Question

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

Answer 1

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

`-2/(3-i) = -3/5 - 1/5i`

Explanation:

The conjugate of a complex number `a+bi` is `a-bi`. 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.

`-2/(3-i) = -2/(3-i)*(3+i)/(3+i)`

`= (-2(3+i))/((3-i)(3+i))`

`= (-6-2i)/(9 +3i -3i +1)`

`= (-6-2i)/10`

`= -3/5 - 1/5i`

Answer 2

`=0.632/_0.3217`. (in polar form)

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

Explanation:

There are 2 methods to do this -
Method 1
First convert everything to polar form then use the formula
`(z_1)/(z_2)=(r_1)/(r_2)cis(theta_1-theta_2)`

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

`=0.632/_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
`(a+ib)*(x+iy)=(ax-by)+i(bx+ay)`.

`therefore (-2+0i)/(3-i)*(3+i)/(3+i)=(-6-2i)/(3^2+1^2)`

`=1/10(-6-2i)`

`=-0.6-0.2i`