Answer
$\dfrac{3-4i}{4+3i}=-i$
Work Step by Step
$\dfrac{3-4i}{4+3i}$
Multiply the numerator and the denominator by the complex conjugate of the denominator:
$\dfrac{3-4i}{4+3i}=\dfrac{3-4i}{4+3i}\cdot\dfrac{4-3i}{4-3i}=\dfrac{(3-4i)(4-3i)}{4^{2}-(3i)^{2}}=...$
$...=\dfrac{12-9i-16i+12i^{2}}{16-9i^{2}}=\dfrac{12-25i+12i^{2}}{16-9i^{2}}=...$
Substitute $i^{2}$ by $-1$ and simplify:
$...=\dfrac{12-25i+12(-1)}{16-9(-1)}=\dfrac{12-12-25i}{16+9}=\dfrac{-25i}{25}=-i$
