© Teachers Teaching with Technology (Scotland)
Teachers Teaching with Technology
T3 Scotland
Complex Numbers
©Teachers Teaching with Technology (Scotland)
COMPLEX NUMBERS
Aim
To demonstrate how the TI-83 can be used to explore the concept of and solve problems
involving imaginary and complex numbers.
Objectives
Mathematical objectives
By the end of this topic you should
• Know when an imaginary number exists.
• Know how to add, subtract, multiply and divide complex numbers.
• Know how to simplify complex numbers involving powers of i
• Understand the nature and uses of the conjugate complex.
• Be able to find the modulus and argument of a complex number.
Calculator objectives
By the end of this topic you should
• Be able to set the MODE screen to allow non-real answers to be displayed.
• Be able to carry out complex calculations on the TI-83
• Be able to store complex numbers as variables for ease of calculation.
• Be able to navigate to various functions on the MATHS: CPX Menu
T3 Scotland
Complex Numbers
Page 1 of 11
COMPLEX NUMBERS
Calculator Skills Sheet
Set the MODE on the calculator to the screen shown.
− 1 and give a reason
Write down what you expected to happen when you type in
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
Type in
−1
and write down the answer ___________________________________
i
Find the key
i
and type in:
2nd
x2
to get the screen shown.
The square root of -1 is a number whose symbol is i
i
for imaginary since numbers which use i are called imaginary numbers although they do
exist.
Working with imaginary numbers.
From the laws of surds we can use − 1
Examples
to find other negative square roots
− 2 = 2 × −1 = 2 × − 1 = 2.i
− 3 = 3 × −1 = 3 × − 1 = 3.i
Try these.
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a) − 4
b) − 9
c) − 12
d) − 8
Complex Numbers
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On the TI-83 it is possible to do all of the above, unfortunately it can not give answers in
exact (surd form).
Try
3 you should get
Try
− 3 you should get (notice the use of brackets)
You can see that the calculator returns the decimal equivalent of the exact square root and i
Try the following on the calculator and check them with your answers above.
a) − 4
b) − 9
c) − 12
d) − 8
Powers of i
Powers of i can be simplified as shown
Arithmetic with imaginary numbers
Can imaginary numbers be added and subtracted ?
Try these on the TI-83
3i + 9i = ________________
12i + 4i + 6i = ________________
9i - 4i = ________________
7i - 2i = ________________
Make a statement about how imaginary numbers.are added or subtracted.
__________________________________________________________________________
__________________________________________________________________________
__________________________________________________________________________
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Complex Numbers
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Can imaginary numbers be multiplied ?
Try the following on the TI-83
2i × 5i = ____________________
6i × 5i = ____________________
Explain why the answer is always a “real” number and negative
___________________________________________________________________________
___________________________________________________________________________
___________________________________________________________________________
___________________________________________________________________________
Can imaginary numbers can be divided ?
Try the following on the TI-83
6i ÷ 2i = _____________________
12i ÷ 4i = _____________________
3i ÷ 7i = _____________________
Explain why the answer is always a “real” number.
___________________________________________________________________________
___________________________________________________________________________
___________________________________________________________________________
___________________________________________________________________________
___________________________________________________________________________
___________________________________________________________________________
___________________________________________________________________________
Now do these on the TI-83:
a) 7i + 12i = _________
c) 4i + 6i - 10i = ___________
e) (2i ) 2 =___________
b) 5i + 8i - 20i = _________
d) 2i × (3i + 12i ) =__________
f) (4i + 6i ) 3 = ______________
Try this on the TI-83:
2 + 3i = __________
Comment on the result:________________________________________________________
___________________________________________________________________________
___________________________________________________________________________
T3 Scotland
Complex Numbers
Page 4 of 11
Complex numbers on the TI-83
When a real number and an imaginary number are added together (or subtracted) the
expression which is obtained cannot be simplified and is called a complex number.
e.g
2 + 3i, 4 - 7i and
-1 + 4i
are complex numbers.
A general complex number can be represented in the form a + bi
where a and b can have any real value including zero.
_
_
_
_
If a = 0 we have a number of the form bi i.e. an imaginary number
If b = 0 we have a number of the form a i.e. a real number
Algebra of complex numbers on the TI-83
Addition and subtraction.
Real and imaginary parts of complex numbers can be collected separately in two groups,
e.g. 1.
(2 + 4i) - (3 - i) = (2 + 3) + (4i - i) = 5 + 3i
e.g. 2.
(4 - 2i) - (5 + 6i) = (4 - 5) + (-2 - 6i) = -1+-8i = -1 - 8i
The above two examples can be done on the TI-83 as shown
The trick is to remember the brackets, to keep the signs right.
_
_
_
_
_
_
_
Multiplication
The distributive law of multiplication applies to complex numbers:
e.g. 1.
i (4 - 3i) = 4i - 3i 2 = 4i -3(-1) = 3 + 4i
e.g. 2.
(2 + 3i) (4 - i) = 8 - 2i + 12i - 3i 2 = 8 - 10i - 3(-1) = 11 + 10i
e.g. 3.
