# How do you find abs(1 - 5i)?

$\sqrt{26}$.
Any complex number has the form $z = a + b i$, where $a$ is its real part and $b$ is its imaginary part. The norm of a complex number, $z$, is defined as $| z | = \sqrt{{a}^{2} + {b}^{2}}$.
In your case, the complex number $1 - 5 i$ defines $a = 1$ and $b = - 5$. So, ${a}^{2} + {b}^{2} = 1 + 25 = 26$