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

Nov 29, 2016

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

#### Explanation:

The standard form is $\textcolor{b l u e}{a} + \textcolor{red}{b} i$

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

Multiply the numerator and the denominator by the conjugate of the denominator

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

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

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

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

${i}^{2} = - 1$

$= \frac{a c + \left(c b - a d\right) i - b d \left(- 1\right)}{{c}^{2} - {d}^{2} \left(- 1\right)}$

$= \frac{a c + \left(c b - a d\right) i + b d}{{c}^{2} + {d}^{2}}$

$= \frac{a c + b d + \left(c b - a d\right) i}{{c}^{2} + {d}^{2}}$

$= \frac{a c + b d}{{c}^{2} + {d}^{2}} + \frac{\left(c b - a d\right) i}{{c}^{2} + {d}^{2}}$

$= \textcolor{b l u e}{\frac{a c + b d}{{c}^{2} + {d}^{2}}} + \textcolor{red}{\frac{c b - a d}{{c}^{2} + {d}^{2}}} i$