# How do you find abs( 2sqrt3 - πi )?

For a complex number, $\left\mid a + b i \right\mid = \sqrt{{a}^{2} + {b}^{2}}$, or the distance of the segment.
So, $\left\mid 2 \sqrt{3} - \pi i \right\mid = \sqrt{{\left(2 \sqrt{3}\right)}^{2} + {\left(- \pi\right)}^{2}} = \sqrt{4 \left(3\right) + {\pi}^{2}} = \sqrt{12 + {\pi}^{2}}$.