I'm trying to understand the Hochschild-Serre spectral sequence by an example. Consider the short exact sequence of groups: $1\to N\to G\to G/N\to 1$ where $G\cong \mathbb{Z}_4$, $N\cong\mathbb{Z}_2$. Assume $G$ acts on the module $\mathbb{C}^*$ by conjugation (the generator of $G$ sends $a+bi$ to $a-bi$). Note that $N$ acts trivially on the module. The inflation-restriction exact sequence that comes from the Hochschild-Serre spectral sequence is:

$1\to H^1(G/N,\mathbb{C}^*)\to H^1(G,\mathbb{C}^*)\to H^1(N,\mathbb{C}^*)^{G/N}\to H^2(G/N,\mathbb{C}^*)\to H^2(G,\mathbb{C}^*)$

I computed these groups explicitly and got that they are:

$1\to 1\to 1\to \mathbb{Z}_2\to \mathbb{Z}_2\to H^2(G,\mathbb{C}^*)$

and the last morphism is trivial. This means that $d_2:H^1(N,\mathbb{C}^*)^{G/N}\to H^2(G/N,\mathbb{C}^*)$ is nontrivial. But I fail to compute it explicitly.

I begin with a cocycle $\alpha\in H^1(N,\mathbb{C}^*)^{G/N}$. Observe that $d_1(\alpha)$ should be trivial as an element of $E_1^{1,1}$. Therefore $d_1(\alpha)$ is in the image of $d_0:E_0^{1,0}\to E_0^{1,1}$ (a coboundary). Deriving a preimage of this coboundary should give me the desired element of $H^2(G/N,\mathbb{C}^*)$. But when I try to follow these steps with a specific cocycle (the one that sends the nontrivial element of $N$ to $-1$), I see that $d_1(\alpha)$ is trivial already in $E_0^{1,1}$ because $\alpha=\bar{\alpha}$. Moreover, $d_0:E_0^{1,0}\to E_0^{1,1}$ is trivial since $N$ acts trivially on the module. I conclude that $d_2$ is trivial, contradicting the inflation-restriction exact sequence.

What am I doing wrong?