Computations with Complex Numbers
You can perform many computations with complex numbers. Here are
some examples of these computations:
Addition
(a + bi) + (c + di) = (a + c) + (b + d)i
Example
Simplify (4 – 2i) + (1 + 5i).
Answer: To add complex numbers, add the real parts together and
add the imaginary parts together. So, (4 – 2i) + (1 + 5i) can be
written as 4 + 1 + (-2i + 5i), which simplifies to 5 + 3i.
Subtraction
(a + bi) – (c + di) = (a – c) + (b – d)i
Example
(5 + i) – (2 + 4i) = ?
Answer: To subtract complex numbers, subtract the real parts and
subtract the imaginary parts. So, (5 + i) – (2 + 4i) can be written as
5 – 2 + (i – 4i), which simplifies to 3 – 3i.
Multiplication
(a + bi)(c + di) = (ac – bd) + (ad + bc)i
Example
Multiply (3 + 4i)(2 – 5i).
Answer: Using FOIL, the following terms are produced:
First: 3 · 2 = 6
Outer: 3 · -5i = -15i
Inner: 4i · 2 = 8i
Last: 4i · -5i = -20i2 = -20(-1) = 20
Combining the like terms 6 + 20 – 15i + 8i yields 26 – 7i.
Division
a + bi
c + di
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When dividing by a complex number, remember that i = − 1 , so it is
as if there is a radical in the denominator. Therefore, the radical must
be mathematically removed. To remove the imaginary part of the
denominator, multiply the denominator by its complex conjugate. For
example, c + di and c – di are complex conjugates. Then, multiply the
numerator by the same complex conjugate.
a + bi  c − di 


c + di  c − di 
Example
Simplify the expression.
3−i
1+ i
Answer: To simplify, you need to multiply the numerator and the
denominator by the conjugate of the complex denominator.
3 − i  1 − i  3 − 3i − i + i 2 3 − 3i − i + (−1) 2 − 4i 2 4i
=
=
=
= −
=1 − 2i
1 + i  1 − i 
1 − i + i − (−1)
2
2 2
1 − i + i − i2
© LaurusSoft, Inc. 2010