Zeros of Eisenstein series for full modular group $SL_{2}(\mathbb{Z})$.

Question

I am reading the paper "On the zeros of Eisenstein series" link. In the third paragraph of which it says that, "picking out the four terms with $c^{2}+d^{2}= 1$, we obtain $ F_{k}(\theta)=2 \cos(k \theta/2)+R$". I am not able to conclude the above observation. $$ $$ Here is my effort: $ $ Since $E_{k}(\theta)$ is uniformly and absolutely convergent ( hence $F_{k}(\theta)$ ). So we can rearrange the double sum by dividing the sum into two parts, first part $c^{2}+d^{2}=1$ and second part $c^{2}+d^{2}>1$. Then, $$ F_{k}(\theta)= \frac{1}{2}\left[ \frac{1}{(e^{i \theta/2}+e^{-i \theta/2})^{k}}+\frac{1}{(e^{i \theta/2}-e^{-i \theta/2})^{k}}+\frac{1}{(-e^{i \theta/2 }+e^{-i \theta/2})^{k}}+\frac{1}{(-e^{i \theta/2}-e^{-i \theta/2})^{k}} \right]+R$$ Since $k$ is even we have that $$ F_{k}(\theta)=\frac{1}{(e^{i \theta/2}+e^{-i \theta/2})^{k}}+\frac{1}{(e^{i \theta/2}-e^{-i \theta/2})^{k}}+R$$ Now using $ \cos(a)=(e^{ia}+e^{-ia})/2 $ and $ \sin(a)=(e^{ia}-e^{-ia})/2i $, we get $$ F_{k}(\theta)=\frac{1}{(2\cos(\theta/2))^{k}}+\frac{1}{(2i \sin(\theta/2))^{k}}+R$$ After this point I got stuck. Any help would be appreciated! Thanks in advance!

Answer 1

If $c^2+d^2=1$ then either $c=\pm 1$ and $d=0$ or $c=0$ and $d=\pm 1$. Moreover, since $k$ is even the term with $c=1,d=0$ is equal to the term with $c=-1,d=0$, and similarly for $c=0,d=\pm 1$.

From the terms $c=\pm 1$, $d=0$ we get $e^{-\frac{ik\theta}{2}}$, and from the terms $c=0,d=\pm 1$ we get $e^{\frac{ik\theta}{2}}$, hence the total is $$ e^{-\frac{ik\theta}{2}}+e^{\frac{ik\theta}{2}}=2\cos\frac{k\theta}{2}$$