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I'm referring to this question : Finding a trigonometric polynomial

The OP says :

On the unit circle $$f(\theta) = F(e^{i\theta}) = c \prod_{j=1}^n\frac{(e^{i\theta}-\beta_j)(1-\overline\beta_j e^{i\theta})}{(e^{i\theta}-\gamma_j)(1-\overline\gamma_j e^{i\theta})}$$ The terms in the product simplify to $$\left| \frac{e^{i\theta} - \beta_j}{e^{i\theta} - \gamma_j} \right|^2$$

I've developed the terms in the product but it's not obvious how to derive $\left| \frac{e^{i\theta} - \beta_j}{e^{i\theta} - \gamma_j} \right|^2$. A bit of help would be welcome. Thank you guys.

Community
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user322240
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1 Answers1

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Notice \begin{align*} \frac{(e^{i\theta}-\beta_j)(1-\overline\beta_j e^{i\theta})}{(e^{i\theta}-\gamma_j)(1-\overline\gamma_j e^{i\theta})} & = \frac{e^{i\theta}(1-\beta_je^{-i\theta})(1-\overline\beta_j e^{i\theta})}{e^{i\theta}(1-\gamma_je^{-i\theta})(1-\overline\gamma_j e^{i\theta})}\\ & = \frac{(1-\beta_je^{-i\theta})\overline{(1-\beta_je^{-i\theta})}}{(1-\gamma_je^{-i\theta})\overline{(1-\gamma_je^{-i\theta})}} \\ & =\left| \frac{1-\beta_je^{-i\theta}}{1-\gamma_je^{-i\theta}} \right|^2\\ & =\left| \frac{e^{-i\theta}}{e^{-i\theta}} \right|^2 \left| \frac{e^{i\theta}-\beta_j}{e^{i\theta}-\gamma_j} \right|^2\\ & =\left| \frac{e^{i\theta}-\beta_j}{e^{i\theta}-\gamma_j} \right|^2. \end{align*}

C. Dubussy
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