# How do you write (a-bi)/(c-di) in standard form?

Sep 21, 2016

$\frac{a - b i}{c - \mathrm{di}} = \frac{a c + b d}{{c}^{2} + {d}^{2}} + i \frac{a d - b c}{{c}^{2} + {d}^{2}}$

#### Explanation:

To write $\frac{a - b i}{c - \mathrm{di}}$ in standard form, what is needed is first rationalize the denominator.

To rationalize denominator, we need to multiply it by the conjugate complex of denominator $c - \mathrm{di}$, which is $c + \mathrm{di}$. Hence,

$\frac{a - b i}{c - \mathrm{di}}$

= $\frac{\left(a - b i\right) \left(c + \mathrm{di}\right)}{\left(c - \mathrm{di}\right) \left(c + \mathrm{di}\right)}$.

= $\frac{a \left(c + \mathrm{di}\right) - b i \left(c + \mathrm{di}\right)}{c \left(c - \mathrm{di}\right) + \mathrm{di} \left(c - \mathrm{di}\right)}$

= $\frac{a c + a \mathrm{di} - b c i - b {\mathrm{di}}^{2}}{{c}^{2} - c \mathrm{di} + c \mathrm{di} - {d}^{2} {i}^{2}}$

= $\frac{a c + a \mathrm{di} - b c i + b d}{{c}^{2} \pm {d}^{2}}$

= $\frac{a c + b d}{{c}^{2} + {d}^{2}} + i \frac{a d - b c}{{c}^{2} + {d}^{2}}$