#### Answer

$\dfrac{4-3i}{4+3i}=\dfrac{7}{25}-\dfrac{24}{25}i$

#### Work Step by Step

$\dfrac{4-3i}{4+3i}$
Begin the evaluation of the quotient by multiplying the numerator and the denominator by the complex conjugate of the denominator:
$\dfrac{4-3i}{4+3i}\cdot\dfrac{4-3i}{4-3i}=\dfrac{(4-3i)^{2}}{4^{2}-(3i)^{2}}=...$
Evaluate the operations indicated in the numerator and in the denominator:
$...=\dfrac{16-2(4)(3i)+(3i)^{2}}{16-9i^{2}}=\dfrac{16-24i+9i^{2}}{16-9i^{2}}=...$
Substitute $i^{2}$ by $−1$ and simplify:
$...=\dfrac{16-24i+9(-1)}{16-9(-1)}=\dfrac{16-24i-9}{16+9}=\dfrac{7-24i}{25}=...$
$...=\dfrac{7}{25}-\dfrac{24}{25}i$