(2 + 3i) (2 - 3i) = 4 - 6i + 6i - 9i 2 = 4 + 9 = 13
Again this can be done on the TI-83
_
_
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Complex Numbers
Page 5 of 11
Conjugate complex numbers
The answer in this last example is very important.
Here we see that two complex numbers multiplied can give gave an entirely real number result.
Any two such complex numbers are called complex conjugates.
a2 - abi + abi - b2i 2
=
a2 + b2
(a + bi) (a - bi) =
Generally:
If a complex number, (a + bi ), is denoted by z
then its conjugate, (a - bi ), is denoted by z or z*
Conjugates are easily found on the TI-83.
Press
MATH
and come across to CPX to see the complex numbers menu shown here
ENTER
Press 1 (or
).
This pastes the conjugate command to the home screen
Then type in the complex number and press
ENTER
Use the TI-83 to find the conjugates of the following:
a ) 2 + 5i
b) - 7 - 4i
d ) (7 + 3i) + (2 - 7i)
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c)
1 3
− i
4 2
e) (7 + 3i) × (2 - 7i)
Complex Numbers
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Division
t.
Division of complex numbers cannot be carried out directly because the denominator is made
up of two independent terms.
We can do it, however, if we can make the denominator real and we know how to do this using
the conjugate.
2 + 9i (2 + 9i )(5 + 2i ) 10 + 49i + 18i 2
=
=
5 − 2i (5 − 2i )(5 + 2i )
25 − 4i 2
− 8 + 49i
29
8 49
=− + i
29 29
=
This can be done on the TI-83 as shown
MATH
We can get fractional values by using
to get the convert to fraction command
Press
and
ENTER
ENTER
menu
to see
again to see the answer
Use the TI-83 to find the following simplifying your answer where possible
a) ( 3+i) ÷ ( 4-2i)
d)
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b) 4i ÷ ( 4+i)
2+5i
1-i
Complex Numbers
c) (-2+3i) ÷ i
e)
21+10i
( 4+i)2
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Square roots
Use the TI-83 to find the square root of 15 + 8i
Is this the whole story ?
Try squaring the answer!
Now try squaring -( 4 + i )
It is clear that as with real numbers there are “two” square roots but the calculator only gives us
one.
Finding square roots by hand is a little more complex !
(15 + 8i = a + bi
15 + 8i = (a + bi ) 2
= a 2 + b2 + 2abi
Equating real and imaginary parts we get
a 2 − b2 = 15
2ab = 8
These can be solved to give the answers above and it is clear that because of the squared terms
we would expect “two” answers. TRY IT!
Do these on the calculator and algebraically
T3 Scotland
a)
(3 + i)
d)
2 + 5i
1- i
b)
4i
e)
Complex Numbers
c) (-2 + 3i)
2 + 16i
(4 + 3i)
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Modulus and argument of a complex number
We can show that a complex number can be drawn on a diagram (a little like a vector) with the
real part measured along the x axis and the imaginary part measured “up” the y axis.
Such a diagram is known as an Argand Diagram.
Consider the point A(a , b), representing the complex number a + bi shown on the Argand
Diagram below.
i
A(a,b)
r
θ
Real Numbers
The point A can be found in two ways:
Go “a” along the real number axis and “b” up the i axis
Go “r” along a straight line which makes the angle
with the real numbers axis.
We can relate these two by considering Pythagoras and the tangent ratio.
It is clear that
r = (a 2 + b 2 this is usually refered to as the modulus of a + bi
and is denoted by
The angle
a + bi
−1
 b
 a
with the positive direction of the real numbers axis can be found fromtan  
This is usually called the argument and denoted “arg”
So
and
a + bi = r = (a 2 + b2
 b
arg(a + bi ) = α = tan −1  
 a
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Complex Numbers
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Both can be easily found on the TI-83
MATH
and come across to CPX to see the complex
Press
numbers menu shown.
To find the angle use number 4 then
to paste angle( to the home screen.
ENTER
Type in the complex number and press
(try it by hand to check the angle).
ENTER
To find the modulus we use the abs (the American term)
number 5 as above to give the screen shown.
T3 Scotland
Complex Numbers
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Exercise
1. Simplify
i 2 , i -3, i 9, i -5, i 4n, i 4n+1
2. Add the following pairs of complex numbers:
a ) 3 + 2i and 4 + 6i
b) 5 - i and 3 + i
c) 2 - i and
d) i
- 4 -i
and
2 - 2i
3. Subtract the following pairs of complex numbers:
a ) 3 + 2i and 4 + 6i
b) 5 - i and 3 + i
c) 2 - i and
d) i
- 4 -i
and
2 - 2i
4. Express the following complex numbers in the form
a)
2
3 + 2i
b)
5-i
1+i
c)
4 + i a + bi 1- i
d)
3+i
i
e)
- 2 + 3i
−i
f)
i
-2 -i
5. Solve the following equations for x and y
a ) x + yi = (3 + 2i )(2 − 3i )
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b) x + yi = 2
Complex Numbers
c) x + yi =
3+i
7+i
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