Question

How do you divide `(4+5i)/(2-3i)`?

Answer 1

Multiply both numerator and denominator by the Complex conjugate of the denominator, then simplify to find:

`(4+5i)/(2-3i)=-7/13 + 22/13i`

Explanation:

Multiply numerator and denominator by `2+3i` (using FOIL if it helps) and simplify:

`(4+5i)/(2-3i)`

`=((4+5i)(2+3i))/((2-3i)(2+3i))`

`=(stackrel "First" overbrace((4*2))+stackrel "Outside" overbrace((4*3i))+stackrel "Inside" overbrace((5i*2))+stackrel "Last" overbrace((5i*3i)))/(2^2+3^2)`

`=(8+12i+10i-15)/(4+9)`

`=(-7+22i)/13`

`=-7/13 + 22/13i`

Answer 2

`=-7/13 + 22/13i`

Explanation:

`(4+5i)/(2-3i)`

Whenever we divide complex numbers we multiply both numerator and denominator with the complex conjugate of the denominator, this makes the denominator a real number

If the complex number is `a+ib` then the complex conjugate is `a-ib`

For example
`(a+ib)(a-ib)`
`= (a)^2 - (ib)^2`
`=a^2-i^2b^2`
`=a^2-(-1)b^2`
`=a^2+b^2` this is a real number.

Now back to our problem.

`(4+5i)/(2-3i)`

`=(4+5i)/(2-3i)*(2+3i)/(2+3i)`

`=((4+5i)(2+3i))/(2^2+3^2)`

`=(4(2)+4(3i)+5i(2) + 5i(3i))/(4+9)`

`=(8+12i+10i+15i^2)/13`

`=(8+22i+15(-1))/13`

`=(8-15+22i)/13`

`=(-7+22i)/13` Answer

`=-7/13 + 22/13i quad` Answer in `a+ib` form