# Complex analysis question

http://www.math.harvard.edu/~siu/math113/solution_homework_feb5a.pdf

Can someone explain me the solutions to the questions on page 1.

I am not sure how $|w-z| / |1-w \bar z|<1$ becomes $|w-z|^2 < |1-w\bar z|^2$

and later how the equation once you expand it has wz(bar)

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For the former, multiply by $|1-w\bar{z}|$. Then square, which you're allowed to do with inequalities if both sides are positive (which absolute values are), since squaring is a monotone function. – fgp Oct 20 '12 at 17:41

It is not that the first inequality "becomes" the second one. What they're saying is: if the second inequality is true then so is the first one. Why? Because

$$|w-z|^2<|1-w\overline z|^2\Longrightarrow\left(\frac{|w-z|}{|1-w\overline z|}\right)^2<1\Longrightarrow \frac{|w-z|}{|1-w\overline z|}<1$$

as everything's positive here.

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