imaginary number help?

Question

3 + 2i
--------
3 - 2i.

its a fraction. please help

Answer 1

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

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

((3+2i)*(3+2i))/
(9+6i-6i-4i^2)=

((3+2i)*(3+2i))/(9-4i^2)=
((3+2i)*(3+2i))/(9-4*-1)=
((3+2i)*(3+2i))/(9+4)=
((3+2i)*(3+2i))/13=
(9+12i+4*i^2)/13=
(9+12i+4*-1)/13=
(9+12i-4)/13=
(5+12i)/13

Answer 2

=(3+2i)^2/13 =1/13( 9+12i-4) = 5/13+12/13 i
I multiplied both terms by 3+2i conjugate of 3-2i

Answer 3

i have hoping something like this would come up
to show a neat trick

(-1)^.5 is called imaginary because up to infinity it cannot be solved

Sort of wrong

let 17=infinity
use central pairs
4,13
and
i have hoping something like this would come up
to show a neat trick

(-1)^.5 =4 or 13

so for i use 4

we still need to multiply top and bottom
by (3+2i)
(3+2i)(3+2i)=9+12i-4
-----------------------------------------------
(3-2i)(3+2i)=9+4
---------------------------------------
5+12i
--------
13

now check with z=17 and cp=4
orginal=3+8=11
3-8=-5=12
1/12=10
11*10=110 in z (are infinity)=8

answer=5+48 in z=2
2/13=4*2=8

thus we are correct

Answer 4

multiply both top and bottom by (3 + 2i) and remember that i squared is (-1)