Trying to prove that $\left|z-(z^2+1)^{\frac{1}{2}}\right| < 1$ for $Re(z)>0$.

Question

I'm trying to prove that $$\left|z-(z^2+1)^{\frac{1}{2}}\right| < 1$$ for $Re(z)>0$, but struggling to do so. Any thoughts on this much appreciated.

Answer 1

Hint: $ z - \sqrt{z^2+1} = \frac{1 }{ z + \sqrt{z^2 + 1}} $.

Why is the norm of the denominator greater than 1?