# A Modular Forms: Non-holomorphic Eisenstein Series E2 identity

#### binbagsss

Hi,

As part of showing that $E^*_{2}(-1/t)=t^{2}E^*_{2}(t)$

where $E^*_{2}(t)= - \frac{3}{\pi Im (t) } + E_{2}(t)$

And since I have that $t^{-2}E_{2}(-1/t)=E_{2}(t)+\frac{12}{2\pi i t}$

I conclude that I need to show that $\frac{-1}{Im(-1/t)}+\frac{2t}{i} = \frac{-t^{2}}{Im(t)}$ 

where $t$ inside the upper half plane
Im=imaginary

I also have the identification $Im(\gamma.t)=\frac{Im(t)}{|ct+d | ^{2}}$ ,

where $\gamma$ is inside $SL_{2}(Z)$, the modular group of 2x2 matrices with integer numbers and determinant 1, (apologies I'm unsure how you do a matrix in latex), with components $\gamma_{11}=a,\gamma_{12}=b, \gamma_{21}=c, \gamma_{22}=d$ .

So this is part of a question where I am showing that $E_{2}(t)^*$ is weakly modular via showing that for the generators $T$ and $S$ the relevant identifications hold,$S$ and $T$ inside $SL_{2}(Z)$, so here, the one for $S$ being that :

$f(-1/t)=t^{k}f(t)$, $S$ is the matrix with components $a=0,b=-1,c=1,d=0$

Now I am looking at  and, using $S$ as $\gamma$ that
$Im(S.t)=\frac{Im(t)}{t^{2}}$, but we also know $S.t=-1/t$ and therefore I have $Im(-1/t)=\frac{Im(t)}{t^{2}}$, and so I have no idea how I'm going to get a $\frac{-2t}{i}$ term in 

Many thanks in advance.

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