The power is $1-\beta$, where $\beta$ is the probability of Type II error, that is, the probability of not rejecting the null hypothesis. For a normal population with known variance, the power function would be
$\pi(\mu) = 1 - P(-z_{\alpha/2}-\dfrac{\mu-\mu_0}{\sigma/\sqrt{n}},
where $Z\sim N(0,1)$, $n$ is the size of the sample, $\sigma$ the standard deviation (sq. root of the variance) and $z_{\alpha/2}=\Phi^{-1}(1-\alpha/2)$, the inverse of standard normal cdf. Also, in your case your hypothesis is that $\mu_0=0$.
The probability is calculated by means of the standard normal cdf $\Phi(\cdot)$ :
$\pi(\mu) = 1 - \Phi(z_{\alpha/2}-\dfrac{\mu-\mu_0}{\sigma/\sqrt{n}}) + \Phi(-z_{\alpha/2}-\dfrac{\mu-\mu_0}{\sigma/\sqrt{n}})$.
If $\alpha=0.05$, that is, confidence intervals at 95%, you have $z_{\alpha/2}=1.9599639845\approx 1.96$