We have the elliptic lambda function: $$\lambda(\tau)=\frac{e_3-e_2}{e_1-e_2}$$ We want to look at how $\lambda$ changes under a modular transformation: $$\omega'_2=a\omega_2+b\omega_1$$ $$\omega'_1=c\omega_2+d\omega_1$$ Now for $a$ and $d$ odd and $b$ and $c$ even we have $$\lambda\left (\frac{a\tau+b}{c\tau+d}\right )=\lambda(\tau)$$ Taking matrices outside of this we can just look at $\begin{pmatrix} 1&1\\0 &1 \end{pmatrix}$ and $\begin{pmatrix} 0&1\\1&0\end{pmatrix}$ as all other matrices in question can be composed with these. The first gives $$\lambda(\tau+1)=\frac{\lambda(\tau)}{\lambda(\tau)-1}$$ However I am having trouble with the second matrix. I have since $a=0$, $b=1$, $c=1$, $d=0$ that we have $\omega_2'\equiv \omega_1$ and $\omega_1'\equiv \omega_2$. And therefore this means that $e_1$ and $e_2$ are interchanged. Hence we have that $$\tau'=\frac{\omega_2'}{\omega_1'}=\frac{\omega_1}{\omega_2}=\frac{1}{\tau}$$ Or that $$\lambda\left( \frac{a\tau+b}{c\tau+d}\right )=\lambda\left ( \frac{1}{\tau}\right )$$ Therefore $$\lambda\left (\frac{1}{\tau}\right )=\frac{e_3-e_1}{e_2-e_1}=1-\frac{e_3-e_2}{e_1-e_2}=1-\lambda(\tau)$$

However looking at wolfram alpha http://www.wolframalpha.com/input/?i=lambda(-1%2Ftau) and Ahlfors complex analysis text they give the equation as

$$\lambda\left (-\frac{1}{\tau}\right )=1-\lambda(\tau)$$ Where does the negative sign come from? Surely permuting $\omega_1$ and $\omega_2$ just means we have $\frac{1}{\tau}$ without the minus? Sorry for the long explanation.