Question

d

Answer 1

if you multiply the denominator with conjugate
(3-2i)(3+2i)

then the simple i terms will cancel
try it :)

Answer 2

Make sure you also multiply the numerator by the conjugte... \[\Large \frac{ 4 }{( 3-2 i) }*\frac{ (3+2 i) }{( 3+2 i )}\]

Answer 3

(3-2i)(3+2i)

F 9
O 6i
I -6i
L -4 i^2

-4 i^2 = 4 with i^2=-1

9+4=13

you calculated the complex numbers correctly :)

Answer 4

but you also gotta make that multiplication to the numerator
because you're extending the fraction,

just multiplying one of them (numerator, denominator) changes
the value of the fraction

Answer 5

and you're not allowed to change the value, it should be simplified without changing the value, just changing the form

Answer 6

No... your numerator isn't correct. See the pic above i posted.

Answer 7

No, there will be an i in the numerator when you're done.

Make sure you correctly multiply out the numerator here \[\Large \frac{ 4 }{( 3-2 i) }*\frac{ (3+2 i) }{( 3+2 i )}\]i'd help more but gotta teach

Answer 8

yes :)

Answer 9

thnk you