Contour integral around the disk with radius $ r$

Question

In an exercise, we are looking at $\int_0^{2\pi}\frac{1}{re^{i\phi}+w}d\phi$ with $ w \in \Bbb C$. Now I tried to write it as an integral around the disk with radius $1$ and midpoint $1$. However my answer differs by a $-$ sign from the solution. I tried to do the following. $\delta B_r(0): [0, 2\pi]\to \Bbb C,\; \phi \mapsto re^{i\phi}$ and thus I obtain $\int_{\delta B_r(0)}\frac{i}{z(z+w)}\,dz\:$ instead of the solution $\:-\int_{\delta B_r(0)}\frac{i}{z(z+w)}\,dz$. I do not seem to understand where the minus comes into play.

Answer 1

You have the change of variable $\; z=r\mathrm e^{i\phi}$, so that $$\mathrm d z=ir\mathrm e^{i\phi}\,\mathrm d\phi=iz\,\mathrm d\phi\iff \mathrm d\phi=\frac1{i}\frac{\mathrm d z}z,$$ and $\:\dfrac1i=-i$.

Answer 2

If $z=re^{i\theta}$ then ${\rm d}z=ire^{i\theta}{\rm d}\theta=iz\,{\rm d}\theta$, and so

$${\rm d}\theta=\dfrac{1}{iz}{\rm d}z=\color{red}{-}\dfrac{i}{z}{\rm d}z$$