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I am working on this problem:

Let $\alpha$ be a complex number with $0< |\alpha| < 1.$ Prove that the set of all $z$ such that $|z - \alpha| < |1 - \bar\alpha z|$ is the disc $z$ with $|z| < 1$.

I tried squaring both sides, and I got that $|z|^2 + |\alpha|^2 < 1 - |\alpha|^2 |z|^2$. I'm pretty sure that I'm most of the way there, but I can't get the last couple of steps in the proof.

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    probably you'll have to use the fact that $|\alpha|^2 = \alpha\bar{\alpha}$ – 2017-01-29
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    Could you give me some additional information? I don't see how I can apply that to the problem. – 2017-01-29
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    No don't know how to solve it. But you are given the conjugate so probably you need to apply this! – 2017-01-29
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    Hint: Separate out $|z|^2$. And check your signs on the RHS. – 2017-01-29
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    Thanks! I can't believe I didn't see that. – 2017-01-30

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Note that: $$ \require{cancel} \begin{align} |z - \alpha| < |1 - \bar\alpha z| & \iff (z-\alpha)(\bar z - \bar \alpha) \lt (1 - \bar\alpha z)(1 - \alpha \bar z) \\ & \iff z \bar z - \cancel{z \bar \alpha} - \bcancel{\alpha \bar z} + \alpha \bar \alpha \lt 1 - \bcancel{\alpha \bar z} - \cancel{z \bar \alpha} + \alpha \bar \alpha z \bar z \\ & \iff |z|^2(1 - |\alpha|^2) \lt 1 - |\alpha|^2 \\ & \iff |z|^2 \lt 1 \end{align} $$


The error in your proof attempt is here, you got the sign wrong on the RHS (it's $+|\alpha|^2|z|^2$):

I tried squaring both sides, and I got that $|z|^2 + |\alpha|^2 < 1 - |\alpha|^2 |z|^2$

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    I know that. When I squared it, I cancelled, the $2Re(\bar \alpha z)$ from both sides. – 2017-01-30
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    @RealMath Right, thanks. I corrected the respective comment. – 2017-01-30
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    You're right. Thanks for correcting it. – 2017-01-30
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Let $z=x+iy$ and $\alpha=a+ib$, substitute these into the equation. Now totally cheat & get reduce to do the algebra ... enter image description here

The first factor is non zero by hypothesis. So the second factor is less than zero.