Question

If (A+iB)=(2+i)/(2-3i), then A^2+B^2=?

Answer 1

pout the values

Answer 2

values of what?

Answer 3

what is the answer?

Answer 4

We have this, factored by using complex numbers
\[\large a^2 + b^2 = (a + bi)(a - bi)\]

Answer 5

You have \((a+bi)\) and \((a-bi)\) is its conjugate.

Answer 6

ok, the answer i m getting is 5/13. is this correct?

Answer 7

\[(\frac{2+i}{2-3i})(\frac{2-i}{2+3i})\]

I believe it's something like this, someone can confirm? I don't work often with complex numbers :[

Answer 8

i got the 5/13

@zepp numerator also 2+3i... rationalizing denom right ?

Answer 9

ya right, the answer is 5/13 then..

Answer 10

If I'm correct, it would become \[\large \frac{2^2-i^2}{2^2-(3i
)^2}=\frac{4-(-1)}{4-9(-1)}=\frac{4+1}{4+9}=\frac{5}{13}\]
That's my work

Answer 11

Great! Thnx!

Answer 12

You are welcoe :)

Answer 13

welcome*

Answer 14

no need to rationalize

Answer 15

yeah... i see that lol i was thinking of rationalizing and comparing i terms & real ones...

Answer 16

its asking you to take modulus of both sides :
use them :
1)|a + ib| = sqrt(a^2 + b^2)
2)|p/q| = |p|/|q|
where p and q are complex nos..

Answer 17

rather than these answers i will prefer u to go to these links also :

1) for practice : http://openstudy.com/study#/updates/4fe5a252e4b06e92b873c296

2) for learning : http://openstudy.com/study#/updates/4fe196d3e4b06e92b870a14a

best of luck @Hesan