How do I find the product of two imaginary numbers?

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Answer 1 · 2014-09-04T21:57:23

First, complex numbers can come in a variety of forms!

Ex: multiply $3 i \cdot - 4 i =$ $3i*-4i =$

Remember, with multiplication you can rearrange the order (called the Commutative Property):

$3 \cdot - 4 \cdot i \cdot i = - 12 i^{2}$ $3*-4*i*i =-12i^2$

... and then always substitute -1 for $i^{2}$ $i^2$ :

$- 12 \cdot - 1 = 12$ $-12*-1 = 12$

Ex: the numbers might come in a radical form:

$\sqrt{- 3} \cdot 4 \sqrt{- 12} =$ $sqrt(-3)*4sqrt(-12) =$

You should always "factor" out the imaginary part from the square roots like this:

$\sqrt{- 1} \sqrt{3} \cdot 4 \cdot \sqrt{- 1} \sqrt{4} \sqrt{3} =$ $sqrt(-1)sqrt(3)*4*sqrt(-1)sqrt(4)sqrt(3) =$

and simplify again:

$= i \cdot 4 \cdot \sqrt{3} \cdot \sqrt{3} \cdot \sqrt{4}$ $=i*4*sqrt(3)*sqrt(3)*sqrt(4)$
$= i \cdot 4 \cdot 3 \cdot 2 = 24 i$ $=i*4*3*2 = 24i$

Ex: what about the Distributive Property? $3 i (4 i - 6) =$ $3i(4i - 6) =$

$= 12 i^{2} - 18 i$ $=12i^2- 18i$
$= 12 (- 1) - 18 i$ $=12(-1) - 18i$
$= - 12 - 18 i$ $= -12 - 18i$

And last but not least, a pair of binomials in a + bi form:

Ex: (3 - 2i)(4 + i) =

=12 + 3i - 8i - $2 i^{2}$ $2i^2$
= 12 - 2(-1) + 3i - 8i
= 12 + 2 - 5i
= 14 - 5i