# How do you simplify (3-i)/(4-i)?

Nov 23, 2015

$\frac{13}{17} - \frac{1}{17} i$

#### Explanation:

Multiply by the complex conjugate of the denominator.

$\frac{3 - i}{4 - i} \left(\frac{4 + i}{4 + i}\right) = \frac{12 + 3 i - 4 i - {i}^{2}}{16 + 4 i - 4 i - {i}^{2}} = \frac{12 - i - {i}^{2}}{16 - {i}^{2}}$

Remember that $i = \sqrt{- 1}$, so ${i}^{2} = - 1$.

$\frac{12 - i - \left(- 1\right)}{16 - \left(- 1\right)} = \frac{13 - i}{17} = \frac{13}{17} - \frac{1}{17} i$

Notice that the answer is written in the form of a complex number $a + b i$.