Question

-3-6i divided by 3+2i please help

Answer 1

To divide complex numbers, you multiply by the conjugate of the denominator:

`(-3-6i)/(3+2i)\cdot(3-2i)/(3-2i)`

We do this to turn the denominator into a real number.

`\begin{aligned}(-3-6i)/(3+2i)\cdot(3-2i)/(3-2i) &= (-9+6i-18i+12i^2)/(9-4i^2)\\[0.5em] &= (-9-12i-12)/(9+4)\\[0.5em] &= (-21-12i)/(13)\\[0.5em] &= -(21)/(13)-(12)/(13)i\end{aligned}`

Answer 2

`-(21)/(13)-(12)/(13)i`

Explanation:

Given rational expression:

`(-3-6i)/(3+2i)`

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

`\implies (-3-6i)/(3+2i) \cdot (3-2i)/(3-2i)`

Simplify:

`\implies ((-3-6i)(3-2i))/((3+2i)(3-2i))`

`\implies (-9+6i-18i+12i^2)/(9-4i^2)`

`\implies (-9-12i+12i^2)/(9-4i^2)`

Apply the imaginary number rule
`i^2=-1` :

`\implies (-9-12i+12(-1))/(9-4(-1))`

`\implies (-9-12i-12)/(9+4)`

`\implies (-21-12i)/(13)`

`\textsf{Apply the fraction rule} \quad (a-b)/(c)=(a)/(c)-(b)/(c):`

`\implies -(21)/(13)-(12)/(13)i`