Answer
$|4-3i|=5$
Work Step by Step
The absolute value of the complex number $a+bi,$ written as $|a+bi|,$ is given by $\sqrt{a^2+b^2}.$ Then, the absolute value of $
4-3i
,$ is
\begin{align*}
&
\sqrt{(4)^2+(-3)^2}
\\&=
\sqrt{16+9}
\\&=
\sqrt{25}
\\&=
5
.\end{align*}
Hence, $
|4-3i|=5
.$