The pdf of the zero-inflated beta distribution is as follows:
\begin{equation} f(y; \mu, \sigma, \nu) = \begin{cases} \nu & \text{if $y=0$}\\ (1-\nu)f(y;\mu, \sigma) & \text{if $y \in (0,1)$} \end{cases} \label{eqn:5} \end{equation}
where $f(y;\mu, \sigma)$ is $\frac{\Gamma(\sigma)}{\Gamma(\mu\sigma)\Gamma((1-\mu)\sigma))}*y^{\mu\sigma-1}*(1-y)^{(1-\mu)\sigma-1}$
The variance is claimed to be:
$$(1-\nu)*\frac{\mu*(1-\mu)}{\sigma+1} + \nu*(1-\nu)*\mu^2$$
How is that variance derived?
If $X \sim \operatorname{Beta}(\mu, \sigma)$ with the parametrization you have specified, and $B \sim \operatorname{Bernoulli}(1-\nu)$ with $\Pr[B = 1] = 1-\nu$, where $B$ and $X$ are independent, then $Y = BX$ is zero-inflated beta with the density you have also specified. Then by the law of total variance, $$\begin{align} \operatorname{Var}[Y] &= \operatorname{E}[\operatorname{Var}[BX \mid B]] + \operatorname{Var}[\operatorname{E}[BX \mid B]] \\ &= \operatorname{E}[B^2 \operatorname{Var}[X]] + \operatorname{Var}[B\operatorname{E}[X]] \\ &= \operatorname{Var}[X]\operatorname{E}[B^2] + \operatorname{E}[X]^2\operatorname{Var}[B]. \end{align}$$ Since $\operatorname{E}[B^2] = \operatorname{E}[B] = 1-\nu$ and $\operatorname{Var}[B] = \nu (1-\nu)$, the rest is straightforward; all that is needed is to compute the mean and variance of $X$ for the choice of parametrization you used.