$Re\left((1-z^2)^{\frac{1}{2}}+iz\right) > 0$ for $z\in\mathbb{C}\setminus\{z: Im(z) =0, |z|\ge1\}$ I need to prove that $$Re\left((1-z^2)^{\frac{1}{2}}+iz\right)&gt;0,\quad Re\left(-(1-z^2)^{\frac{1}{2}}+iz\right)&lt;0$$
if $z\in\mathbb{C}\setminus \{z: Im(z) =0, |z|&gt;=1\}$. 

The expontential of power $1/2$ means the principal square-root. The two inequalities can be verified easily when $Im(z)=0$ and $|z|&lt;1$. I have no clue about how to handle the case where $Im(z)\ne 0$. I see that

$$\left((1-z^2)^{\frac{1}{2}}+iz\right)\left(-(1-z^2)^{\frac{1}{2}}+iz\right)=-1.$$

This means I just need to prove one of the inequalities and the other holds readily since the arguments of them differ by $\pi$.
Replace $z$ with $z=x+iy$ and simplify the expression. Eventually isolate the square root function and square both sides. Move all terms back to one side and take the real part of it. 

The real part I get is $1-2y^2 &gt; 0$ which when solved for gives $y^2 &lt; \frac{1}{2} $. So my conclusion is that the real part of your expression is greater than zero for any values of $y^2$ that are less than $\frac{1}{2}$.
Replace $z$ with $z=x+iy$ and simplify the expression. Eventually isolate the square root function and square both sides. Move all terms back to one side and take the real part of it. 

The real part I get is $1-2y^2 &gt; 0$ which when solved for gives $y^2 &lt; \frac{1}{2} $. So my conclusion is that the real part of your expression is greater than zero for any values of $y^2$ that are less than $\frac{1}{2}$.