Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and imaginary parts separately.
Example 4: Multiplying a Complex Number by a Real Number
Figure 5
Let’s begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. So, for example,
How To: Given a complex number and a real number, multiply to find the product.
Use the distributive property.
Simplify.
Example 5: Multiplying a Complex Number by a Real Number
Find the product 4(2+5i).
Solution
Distribute the 4.
{4(2+5i)=(4⋅2)+(4⋅5i)=8+20i
Try It 4
Find the product −4(2+6i).
Solution
Multiplying Complex Numbers Together
Now, let’s multiply two complex numbers. We can use either the distributive property or the FOIL method. Recall that FOIL is an acronym for multiplying First, Outer, Inner, and Last terms together. Using either the distributive property or the FOIL method, we get
(a+bi)(c+di)=ac+adi+bci+bdi2
Because i2=−1, we have
(a+bi)(c+di)=ac+adi+bci−bd
To simplify, we combine the real parts, and we combine the imaginary parts.
(a+bi)(c+di)=(ac−bd)+(ad+bc)i
How To: Given two complex numbers, multiply to find the product.
Use the distributive property or the FOIL method.
Simplify.
Example 6: Multiplying a Complex Number by a Complex Number
Division of two complex numbers is more complicated than addition, subtraction, and multiplication because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator. We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called the complex conjugate of the denominator, which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of a+bi is a−bi.
Note that complex conjugates have a reciprocal relationship: The complex conjugate of a+bi is a−bi, and the complex conjugate of a−bi is a+bi. Further, when a quadratic equation with real coefficients has complex solutions, the solutions are always complex conjugates of one another.
Suppose we want to divide c+di by a+bi, where neither a nor b equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply.
a+bic+di where a=0 and b=0
Multiply the numerator and denominator by the complex conjugate of the denominator.
The complex conjugate of a complex number a+bi is a−bi. It is found by changing the sign of the imaginary part of the complex number. The real part of the number is left unchanged.
When a complex number is multiplied by its complex conjugate, the result is a real number.
When a complex number is added to its complex conjugate, the result is a real number.
Example 7: Finding Complex Conjugates
Find the complex conjugate of each number.
2+i5
−21i
Solution
The number is already in the form a+bi//. The complex conjugate is a−bi, or 2−i5.
We can rewrite this number in the form a+bi as 0−21i. The complex conjugate is a−bi, or 0+21i. This can be written simply as 21i.
Analysis of the Solution
Although we have seen that we can find the complex conjugate of an imaginary number, in practice we generally find the complex conjugates of only complex numbers with both a real and an imaginary component. To obtain a real number from an imaginary number, we can simply multiply by i.
How To: Given two complex numbers, divide one by the other.
Write the division problem as a fraction.
Determine the complex conjugate of the denominator.
Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator.
Simplify.
Example 8: Dividing Complex Numbers
Divide (2+5i) by (4−i).
Solution
We begin by writing the problem as a fraction.
(4−i)(2+5i)
Then we multiply the numerator and denominator by the complex conjugate of the denominator.
(4−i)(2+5i)⋅(4+i)(4+i)
To multiply two complex numbers, we expand the product as we would with polynomials (the process commonly called FOIL).
⎩⎨⎧(4−i)(2+5i)⋅(4+i)(4+i)=16+4i−4i−i28+2i+20i+5i2=16+4i−4i−(−1)8+2i+20i+5(−1)=173+22i=173+1722iBecause i2=−1Separate real and imaginary parts.
Note that this expresses the quotient in standard form.
Example 9: Substituting a Complex Number into a Polynomial Function
Let f(x)=x2−5x+2. Evaluate f(3+i).
Solution
Figure 6
Substitute x=3+i into the function f(x)=x2−5x+2 and simplify.
Analysis of the Solution
We write f(3+i)=−5+i. Notice that the input is 3+i and the output is −5+i.
Try It 6
Let f(x)=2x2−3x. Evaluate f(8−i).
Solution
Example 10: Substituting an Imaginary Number in a Rational Function
Let f(x)=x+32+x. Evaluate f(10i).
Solution
Substitute x=10i and simplify.
⎩⎨⎧10i+32+10i3+10i2+10i3+10i2+10i⋅3−10i3−10i9−30i+30i−100i26−20i+30i−100i29−30i+30i−100(−1)6−20i+30i−100(−1)109106+10i109106+10910iSubstitute 10i for x.Rewrite the denominator in standard form.Prepare to multiply the numerator anddenominator by the complex conjugateof the denominator.Multiply using the distributive property or the FOIL method.Substitute −1 for i2.Simplify.Separate the real and imaginary parts.
Try It 7
Let f(x)=x−4x+1. Evaluate f(−i).
Solution
https://youtu.be/XBJjbJAwM1c
Simplifying Powers of i
The powers of i are cyclic. Let’s look at what happens when we raise i to increasing powers.
We can see that when we get to the fifth power of i, it is equal to the first power. As we continue to multiply i by itself for increasing powers, we will see a cycle of 4. Let’s examine the next 4 powers of i.
Since i4=1, we can simplify the problem by factoring out as many factors of i4 as possible. To do so, first determine how many times 4 goes into 35: 35=4⋅8+3.
i35=i4⋅8+3=i4⋅8⋅i3=(i4)8⋅i3=18⋅i3=i3=−i
Q & A
Can we write i35 in other helpful ways?
As we saw in Example 11, we reduced i35 to i3 by dividing the exponent by 4 and using the remainder to find the simplified form. But perhaps another factorization of i35 may be more useful. The table below shows some other possible factorizations.
Factorization of i35
i34⋅i
i33⋅i2
i31⋅i4
i19⋅i16
Reduced form
(i2)17⋅i
i33⋅(−1)
i31⋅1
i19⋅(i4)4
Simplified form
(−1)17⋅i
−i33
i31
i19
Each of these will eventually result in the answer we obtained above but may require several more steps than our earlier method.
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Analysis of the Solution
Although we have seen that we can find the complex conjugate of an imaginary number, in practice we generally find the complex conjugates of only complex numbers with both a real and an imaginary component. To obtain a real number from an imaginary number, we can simply multiply by i.