For 90% confidence level, you need to find a z-value so that
\begin{equation*}
P( -z_{ \alpha/2} \lt z \lt z_{ \alpha/2}) = 0.9 = 1 - 0.1 .
\end{equation*}
Using the symmetry of the normal distribution, this can be rewritten
\begin{equation*}
F(z_{ \frac{0.1}{2}}) = P( z \lt z_{ \frac{0.1}{2}}) = 0.95 = 1 - \frac{0.1}{2} .
\end{equation*}
Using the inverse of the standard normal distribution (on the TI calculator this is InvNorm(0.95)) gives \(z_{ 0.05} \approx 1.645\text{.}\)