Why does $x= \frac{1}{2}(z+\bar{z}) = \frac{1}{2}(z+\frac{r^{2}}{z})$ on the circle?

Question

I some help computing

$$\int_{|z|=r} x \, dz$$

by noting that $x= \frac{1}{2}(z+\bar{z}) = \frac{1}{2}(z+\frac{r^{2}}{z})$ on the circle but I don't understand why this is true. Why does the complex conjugate of $z$ equal this on the circle?

Answer 1

Suppose that $|z|=r$.

Since $x$ is the real part of $z$, we have $$x = \frac{1}{2}(z + \bar{z}).$$

However, $z\bar{z}=|z|^2.$ Assuming that $z\ne 0$, which is equivalent to assuming that $r\ne 0$, we may divide by $z$ to get $$\bar{z} = \frac{|z|^2}{z}.$$

Substituting this into the first equation, and recalling that $|z|=r$, the result follows; $$x = \frac{1}{2}\left(z + \frac{|z|^2}{z}\right) = \frac{1}{2}\left(z + \frac{r^2}{z}\right).$$