How to Multiply Complex Numbers (a+bi Format)

Last Updated: September 24, 2022

If you ever encounter in your degrees of high school math problems dealing with another system of math involving solving for the square root of a negative number, do not fear. A special value - $i$ - can help solve this problem. However, if you have a problem where you have to solve a multiplication problem involving these problems, the solution may look tricky - until you see just how simple it is. This article will give you help to find the right answer.

Method 1

Method 1 of 2:

Multiplying Non-Conjugates

1

Have an understanding of the system of imaginary numbers. If you were trying to calculate the value of the square root of -1, you'd get an error on any calculator. However, when you see that imaginary numbers lie within a set of numbers, not within the rational numbers, an early Italian mathematician - Rafaello Bombelli, came up with a system of numbers to format numbers that couldn't be solved in the real/rational system. His value $i={\sqrt {-1}}$ . ^{[1]
X
Research source
Course III: Integrated Mathematics. Third Edition, ISBN: 9781567655216}
2
Examine your problem. Learn what needs to be solved and write it down.
- Multiply $(4+i)(7+2i)$
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3
Utilize the distributive property of multiplication- if the a and b values aren't the same with different signs in between (which are conjugates discussed below).
- Use FOIL (Firsts, Outsides, Insides, Lasts) or multiply the first term to the entire second term of the first a+bi and multiply it to the entire second term.
- From the example above, you can try to answer this in two ways.
  $(4+i)(7+2i)=4(7+2i)+i(7+2i)=28+8i+7i+2i^{2}$ or
  $(4+i)(7+2i)=4(7)+4(2i)+i(7)+i(2i)=28+8i+7i+2i^{2}$
4
Run some additional cleanup when simplifying the bi values if they have any square root problems in them. When multiplying them, you can combine two square roots into one square root, by multiplying the numbers inside the square root, then re-simplifying - either through finding the common square root answer, or simplifying so there's still a square root (but a smaller square root that's "prime and can't be simplified further" exists.)
- From the example above, you can clean up all those with regular imaginary units and since both $8i+7i=15i$ , you can add bring that into the answer - so far. bringing that to be $28+15i+2i^{2}$ .
5
Realize that the $i^{{2}}$ term will become just the value of $-1$ . The square root of a squared number cancels the whole square root and brings you just the number inside the square root which is $-1$ to become just the number being square rooted, and will ultimately be multiplied by the last term being calculated in your final calculations.
- From the example above, you can utilize and cleanup the imaginary unit squared and multiply it by it's coefficient (the 2) and utilize that.since the imaginary unit squared is just the value of $-1$ , you can utilize multiplying $-1(2)=-2$ .
6
Run some cleanup behind that. Add or subtract the real numbers you found with the -1 (at the tail end), and that should come out to be a real number.
- From the example above, since there are two real numbers (the 28 and -2), you can subtract (or add in some cases, but not in this one), $28-2=26$ .
7

Arrive at a final answer. From the provided example above, bring these forms together, and you'll arrive at $26+15i$

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Method 2

Method 2 of 2:

Multiplying Conjugates

1
Define a conjugate. A conjugate - when it defines a real number and an imaginary unit in the complex number system - acts as the "product of perfect squares", where if a and b are the same yet the operations/signs are switched between addition and subtraction your imaginary units are completely canceled and become the sum of the $a^{{2}}$ and $b^{{2}}$ units. ^{[2]
X
Research source
Course III: Integrated Mathematics. Third Edition, ISBN: 9781567655216}
- Try out a proof of the problem of $(a+bi)(a-bi)$ .
 - $(a+bi)(a-bi)=a^{2}+abi-abi-b^{2}i^{2}=a^{2}+abi-abi+b^{2}$
 $=(a^{2}+b^{2})+(abi-abi)=(a^{2}+b^{2})+(ab-ab)i=(a^{2}+b^{2})+0i$
 $=a^{2}+b^{2}$ ^{[3]
 X
 Research source
 Course III: Integrated Mathematics. Third Edition, ISBN: 9781567655216}
2
Examine your problem. Learn what needs to be solved and write it down.
- Multiply $(4+9i)(4-9i)$ .
3

Bring down the addition sign, since the proof says it'll become a real number that will later be added.
4

Square both the a units and b units. Concern yourself with just the numbers. When $4$ and $9$ are squared (hence $4^{2}=16$ and $9^{2}=81$ ), these values will be the values you'll import into the places for $a^{{2}}$ and $b^{{2}}$ locations.
5
Add these two real numbers together - since the imaginary units break away completely from the problem. Your resultant answer is a combination of two real numbers and can be added together to become your calculated value for your problem.
- In this example, the answer is $16+81=97$ . Your calculated answer is $97$ .
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If you ever run into a problem where you see $i^{3}$ , think of its value as a a $-i$ . ^{[4]
X
Research source
Course III: Integrated Mathematics. Third Edition, ISBN: 9781567655216}

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If you are told in a problem that the $a-bi$ form is to be squared, one sign negates the other and the problem becomes a multiplication of an $a+bi$ problem - because when multiplication of a negative and negative interact, they become a positive value or the complex number becomes a simple addition answer.

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There is quite a bit of resemblance of multiplying complex numbers to other complex numbers based on algebraic methods, so you should not have to worry too much and approach multiplication of these with these methods down pat.

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[2]

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How to Multiply Complex Numbers (a+bi Format)

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Multiplying Non-Conjugates

Multiplying Conjugates

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