# Multiplying Complex Numbers

Imaginary and complex numbers are multiplied the same way you multiply polynomials, with the addition of what you learned in the previous lesson about *i* when raised to varying exponents. Examples are shown below:

Similarly, this idea can be expanded to imaginary numbers found within larger algebraic expressions found in polynomials. For example:

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Starting with (1), you’d expand the factor 3 into the binomial (5 + 2i), giving you: 15 + 6i. In question (2), 3iÃ—2 equals **6i** and 3iÃ—(â€“4i) equals **â€“12iÂ²**. Together, 6i â€“ 12iÂ² = 6i + 12 â‡’ 12 + 6i. For question (3), you’d have to use the foil method (linked), giving you: â€“12 **+ 15i + 8i** â€“ 20iÂ² â‡’ â€“12 + 23i + 20 â‡’ 8 + 23i.

As discussed briefly in the previous lesson, the **conjugate** of a complex number is one that changes the sign of the imaginary portion. In case you’re asked to multiply a complex number by its conjugate, follow the steps outlined in the video below:

# Beware of this common mistake

If you’re asked to multiply, for example:

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Always convert radicals to imaginary numbers first, otherwise contradictions may result:

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This rule only applies to be the radicand is positive. Rather, do this:

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# Dividing Complex Numbers

Dividing complex numbers is relatively easy if the denominator is a **single term**. However, if you’re given an expression where it’s in the form of two binomials, such as ( a Â± bi ) Ã· ( a âˆ“ bi ), this can be reduced by multiplying and dividing the expression by the **conjugate **of the denominator. This demonstrated is illustrated below: