Question

If z₁ = 2 – i and z₂ = – 4 + 3i, find the inverse of z₁, z₂ and `("z"_1)/("z"_2)`

Answer 1

z₁ = 2 – i, z₂ = – 4 + 3i

z₁z₂ = (2 – i)(– 4 + 3i)

= – 8 + 3 + 4i + 6i

= – 5 + 10i

Inverse of z₁z₂ = (z₁z₂)^–1

= `1/("z"_1"z"_2)`

= `1/(- 5 + 10"i") xx (- 5 - 10"i")/(- 5 - 10"i")`

= `(- 5 - 10"i")/((- 5)^2 - (10"i")^2`

= `(- 5 - 10"i")/(25 + 100)`

= `(-5(1 + 2"i"))/125`

= `(-(1 + 2"i"))/25`

`("z"_1)/("z"_2) = (2 - "i")/(-4 + 3"i") xx (- 4 - 3"i")/(-4 - 3"i")`

= `(- 8 - 6"i" + 4"i" + 3"i"^2)/((- 4)^2 - (3"i")^2`

= `(- 8 - 2"i" - 3)/(16 + 9)`

= `(- 11 - 2"i")/25`

= `(-(11 + 2"i"))/25`

Inverse of `("z"_1)/("z"_2) = (("z"_1)/("z"_2))^-1`

= `("z"_2)/("z"_1)`

= `(- 25)/(11 + 2"i") xx (11 - 2"i")/(11 - 2"i")`

= `(- 25(11 - 2"i"))/((11)^2 - (2"i")^2`

= `(- 25(11 - 2"i"))/(121 + 4)`

= `(- 25(11 - 2"i"))/125`

= `(-(11 - 2"i"))/5`

= `(- 11 + 2"i")/5`