How would you multiply these complex numbers?

Question

 Multiply the following complex numbers: (3x+4i)(3x-4i).

Answer 1

The product is the Difference of Squares (from Algebra 1)

(3x + 4i)(3x - 4i)

(3x)² - (4i)²

9x² + 16

Answer 2

Hello,

First you multiply out the brackets as you would normally [(A+B)(C+D)=AC+AD+BC+BD].

So in our case, we would get:

(3x+4i)(3x-4i) = (3x)(3x)+(3x)(-4i)+(4i)(3x)+(4i)(-4i)

Now we simplify our expression:

(3x)(3x)+(3x)(-4i)+(4i)(3x)+(4i)(-4i) = 9x^2-12ix+12xi-16i^2 = 9x^2-16i^2

Finally, we use the fact that i^2=-1 (i is the square root of -1), so

9x^2-16i^2 = 9x^2+16

Answer 3

We can use the FOIL (Firsts, Outers, Inners, Lasts) method to multiply out the terms in the brackets.

(3x + 4i) (3x - 4i) = (3x)(3x) + (3x)(-4i) + (4i)(3x) + (4i)(-4i) = 9x² - 12xi + 12xi - 16i²

= 9x² - 16i² = 9x² + 16

Or, even better, we can recognize this is an example of Difference of Squares (a + b) (a - b) = a² - b²

Hence (3x + 4i) (3x - 4i) = (3x)² - (4i)²

= 9x² - 16i² = 9x² - 16(-1) = 9x² + 16

Or, best of all, we can use the specific Complex Arithmetic Rule (a + bi) (a - bi) = a² + b²

so for 3x + 4i, a = 3x and b = 4 => (3x)² + (4)² = 9x² + 16