Complex Analysis - Bruce P . Palka - Ex: 4.12

Solution 1:

Let's show $\text {Re}\,\bar z (z^2-1)^{1/2}\ge 0,$ where throughout I use the principal branches of $\arg z, z^{1/2}.$ This is obviously true if $z=1,$ so below I'll assume $z\ne 1.$

Fact: If $0 \le \arg w < \pi,$ then $\arg (w-1) \ge \arg w.$ Hopefully that's clear from the geometry. Suppose now $\text {Re}\, z > 0$ and $0\le\arg z < \pi/2.$ Then $\arg z^2 = 2 \arg z.$ Thus $\arg (z^2-1) \ge \arg z^2 = 2\arg z.$ Taking square roots gives $\arg (z^2-1)^{1/2} \ge \arg z.$ Since $\arg \bar z = -\arg z,$ we have $\arg \bar z (z^2-1)^{1/2}\ge 0,$ hence $\text {Re}\,\bar z (z^2-1)^{1/2}\ge 0.$ That takes care of the first quadrant; the proof for the fouth quadrant is similar.