Question

How do you simplify `1/(2+3i)` and write in a+bi form?

Answer 1

`2/13-3/13i`

Explanation:

Multiply by the complex conjugate.

`frac1(2+3i)(frac(2-3i)(2-3i))=frac(2-3i)(4-6i+6i-9i^2)=frac(2-3i)(4-9i^2)`

Recall that `i^2=-1` since `i=sqrt(-1)`.

`=frac(2-3i)(4-(-9))=(2-3i)/13=2/13-3/13i`

Answer 2

`=0,1538-0,23078i`

Explanation:

`1` may be written in rectangular form as `1+0i`, and in polar form as `1/_0^@`.

Similarly, `2+3i` can be written in polar form as
`sqrt(2^2+3^2)/_tan^(-1)(3/2) =sqrt13/_56,31^@`

We may now divide them by using the standard rule for division of complex numbers in polar form :

`(1/_0)/( sqrt13/_56,31^@)=1/(sqrt13)/_(0-56,31)`

This answer may now be converted back to rectangular form using standard trig relationships to get `(1/sqrt13)cos(-56,31)+i(1/sqrt13)sin(-56,31)`

`=0,1538-0,23078i`

(Note : We could have maintained complex rectangular form and multiply top and bottom by the complex conjugate of the denominator `bar z = 2-3i` and then multiply top and bottom. Bottom would then be a real number, and top would require rule of multiplication of complex numbers in rectangular form - quite a complex rule ;) - Hence probably quicker the way I did it by rather converting everything to polar form first)