# [SOLVED] Multiplication with complex numbers

• Oct 3rd 2009, 09:53 AM
Nrt
[SOLVED] Multiplication with complex numbers
I need to derive this equation and get the below result.

\$\displaystyle ((a^2-b^2)^2+a^2b^2)(a^2+b^2)=a^6+b^6\$

\$\displaystyle ===>\$ \$\displaystyle (a^2-iab-b^2)(a+ib)=a^3-ib^3\$

Also a hint teacher gave us:

replace

\$\displaystyle a\$ by \$\displaystyle a^2\$
\$\displaystyle b\$ by \$\displaystyle -ib^2\$

I tried few things (like \$\displaystyle -b^2=-1.b^2=i^2b^2\$) to involve complex numbers in the first equation but im confused and have no clue what to do.
• Oct 3rd 2009, 11:00 AM
awkward
Quote:

Originally Posted by Nrt
I need to derive this equation and get the below result.

\$\displaystyle ((a^2-b^2)^2+a^2b^2)(a^2+b^2)=a^6+b^6\$

\$\displaystyle ===>\$ \$\displaystyle (a^2-iab-b^2)(a+ib)=a^3-ib^3\$

Also a hint teacher gave us:

replace

\$\displaystyle a\$ by \$\displaystyle a^2\$
\$\displaystyle b\$ by \$\displaystyle -ib^2\$

I tried few things (like \$\displaystyle -b^2=-1.b^2=i^2b^2\$) to involve complex numbers in the first equation but im confused and have no clue what to do.

Hi Nrt,

The first equation
\$\displaystyle ((a^2-b^2)^2+a^2b^2)(a^2+b^2)=a^6+b^6\$
is an identity-- it is true for all values of a and b.

Now
\$\displaystyle (a^2-b^2)^2+a^2b^2 = a^4 - a^2b^2 +b^4\$
so
\$\displaystyle (a^4 - a^2b^2 +b^4)(a^2+b^2)=a^6+b^6\$.

Now let \$\displaystyle x = a^2\$ and \$\displaystyle y = b^2 / i\$, so \$\displaystyle b^2 = iy\$.

Then the identity becomes
\$\displaystyle (x^2 - x(iy) + (iy)^2)(x + iy) = x^3 + (iy)^3\$
so
\$\displaystyle (x^2 -i x y -y^2)(x+iy) = x^3 - i y^3\$.

This is the result you were asked to show, except that I used x and y instead of a and b. It seems less confusing to me that way. (I get confused very easily). But you can go back and replace x and y with a and b if you prefer it that way.
• Oct 3rd 2009, 11:52 AM
Nrt
Much appreciated, thank you.