proving complex stuff

• Nov 1st 2009, 11:14 PM
proving complex stuff
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?(Speechless)
• Nov 2nd 2009, 05:31 AM
Prove It
Quote:

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?(Speechless)

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

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

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

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

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

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

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

\$\displaystyle RHS = A^2 + B^2\$

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

\$\displaystyle = (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\$

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

\$\displaystyle = LHS\$.

Q.E.D.