# How do you simplify -2/(3-i)?

Dec 29, 2015

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

$- \frac{2}{3 - i} = - \frac{3}{5} - \frac{1}{5} i$

#### Explanation:

The conjugate of a complex number $a + b i$ is $a - b i$. The product of a complex number and its conjugate is a real number. We will use this property to produce a real number in the denominator of the given expression.

$- \frac{2}{3 - i} = - \frac{2}{3 - i} \cdot \frac{3 + i}{3 + i}$

$= \frac{- 2 \left(3 + i\right)}{\left(3 - i\right) \left(3 + i\right)}$

$= \frac{- 6 - 2 i}{9 + 3 i - 3 i + 1}$

$= \frac{- 6 - 2 i}{10}$

$= - \frac{3}{5} - \frac{1}{5} i$

Dec 29, 2015

$= 0.632 \angle 0.3217$. (in polar form)

$= - 0.6 - 0.2 i$. (in rectangular form)

#### Explanation:

There are 2 methods to do this -
Method 1
First convert everything to polar form then use the formula
$\frac{{z}_{1}}{{z}_{2}} = \frac{{r}_{1}}{{r}_{2}} c i s \left({\theta}_{1} - {\theta}_{2}\right)$

therefore (-2/pi)/((sqrt(3^2+1^2))/_tan^-1(-1/3))=-2/sqrt10 /_ (pi-(-0.32175)

$= 0.632 \angle 3.4633$. (Principle angle arguement is 0.3217 rad).

Method 2
Multiply the quantity by 1, selecting 1 as the complex conjugate of the denominator over itself.
Then the resultant denominator is a real number, and the numerator we multiply in rectangular form using the rule
$\left(a + i b\right) \cdot \left(x + i y\right) = \left(a x - b y\right) + i \left(b x + a y\right)$.

$\therefore \frac{- 2 + 0 i}{3 - i} \cdot \frac{3 + i}{3 + i} = \frac{- 6 - 2 i}{{3}^{2} + {1}^{2}}$

$= \frac{1}{10} \left(- 6 - 2 i\right)$

$= - 0.6 - 0.2 i$