Asymmetric confidence intervals Suppose we have iid data $X_i$ with known variance $\sigma^2$, and wish to write an asymptotic $1-\alpha $ coverage CI for the population mean $\mu$. CLT implies that if $z_q$ represents the $q$ quantile of a standard normal,
$$z_{\alpha/2}=-z_{1-\alpha/2}\leq \frac{\bar X-\mu}{\sigma/\sqrt  n}\leq z_{1-\alpha/2}$$
occurs (asymptotically) with probability $1-\alpha$ and thus implies a CI for $\mu$ of $\bar X\pm z_{1-\alpha/2}\frac{\sigma}{\sqrt n}.$
Any particular reason we take symmetric bounds, or is this just a matter of simplicity? For instance, it seems to me we could have also used
$$ z_{q_1}\leq \frac{\bar X-\mu}{\sigma/\sqrt  n}\leq z_{q_2}$$
for any $q_2-q_1=1-\alpha.$

Update: By "symmetric," I mean using $q_2=1-q_1.$

The symmetric interval minimises interval length in this case
You can find a general exposition of optimal confidence intervals in this related answer.  Here I will show you how to do the relevant optimisation for a confidence interval for the mean with known variance (via the normal approximation from the CLT).  Let $0 \leqslant \theta \leqslant \alpha$ be the upper tail area for the interval and let $z_a$ denote the quantile of the standard normal distribution with upper tail area $a$.  The general interval form is:
$$\text{CI}_\mu (1-\alpha|\theta) = \Bigg[ \bar{x}_n + \frac{z_{1-\alpha+\theta}}{\sqrt{n}} \cdot \sigma, \bar{x}_n + \frac{z_{\theta}}{\sqrt{n}} \cdot \sigma \Bigg].$$
The length of this confidence interval is:
$$\text{Length}(\theta) 
= (z_{\theta} - z_{1-\alpha+\theta}) \times \frac{\sigma}{\sqrt{n}}.$$
In order to obtain the optimal (minimum length) confidence interval of this form, we choose $\theta$ to minimise the length function.  That is, we use the value:
$$\hat{\theta} 
= \underset{0 \leqslant \theta \leqslant \alpha}{\text{arg min}} \ \text{Length}(\theta) 
= \underset{0 \leqslant \theta \leqslant \alpha}{\text{arg min}} \ (z_{\theta} - z_{1-\alpha+\theta}) 
= \frac{\alpha}{2}.$$
Using this optimisation result, we see that the length of the confidence interval is minimised using the equal tail (symmetric) interval.  Using the fact that $z_{1-\alpha/2} = - z_{\alpha/2}$ we can write this optimal confidence interval in the standard form:
$$\text{CI}_\mu (1-\alpha) = \Bigg[ \bar{x}_n - \frac{z_{\alpha/2}}{\sqrt{n}} \cdot \sigma, \bar{x}_n + \frac{z_{\alpha/2}}{\sqrt{n}} \cdot \sigma \Bigg].$$
