Stationarity of the TGARCH I'm going through "GARCH models" by Francq and Zakoian (2010). They define the TGARCH(1,1) as
$$\sigma_t = \omega + \beta_1 \sigma_{t-1} + \alpha_{1,+}\epsilon_{t-1}^+ - \alpha_{1,-}\epsilon_{t-1}^- $$
where $\epsilon_{t-1}^+$ equals $\epsilon_{t-1}$ if positive and 0 if negative (opposite holds for $\epsilon_{t-1}^-$).
On p.252 they write the following TGARCH(1,1) stationarity condition 
$$E[(\alpha_{1,+}z_t^+ - \alpha_{1,-}z_t^- + \beta_1)^2]<1$$
which, assuming $N(0,1)$ innovations $z_t$, simplifies to 
$$\frac{1}{2}(\alpha_{1,+}^2 + \alpha_{1,-}^2)+\frac{2\beta_1}{\sqrt{2 \pi}}(\alpha_{1,+}+\alpha_{1,-}) + \beta_1^2<1$$
I've been trying to derive the second formula from the first one. Whereas I've obtained the last two terms of the equation, I struggle with the first one. More precisely, solving the brackets from the first equation I'm stuck at:
$$\alpha_{1,+}^2 E[z_t^2]+\alpha_{1,-}^2E[z_t^2]-2\alpha_{1,+}\alpha_{1,-}E[z_t^+z_t^-]$$
How do you go from this expression to $$\frac{1}{2}(\alpha_{1,+}^2 + \alpha_{1,-}^2)$$
PS: They also mention that $$E[z_t^+]=-E[z_t^-]=\sqrt{\frac{1}{2\pi}}$$How do you derive this result knowing that $z_t$ is $N(0,1)$ distributed?
 A: First of all, we note that 
\begin{align}
\alpha_{1,+}^2 E[z_t^2]+\alpha_{1,-}^2E[z_t^2]-2\alpha_{1,+}\alpha_{1,-}E[z_t^+z_t^-] 
\end{align}
should be 
\begin{align}
\alpha_{1,+}^2 E[(z_t^+)^2]+\alpha_{1,-}^2E[(z_t^-)^2]-2\alpha_{1,+}\alpha_{1,-}E[z_t^+z_t^-] 
\end{align}
Secondly, we note that $E[z_t^+z_t^-] = 0$ since either $z_t^+$ or $z_t^-$ will be zero (cannot both be non zero at the same time). Now,the only thing we need to show is
$$E[(z_t^+)^2] = E[(z_t^-)^2]=\frac{1}{2}$$
Intuitively, this must be the case since $N(0,1)$ is a symmetric distribution with $E[z_t^2] = 1$. To formally show it, one has to evaluate
$$E[(z_t^+)^2] = \frac{1}{\sqrt{2\pi}}\intop_{0}^{\infty} z^2\exp\left(-\frac{z^2}{2}\right)   dz $$
and
$$E[(z_t^-)^2] = \frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{0} z^2\exp\left(-\frac{z^2}{2}\right)   dz $$
In the same way, you can show that e.g. 
$$E[z_t^-] = \frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{0} z\exp\left(-\frac{z^2}{2}\right)   dz = -\frac{1}{\sqrt{2\pi}}$$
See e.g. this question for the actual steps in relation to evaluating the integrals
