle) of whatever the natural log is taken of (here, $1-z$) therefore has to stay in the branch: $-pi < arg(1-z) le pi$.

A point $z$ on the first contour ($int_infty^1$, running just above the $+Re$ axis), maps to a $1-z$ in the 3rd quadrant (just below or in the $-Im$ direction of the $-Re$ axis--ie an argument of $-pi+epsilon$). To allow integration of the logarithm, we'd like this value of $1-z$ to be put in $re^itheta$ form. For the angle $theta$, it should be easy to see it's $-pi+epsilon$. For the modulus, we get $x-1$, where $x=Re(z)$. This gives $(x-1)e^-i(pi-epsilon)$ for this first contour (where note we've been careful to keep the angle within the stated branch). Similar reasoning gives $(x-1)e^i(pi-epsilon)$ for the third contour (epsilon arbitrary). Hope this helps.