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Multiply Complex Numbers

How to multiply complex numbers, complex conjugates

To multiply to complex numbers such as (4+5i )(3+2i) , you can treat each one as a binomial and apply the FOIL method to find the product.
    FOIL
  • multiply the firsts 4 ×3 = 12
  • multiply the outers 4 × 2i = 8 i
  • multiply the inners 5i ×3 = 15i
  • multiply the lasts 5 i × 2 i = 10i 2= -10
  • Then add:  12 + 4i + 15i -10 = 2 +23i
Video Tutorial on Multiplying Complex Numbers
 Complex Conjugate 
Complex conjugates are any pair of complex number binomials that look like the following pattern: (a + bi) & (a-bi)
    Here are some specific examples. Note that the only difference between the two binomials is the sign.
  • (3 - 2i )( 3 +2i)
  • (5 + 12i )( 5 - 12i)
  • (7 - 33 i )( 7 +33i)
Multiplying complex conjugates causes the middle term to cancel as example 2 illutrates.
Example 1 Let's multiply the following 2 complex numbers
(5 + 2i ) (7 + 12i)
Step 1) Foil the binomials.
F : (5 + 2i ) (7 + 12i) 5 × 7 35
O : (5 + 2i ) (7 + 12i) 5 × 12i 60i
I : (5 + 2i ) (7 + 12i) 2i × 7 14i
L : (5 + 2i ) (7 + 12i) 2i × 12i 24i2 = -24 (remember i2 = -1)


Step 2) Simplify by adding the terms
  35
  60i
  14i
+ -24
  11+ 74i
Example 2 Let's multiply 2 complex conjugates
(4 + 6i ) (4 - 6i)
Step 1) Foil the binomials.
F : (4 + 6i ) (4 - 6i) 4 × 4 16
O : (4 + 6i ) (4 - 6i) 4 × -6i -24i
I : (4 + 6i ) (4 - 6i) 6i × 4 24i
L : (4 + 6i ) (4 - 6i) 6i × 6i 36i2 = -36 (remember i2 = -1)


Step 2) Simplify by adding the terms
  16
  -24i
  24i
+ -36
  52 (notice how the imaginary terms are additive inverses or 'cancel' each other)
Practice Problems
Problem 1)Let's multiply the following complex numbers
(5 + 4i ) (6 + 4i)
Step 1
Problem 2) What is the product of the following complex numbers?
(9 + 7i ) (6 + 8i)
Step 1
Problem 3) multiply the following complex numbers
(5 + 4i ) (5 - 4i)
Step 1
Problem 4) multiply the following complex numbers
(6 + 2i ) (6 - 2i)
Step 1