How do you simplify #(3 + 4i) (3 + 4i)#?

1 Answer
Jun 7, 2018

#-7+24i#

Explanation:

Your first step would be just like if you multiplied together two binomials: factor out each to get a trinomial. If you use FOIL, it's easy: multiply the First terms, the Outer terms, the Inner terms, and the Last terms:

F: #3*3=9#
O: #3*4i=12i#
I: #4i*3=12i#
L: #4i*4i=16i^2#

Sum these together, and you can get your halfway-simplified version:

#16i^2+12i+12i+9=16i^2+24i+9#

We know that #i=sqrt(-1)#, so if we have #i^2#, we're really just saying #(sqrt(-1))^2# which simplifies to #-1#. Knowing this, we can simplify farther yet:

#16*(-1)+24i+9#
#-16+24i+9#
#24i-7#

Since standard imaginary form is #a+bi#, we can swap it around to

#-7+24i# to get our final answer.