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As the title says, I am to find the modulus of $f(z) = \frac{z-2a+a^2\bar{z}}{1-a\bar{z}-\bar{a}z + |a|^2}$ where $a,z \in \mathbb{C}$.

I'm a bit confused as to where to start. Would I translate the function into Cartesian form first? If so, I'm not sure how to start.

Thanks anyway.

HoopsMcCann
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    You should recognize the denominator from your other recent question. Is there any additional condition, such that $,|z|=1,$, here too? – dxiv Jun 01 '17 at 18:37
  • @dxiv The denominator is exactly the same from my previous question, I just applied the condition $|z|=1$ to it. I am to show that if $|z|=1$ and $1-\bar{a}z \neq 0$ $\Rightarrow$ $|f(z)|=1$ – HoopsMcCann Jun 01 '17 at 18:48
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    if |z|=1 The question does not mention that condition. You should edit it in. – dxiv Jun 01 '17 at 18:51
  • @dxiv after applying all the conditions I am left with the function above, I just need to find the modulus of it – HoopsMcCann Jun 01 '17 at 18:56
  • If the problem specifies $|z|=1$ and you are working on that assumption, then you must spell that out in the body of the question. – dxiv Jun 01 '17 at 19:02

1 Answers1

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Assuming (as mentioned by OP in a comment) that $|z|^2=1 \iff z \bar z = 1 \iff \bar z = \cfrac{1}{z}\,$:

$$\require{cancel} f(z)=\frac{\cfrac{1}{\bar z}-2a+a^2 \bar z}{1-a\bar{z}-\bar{a}z + |a|^2|z|^2}=\frac{\left(\cfrac{1}{\bar z}-a\right)\bcancel{\left(1-a \bar z\right)}}{(1- \bar a z)\bcancel{(1-a \bar z)}}=\frac{1}{\bar z}\cdot\frac{1 - a \bar z}{1-\bar a z} $$

Then, using that $\,\overline{1 - \bar a z}=1-\overline{\bar a z}=1-a \bar z\,$:

$$ |f(z)|=\left|\frac{1}{\bar z}\right|\cdot\left|\frac{1 - a \bar z}{1-\bar a z}\right|=1 \cdot \left|\frac{\overline{1 - \bar a z}}{1-\bar a z}\right|=1 $$

dxiv
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