Up to this point the square root of a negative number has been left undefined. For example, we know that \(\sqrt < - 9 >\) is not a real number.
There is no real number that when squared results in a negative number. We begin to resolve this issue by defining the imaginary unit 26 , \(i\), as the square root of \(−1\).
\(i = \sqrt < - 1 >\quad \text < and >\quad i ^ < 2 >= - 1\)
To express a square root of a negative number in terms of the imaginary unit \(i\), we use the following property where \(a\) represents any non-negative real number:
With this we can write.
\(\sqrt < - 9 >= \sqrt < - 1 \cdot 9 >= \sqrt < - 1 >\cdot \sqrt < 9 >= i \cdot 3 = 3 i\)
If \(\sqrt < - 9 >= 3 i\), then we would expect that \(3i\) squared will equal \(−9\):
In this way any square root of a negative real number can be written in terms of the imaginary unit. Such a number is often called an imaginary number 27 .
Rewrite in terms of the imaginary unit \(i\).
Solution
When an imaginary number involves a radical, we place \(i\) in front of the radical. Consider the following:
\(6 i \sqrt < 2 >= 6 \sqrt < 2 >i\)
Since multiplication is commutative, these numbers are equivalent. However, in the form \(6 \sqrt < 2 >i\), the imaginary unit \(i\) is often misinterpreted to be part of the radicand. To avoid this confusion, it is a best practice to place \(i\) in front of the radical and use \(6 i \sqrt < 2 >\)
A complex number 28 is any number of the form,
where \(a\) and \(b\) are real numbers. Here, a is called the real part 29 and \(b\) is called the imaginary part 30 . For example, \(3 − 4i\) is a complex number with a real part of \(3\) and an imaginary part of \(−4\). It is important to note that any real number is also a complex number. For example, \(5\) is a real number; it can be written as \(5 + 0i\) with a real part of \(5\) and an imaginary part of \(0\). Hence, the set of real numbers, denoted \(ℝ\), is a subset of the set of complex numbers, denoted \(ℂ\).
Complex numbers are used in many fields including electronics, engineering, physics, and mathematics. In this textbook we will use them to better understand solutions to equations such as \(x^ + 4 = 0\). For this reason, we next explore algebraic operations with them.
Adding or subtracting complex numbers is similar to adding and subtracting polynomials with like terms. We add or subtract the real parts and then the imaginary parts.
Add \(( 5 - 2 i ) + ( 7 + 3 i )\).
Solution
Add the real parts and then add the imaginary parts.
\(\begin ( 5 - 2 i ) + ( 7 + 3 i ) & = 5 - 2 i + 7 + 3 i \\ & = 5 + 7 - 2 i + 3 i \\ & = 12 + i \end\)
Answer
To subtract complex numbers, we subtract the real parts and subtract the imaginary parts. This is consistent with the use of the distributive property.
Subtract \(( 10 - 7 i ) - ( 9 + 5 i )\).
Solution
Distribute the negative sign and then combine like terms.
\(\begin ( 10 - 7 i ) - ( 9 + 5 i ) & = 10 - 7 i - 9 - 5 i \\ & = 10 - 9 - 7 i - 5 i \\ & = 1 - 12 i \end\)
Answer:
In general, given real numbers \(a\), \(b\), \(c\), and \(d\):
Simplify \(( 5 + i ) + ( 2 - 3 i ) - ( 4 - 7 i )\).
Solution
\(\begin ( 5 + i ) + ( 2 - 3 i ) - ( 4 - 7 i ) & = 5 + i + 2 - 3 i - 4 + 7 i \\ & = 3 + 5 i \end\)
Answer:
In summary, adding and subtracting complex numbers results in a complex number.
Multiplying complex numbers is similar to multiplying polynomials. The distributive property applies. In addition, we make use of the fact that \(i^ = −1\) to simplify the result into standard form \(a + bi\).
Multiply \(- 6 i ( 2 - 3 i )\).
Solution
We begin by applying the distributive property.
Answer:
Multiply \(( 3 - 4 i ) ( 4 + 5 i )\).
Solution
Answer:
In general, given real numbers \(a\), \(b\), \(c\), and \(d\):
\(\begin ( a + b i ) ( c + d i ) & = a c + a d i + b c i + b d i ^ < 2 >\\ & = a c + a d i + b c i + b d ( - 1 ) \\ & = a c + ( a d + b c ) i - b d \\ & = ( a c - b d ) + ( a d + b c ) i \end\)
Simplify: \(( 3 - 2 i ) ^ < 2 >\).
