Consider the function $w=\frac{1}{z^2}$ on the unit circle in $z$. If we follow the unit circle in the usual positive direction once around the center, what is the net change in $\log(z)$? How about the net change in $\log(w)$?
I'm not quite sure on how to approach this. If $w$ is a rational function of a complex variable, what would it mean to take a logarithm of this function?
If $z=re^{i\theta}$ then $$\ln(z)=\ln(r)+i\theta$$ The change net change in $\log(z)$ can then be described as some change between $z_0$ and $z_1$ so $$\Delta \ln(z)=\ln(z_1)-\ln(z_0)$$
Using log rules simplifies this to $$\Delta \ln(z)=\ln\frac{z_1}{z_0}.$$ Plugging the polar value for $z$ gives $$\Delta \ln(z)=\ln(\frac{re^{i(\theta+2\pi)}}{re^{i\theta}})=i2\pi.$$
Next, defining $w=\frac{1}{z^2}$ the net change in $\log(w)$ can be expressed the method as above, $$\Delta\ln(w)=\ln(\frac{1}{z_1^2})-\ln(\frac{1}{z_0^2})$$
Using log subtraction rules $$\Delta\ln(w)=\ln((\frac{z_0}{z_1})^2)$$
Using log rules and factoring out a -2 in the exponent $$\Delta\ln(w)=-2\ln(\frac{z_1}{z_0})$$
We defined $\ln(\frac{z_1}{z_0}$ as $i2\pi$ so our net change in $\log(w)$ is $-4\pi$.