proving complex stuff

purebladeknight · #1 $Old$ November 2nd, 2009, 01:14 AM

if A+iB = (a.+ib.) (a,+ib,), prove (a.^2+b.^2) (a,^2 +b,^2 ) = (A^2 + B^2)
a. & b. is suppose to be subcript 1 and a, & b, is suppose to be subscript 2.

i dont see any short way of proving this, i have tried squaring A+iB and re-arranging to prove (A^2 + B^2) but hit a dead end.

any suggestions on how to attemp?

Prove It · #2 $Old$ November 2nd, 2009, 07:31 AM

Quote:

Originally Posted by purebladeknight $View Post$

if A+iB = (a.+ib.) (a,+ib,), prove (a.^2+b.^2) (a,^2 +b,^2 ) = (A^2 + B^2)
a. & b. is suppose to be subcript 1 and a, & b, is suppose to be subscript 2.

i dont see any short way of proving this, i have tried squaring A+iB and re-arranging to prove (A^2 + B^2) but hit a dead end.

any suggestions on how to attemp?

$A + iB = (a_1 + ib_1)(a_2 + ib_2)$

$= a_1a_2 + ia_1b_2 + ia_2b_1 + i^2b_1b_2$

$= a_1a_2 - b_1b_2 + i(a_1b_2 + a_2b_1)$

Therefore $A = a_1a_2 - b_1b_2$ and $B = a_1b_2 + a_2b_1$ .

Now let's try to prove the statement that $(a_1^2 + b_1^2)(a_2^2 + b_2^2) = A^2 + B^2$ .

$LHS = (a_1^2 + b_1^2)(a_2^2 + b_2^2)$

$= (a_1a_2)^2 + (a_1b_2)^2 + (a_2b_1)^2 + (b_1b_2)^2$ .

$RHS = A^2 + B^2$

$= (a_1a_2 - b_1b_2)^2 + (a_1b_2 + a_2b_1)^2$

$= (a_1a_2)^2 - 2a_1a_2b_1b_2 + (b_1b_2)^2 + (a_1b_2)^2 + 2a_1a_2b_1b_2 + (a_2b_1)^2$

$= (a_1a_2)^2 + (a_1b_2)^2 + (a_2b_1)^2 + (b_1b_2)^2$

$= LHS$ .

Q.E.D.