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

The expontential of power $1/2$ means the principal square-root. The two inequalities can be verified easily when $Im(z)=0$ and $|z|<1$. 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 $\pi$.

Hua
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1 Answers1

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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 $1-2y^2 > 0$ which when solved for gives $y^2 < \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 $\frac{1}{2}$.

Icarus 369
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Erock Brox
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  • @Erok Brox Thanks for your idea but I think you misunderstand something. What do you mean by "isolate the square root function". Actually there is no equation here; I don't understand how we can move terms from one side to the other. And the result is not desirable. There should be no restriction on imaginary part of $z$ – Hua Aug 15 '16 at 09:05
  • The expression is set greater than zero and this is why I consider it as if it were an equation. – Erock Brox Aug 15 '16 at 21:31
  • @Erok Brox Er, I see. But my question is that the real prat of the expression should be greater than zero, instead of the whole expression. So I don't think we could do this before we get the explicit formula of the real part. This is the hardest step, since the formula is really complicated. – Hua Aug 16 '16 at 03:08