S k i l l
30 COMPLEX

a)  i^{2} = −1  b)  i · 2i = 2i^{2} = 2(−1) = −2  
c)  (3i)^{2} = 3^{2}i^{2} = −9  d)  −5i · 4i = −20i^{2} = 20 
Example 3.  =  i  
=  i = 2i  
=  i = 2i 
The square root of −a is equal to i times the square root of a.
Problem 2. Express each of the following in terms of i.
a)  = i  b)  = 3i  c)  = 7i  
d)  = i  e)  = i  f)  = 2i  
g)  = 3i  h)  = 5i  i)  = 7i 
Note: To multiply complex numbers , we must first express them in terms of i.
= i·i = −.
The rule does not hold when both a and b are negative.
Powers of i
Let us begin with i^{0}, which is 1. (Any number with exponent 0 is 1.) Each power of i can be obtained from the previous power by multiplying it by i. We have:
i^{0}  =  1 
i^{1}  =  i 
i^{2}  =  −1 
i^{3}  =  −1· i = −i 
i^{4}  =  −i· i = −i^{2} = −(−1) = 1 
Since we are back at 1, the cycle of powers will repeat. Any power of i will be either
1, i, −1, or −i
—according to the remainder upon dividing the exponent n by 4.
Example 4.
i^{9}  =  i, because on dividing 9 by 4, the remainder is 1.  
i^{9} = i^{1}.  
i^{18}  =  −1, because on dividing 18 by 4, the remainder is 2.  
i^{18} = i^{2}.  
i^{35}  =  −i, because on dividing 35 by 4, the remainder is 3.  
i^{35} = i^{3}.  
i^{40}  =  1, because on dividing 40 by 4, the remainder is 0.  
i^{40} = i^{0}. 
Note: Even powers of i will be either 1 or −1, according as the exponent is a multiple of 4, or 2 more than a multiple of 4. While odd powers will be either i or −i.
Problem 3. Evaluate each power of i.
a) i^{3} = −i  b) i^{4} = 1  c) i^{6} = i^{2} = −1  
d) i^{9} = i^{1} = i  e) i^{12} = i^{0} = 1  f) i^{17} = i^{1} = i  
g) i^{27} = i^{3} = −i  h) i^{30} = i^{2} = −1  i) i^{100} = i^{0} = 1 
Problem 4. 
Algebra with complex numbers
Complex numbers follow the same rules as real numbers. For example, to multiply
(2 + 3i)(2 − 3i)
the student should recognize the form (a + b)(a − b)  which will produce the difference of two squares. Therefore,
(2 + 3i)(2 − 3i)  =  4 − 9i^{2} 
=  4 − 9(−1)  
=  4 + 9  
=  13. 
Again, the factor i^{2} changes the sign of the term.
Problem 5. Multiply.
a) (1 + i)(1 − i) = 1 − 2i^{2} = 1 + 2 = 3
b) (3 − i)^{2}  =  9 − 6i + 2i^{2},  upon squaring the binomial, 
=  9 − 6i − 2  
=  7 − 6i 
c) (2 + 3i)(4 − 5i)  =  8 − 10i + 12i − 15i^{2} 
=  8 + 2i + 15  
=  23 + 2i 
Problem 6. (x + 1 + 3i)(x + 1 − 3i)
a) What form will that produce? The difference of two squares.
b) Multiply out.
(x + 1 + 3i)(x + 1 − 3i)  =  (x + 1)^{2} − 9i^{2} 
=  x^{2} + 2x + 1 + 9  
=  x^{2} + 2x + 10 
c)  (x − 2 − i)(x − 2 + i)  =  (x − 2)^{2} − 2i^{2} 
=  x^{2} − 4x + 4 + 2  
=  x^{2} − 4x + 6 
The real and imaginary components
Here is what is now called the standard form of a complex number:
a + bi.
It is the real number a plus the complex number .
For example,
3 + 2i.
a—that is, 3 in the example—is called the real component (or the real part). b (2 in the example) is called the imaginary component (or the imaginary part).
The components are real.
Problem 7. Name the real component a and the imaginary component b.
a)  3 − 5i a = 3, b = −5.  b)  1 + i a = 1, b = .  
c)  i a = 0, b = 1.  d)  −6 a = −6, b = 0. 
Complex conjugates
The complex conjugate of a + bi is a − bi. The point about a conjugate pair is that when they are multiplied—
(a + bi)(a − bi)
—the product is a positive real number. That form is the difference of two squares:
(a + bi)(a − bi) = a^{2} − b^{2}i^{2} = a^{2} + b^{2}
The product of a complex conjugate pair
is equal to the sum of the squares of the components.
Problem 8. Calculate the positive real number that results from multiplying each of the following with its complex conjugate.
a) 2 + 3i
(2 + 3i)(2 − 3i) = 2^{2} + 3^{2} = 4 + 9 = 13.
b) 3 − i.
(3 − i)(3 + i) = 3^{2} + ()^{2}= 9 + 2 = 11.
c) u + iv. (u + iv)(u − iv) = u^{2} + v^{2}.
d) 1 + i. (1 + i)(1 − i) = 1^{2} + 1^{2} = 2.
e) −i. (−i)(i) = −i^{2} = 1.
Next Lesson: Rectangular coördinates
Please make a donation to keep TheMathPage online.
Even $1 will help.
Copyright © 2020 Lawrence Spector
Questions or comments?
Email: themathpage@yandex.com