Answer
$\dfrac{7}{4+3i}=\dfrac{28}{25}-\dfrac{21}{25}i$
Work Step by Step
$\dfrac{7}{4+3i}$
Multiply the numerator and the denominator of this expression by the complex conjugate of the denominator:
$\dfrac{7}{4+3i}=\dfrac{7}{4+3i}\cdot\dfrac{4-3i}{4-3i}=\dfrac{7(4-3i)}{4^{2}-(3i)^{2}}=\dfrac{7(4-3i)}{16-9i^{2}}=...$
Substitute $i^{2}$ with $-1$ and simplify:
$...=\dfrac{7(4-3i)}{16-9(-1)}=\dfrac{7(4-3i)}{16+9}=\dfrac{28-21i}{25}=\dfrac{28}{25}-\dfrac{21}{25}i$
