# Prove that $w=\frac{\sqrt{z^2+z+1}}{z^2-3z+2}$ analytic on $|z|<1$

I've self-learning of a complex analysis course. My first difficulties are in analytic functions. In that book, that I learning from, there is no even one example on how proving such problem, like in the title:

Prove that $w=\frac{\sqrt{z^2+z+1}}{z^2-3z+2}$ analytic on $|z|<1$

Could you direct me by providing a full solution of the above?

Thank you.

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1
– Pragabhava Oct 16 '12 at 22:06

## 1 Answer

One can use a few general facts about how one can compose analytic functions:

• The sum and product of two analytic functions $f,g$ is analytic. This is clear from $(f+g)'=f'+g'$ and $(fg)'=fg'+f'g$.
• If $f(z)$ is analytic on $\Omega$ and has no zero there, then $\frac1{f(z)}$ is analytic as well. This is clear from $\frac d{dz}\frac1{f(z)}=\frac{f'(z)}{f^2(z)}$.
• If $f(z)$ is analytic on $\Omega$ and $\Omega$ is simply connected and $f(z)$ has no zero there, then $\sqrt{f(z)}$ is analytic on $\Omega$ as well. This follows by monodromy.
• The open unit disc given by $|z|<1$ is simply connected: The disc and its complement are clearly connected, it is not even difficult to give explicit paths between points.

If you know these four facts, you can combine them to conclude that $f(z)=\frac{\sqrt{z^2+z+1}}{(z-2)(z-1)}$ is analytic on $|z|<1$: The denominator is zero only at $z=1$ and $z=2$, hence not in the open disc; the radicand in the numerator has two complex root of absolute value $1$, hence none inside the disc.

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nicely done (+1) – robjohn♦ Oct 20 '12 at 14:16
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