Mathematical Investigations III
Name:
Mathematical Investigations III - A View of the World
An Introduction to Complex Numbers
The system of real numbers is not quite algebraically “complete” since not all
polynomials have zeroes in the real numbers. For example, the quadratic polynomial
f ( x)  x 2  1 has no real number zeroes. In order to be able to solve all polynomial
equations, we need to expand our system of numbers.
Let i  1 , so i 2  1 . With this definition of the imaginary unit, all
polynomials whose coefficients are real numbers have zeroes. For example, the
function f ( x) above has zeroes x  i and x  i , since
f (i)  i 2  1  1  1  0
and
f (i )  (i ) 2  1  (1 i ) 2  1   1 i 2  1  1  1  0 .
2
In fact, with this definition, all negative real numbers have square roots. For
instance, 25  25  1  5  i .
Complex numbers are numbers that can be written as a  bi where a and b are
real numbers. All real numbers are complex numbers, since the real number
r can be written as r  0  i . Therefore, the real numbers are a subset of the
complex numbers. Other examples of complex numbers include 5i (which can
be written as 0  5  i ) and 3  4  i .
Since the definition of complex numbers is based on real numbers (a and b),
operations with complex numbers are natural extensions of operations with real
numbers and so are fairly intuitive. Perform each of the following operations and
write all complex answers in this a  bi form.
1.
Simplify.
27
45 +
20  5
80
2.
Simplify.
( 4  3i )  ( 2  7 i )
3(2  6 i )  2(3  i )
3.
Multiply and simplify.
2 i ( 35 i )
(4  i )(3  2 i )
Poly 7.1
Rev. S11
Mathematical Investigations III
Name:
3. (cont.)
(3  2 i )2
(5  2 i )(5  2 i )
(a  b i )2
(a  b i )(a  b i )
1
3 
 2  2 i


2
 2
2 
 2  2 i


2
5
by multiplying the numerator
42 3
and denominator by the conjugate of the denominator, namely 4  2 3 . Do that now:
Previously, you learned to rationalize the denominator of
5

42 3
5
. Multiply the numerator and denominator by
4  2i
4  2i so that its denominator is rationalized.
5

4  2i
Now apply that same technique to rewrite
We call 4  2i the complex conjugate of 4  2i .
Complex Conjugates
If z  a  bi is a complex number, the complex conjugate of z is a  bi and is denoted z .
Remark: We will mostly use the simpler term “conjugate” instead of “complex conjugate”.
Remark: We usually pronounce z as “z bar”.
Poly 7.2
Rev. S11
Mathematical Investigations III
Name:
4. Use the conjugate of each denominator to rationalize the following:
3i
5i 2
2i
3i 2
3
2i
6
8

2i 3i
5. Solve each of the following equations for z. Express your answer in a + bi form.
(b) 3z  4i  5  zi
(a) (2  3i)z + (7  4i) = 6  5i
Poly 7.3
Rev. S11
Mathematical Investigations III
Name:
Poly 7.4
Rev. S11
Mathematical Investigations III
Name:
6. Now try your hand at simplifying powers of the imaginary unit, i. Look for a pattern.
i1 = i (wow!)
i2 
i3 
i4 
i5 
i6 
i7 
i8 
7. What pattern do you notice about the powers of i?
8. Use the pattern you found to evaluate.
i15 
i 29 
i 437 
i 2543 
We have seen how to solve linear equations involving complex numbers. We now shift our
attention to quadratic equations.
9. Solve each quadratic equation below by completing the square:
(a)  x  5   49
2
(b) x 2  6 x  10  0
(c) 2 x 2  10 x  21  0
10. What do you notice about the pairs of solutions to the quadratic equations in Question 9?
Poly 7.5
Rev. S11
Mathematical Investigations III
Name:
11. Solve each of the following equations for z over the complex numbers. The quadratic
formula may be useful.
(a) z 2  6iz  8  0
(b) iz 2  2 z  5i  0
12. Does your observation from Question 11 hold with the pairs of solutions to the quadratic
equations in Question 11? Explain.
13. What was different about the coefficients in the quadratic equations between Question 9
and Question 11?
14. Summarize the observation about the pair of solutions to a quadratic equation: if the
coefficients are real numbers and one solution is a non-real complex number, then the
solutions are…
15. (Optional problem—come back to problem 15 if you have time after completing the
rest of this packet.) Use the quadratic formula to prove your answer to Question 14
about the solutions of the quadratic equation
ax 2  bx  c  0
if a, b, and c are all real numbers.
Poly 7.6
Rev. S11
Mathematical Investigations III
Name:
16. Let f ( x)  6 x 2  4 x  3 . Enter this as y1(x) in your calculator. Then use your calculator
to evaluate f ( x) at the following pairs of points (you may simply type y1(1 + i); the “i”
key is 2ND-Catalog on the TI-89):
(a) 1  i , 1  i
(b) 2  3i , 2  3i
(c) 3  i , 3  i
17. Describe the relationship demonstrated by your answers to the previous Question. Your
response should probably have a z in it somewhere!
Poly 7.7
Rev. S11