Question

Multiply. Write the result in the form a + bi
3i(2-3i)

Answer 1

\(6i-9i^2\) is a start
then since \(i^2=-1\) you get \(6i+9\) which in standard form is \(9+6i\)

Answer 2

soo would (4 +5i) (2+i) be the same way..
I would multiply the two complex numbers together first right?

Answer 3

no you can do this in one of two ways
either multiply out using that loathsome "foil" i .e. do 4 multiplications, or just know that
\[(a+bi)(c+di)=(ac-bd)+(ad+bc)i\]

Answer 4

so it would be 40i .. or am I just doing this whole thing wrong?

Answer 5

i think probably you are

Answer 6

\((8-5)+(4+10)i\) is a start

Answer 7

where did you get those numbers?

Answer 8

lets do it the slow way first, then i will show you where they came from

Answer 9

oh okay. sorry about this.. i'm terrible at math..

Answer 10

no problem

Answer 11

suppose i had to multiply \[(4 +5x) (2+x) \] you would ge t
\[8+4x+10x+5x^2=8+14x+5x^2\]
clear or no?

Answer 12

clear.

Answer 13

ok lets repeat replacing \(x\) by \(i\) and go right to the answer
\[8+14i+5i^2\]
is that ok?

Answer 14

let me know if it is not
i replaced \(x\) by \(i\) and did the same thing as before, just didn't write down the steps this time, because they are identical

Answer 15

yeah I got it

Answer 16

ok one more step
since \(i^2=-1\) you actually have
\[8+14i-5\]
ok with that?

Answer 17

yes I am.

Answer 18

and finally since \(8-5=3\) the "final answer" is \(3+14i\)

Answer 19

now maybe it is clear where \[(8-5)+(4+10)i\] came from

Answer 20

oops typo
for \((a+bi)(c+di)\) when you multiply out you get \(bdi^2=-bd\)
if this is confusing, you can always do it the way that we just did

Answer 21

oh okaay thankyou!!!

Answer 22

yw