Simplify $|1 - \frac{5}{6}e^{-i2\pi f}+\frac{1}{6}e^{-i2\pi f2}|^2$ to real valued form easily

Question

How can I simplify $$|1 - \frac{5}{6}e^{-i2\pi f}+\frac{1}{6}e^{-i2\pi f2}|^2$$ to real-valued form easily?

Answer 1

$\def\ca{\cos(\alpha)} \def\caa{\cos(2\alpha)} \def\sa{\sin(\alpha)} \def\saa{\sin(2\alpha)} \def\fa{\tfrac{5}{6}} \def\fb{\tfrac{1}{6}}$

Let $\alpha=2\pi f$ which gives \begin{align*} z &=1-\fa e^{i\,(-2\pi f)}+\tfrac{1}{6}e^{i\,(-4\pi f)} \\&=1-\fa e^{i\,(-\alpha)}+\fb e^{i\,(-2\alpha)} \\&=1-\fa\bigl(\ca-i\sa\bigr)+\fb\bigl(\caa-i\saa\bigr) \\&=1-\fa\ca+i\fa\sa+\fb\caa-i\fb\saa \\&=1-\fa\ca+\fb\caa+i\bigl(\fa\sa-\fb\saa\bigr). \end{align*}

We have \begin{align*} \lvert z \rvert^2 &=z\cdot\overline{z} \\&=\Bigl(1-\fa\ca+\fb\caa+i\bigl(\fa\sa-\fb\saa\bigr)\Bigr) \\&\qquad\cdot\Bigl(1-\fa\ca+\fb\caa-i\bigl(\fa\sa-\fb\saa\bigr)\Bigr) \\&=\bigl(1-\fa\ca+\fb\caa\bigr)^2+\bigl(\fa\sa-\fb\saa\bigr)^2 \\&=\tfrac{1}{36}\bigl(6-5\ca+\caa\bigr)^2+\tfrac{1}{36}\bigl(5\sa-\saa\bigr)^2 \end{align*} which is equivalent to \begin{align*} 36\lvert z \rvert^2 &=\bigl(6-5\ca+\caa\bigr)^2+\bigl(5\sa-\saa\bigr)^2 \\&=\bigl(6-5\ca\bigr)^2+2\bigl(6-5\ca\bigr)\caa+\caa^2 \\&\qquad{}+25\sa^2-10\sa\saa+\saa^2 \\&=36-60\ca+25\ca^2 \\&\qquad{}+12\caa-10\ca\caa+\caa^2 \\&\qquad{}+25\sa^2-10\sa\saa+\saa^2 % \\&=36-60\ca+25(\ca^2+25\sa^2) \\&\qquad{}+(\caa^2+\saa^2)+12\caa \\&\qquad{}-10\ca\caa-10\sa\saa % \\&=36-60\ca+25+1+12\caa \\&\qquad{}-10\ca\caa-10\sa\saa % \\&=62-60\ca+12\caa \\&\qquad{}-10\ca\caa-10\sa\saa % \\&=62-60\ca+12\caa \\&\qquad{}-10\bigl(\ca\caa+\sa\saa\bigr) % \\&=62-60\ca+12\caa-10\ca % \\&=62-70\ca+12\caa \end{align*} since $$ \ca =\cos(2\alpha-\alpha) =\cos(2\alpha)\cos(\alpha)+\sin(2\alpha)\sin(\alpha). $$

Thus $$ 36\lvert z \rvert^2=62-70\ca+12\caa $$ i.e. \begin{align*} \lvert z \rvert^2 &=\tfrac{1}{36}\bigl(62-70\ca+12\caa\bigr) \\&=\tfrac{1}{18}\bigl(31-35\ca+6\caa\bigr) \\&=\tfrac{1}{18}\bigl(31-35\cos(2\pi f)+6\cos(4\pi f)\bigr). \end{align*}