Answer
Given a complex number \(a + bi\), its complex conjugate 31 is \(a − bi\).We next explore the product of complex conjugates.
Multiply \(( 5 + 2 i ) ( 5 - 2 i )\).
Solution
Answer:
In general, the product of complex conjugates 32 follows:
\(\begin ( a + b i ) ( a - b i ) & = a ^ < 2 >- a \cdot b i + b i \cdot a - b ^ < 2 >i ^ < 2 >\\ & = a ^ < 2 >- a b i + a b i - b ^ < 2 >( - 1 ) \\ & = a ^ < 2 >+ b ^ < 2 >\end\)
Note that the result does not involve the imaginary unit; hence, it is real. This leads us to the very useful property
\(( a + b i ) ( a - b i ) = a ^ < 2 >+ b ^ < 2 >\)
To divide complex numbers, we apply the technique used to rationalize the denominator. Multiply the numerator and denominator by the conjugate of the denominator. The result can then be simplified into standard form \(a + bi\).
Solution
In this example, the conjugate of the denominator is \(2 + 3i\). Therefore, we will multiply by \(1\) in the form \(\frac < ( 2 + 3 i ) >< ( 2 + 3 i ) >\).
To write this complex number in standard form, we make use of the fact that 13 is a common denominator.
Answer
Solution
Answer
In general, given real numbers \(a\), \(b\), \(c\) and \(d\) where \(c\) and \(d\) are not both \(0\):
Solution
Here we can think of \(2i = 0 + 2i\) and thus we can see that its conjugate is \(−2i = 0 − 2i\).
Because the denominator is a monomial, we could multiply numerator and denominator by \(1\) in the form of \(\frac\) and save some steps reducing in the end.
Answer
Answer
When multiplying and dividing complex numbers we must take care to understand that the product and quotient rules for radicals require that both \(a\) and \(b\) are positive. In other words, if \(\sqrt [ n ] < a >\) and \(\sqrt [ n ] < b >\) are both real numbers then we have the following rules.
For example, we can demonstrate that the product rule is true when \(a\) and \(b\) are both positive as follows:
However, when \(a\) and \(b\) are both negative the property is not true.
Here \(\sqrt\) and \(\sqrt\) both are not real numbers and the product rule for radicals fails to produce a true statement. Therefore, to avoid some common errors associated with this technicality, ensure that any complex number is written in terms of the imaginary unit \(i\) before performing any operations.
Multiply \(\sqrt < - 6 >\cdot \sqrt < - 15 >\).
Solution
Begin by writing the radicals in terms of the imaginary unit \(i\).
\(\sqrt < - 6 >\cdot \sqrt < - 15 >= i \sqrt < 6 >\cdot i \sqrt < 15 >\)
Now the radicands are both positive and the product rule for radicals applies.
\(\begin \sqrt < - 6 >\cdot \sqrt < - 15 >& = i \sqrt < 6 >\cdot i \sqrt < 15 >\\ & = i \sqrt < 6 \cdot 15 >\\ & = ( - 1 ) \sqrt < 90 >\\ & = ( - 1 ) \sqrt < 9 \cdot 10 >\\ & = ( - 1 ) \cdot 3 \cdot \sqrt < 10 >\\ & = - 3 \sqrt < 10 >\end\)
Answer
Solution
Begin by writing the radicals in terms of the imaginary unit i and then distribute.
\(\begin \sqrt < - 10 >( \sqrt < - 6 >- \sqrt < 10 >) & = i \sqrt < 10 >( i \sqrt < 6 >- \sqrt < 10 >) \\ & = i ^ < 2 >\sqrt < 60 >- i \sqrt < 100 >\\ & = ( - 1 ) \sqrt < 4 \cdot 15 >- i \sqrt < 100 >\\ & = ( - 1 ) \cdot 2 \cdot \sqrt < 15 >- i \cdot 10 \\ & = - 2 \sqrt < 15 >- 10 i \end\)
Answer:
In summary, multiplying and dividing complex numbers results in a complex number.
Simplify: \(( 2 i \sqrt < 2 >) ^ < 2 >- ( 3 - i \sqrt < 5 >) ^ < 2 >\).
Answer
Rewrite in terms of imaginary unit \(i\).
