# How do you find abs(4-3i)?

Nov 5, 2016

$5$

#### Explanation:

$| 4 - 3 i |$

for complex numbers the defn. is:

$| a \pm i b | = \sqrt{{a}^{2} + {b}^{2}}$

so we have:

$| 4 - 3 i | = \sqrt{{4}^{2} + {3}^{2}}$

$= \sqrt{16 + 9} = \sqrt{25} = 5$