Teachers Teaching with Technology T · T3 Scotland Complex Numbers Page 6 of 11 Conjugate complex...

© Teachers Teaching with Technology (Scotland)

©Teachers Teaching with Technology (Scotland)

Teachers Teaching with Technology

T3 Scotland

Complex Numbers

T3 Scotland Complex Numbers Page 1 of 11

COMPLEX NUMBERSAim

To demonstrate how the TI-83 can be used to explore the concept of and solve problems involving imaginary and complex numbers.

ObjectivesMathematical objectivesBy 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 objectivesBy 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


COMPLEX NUMBERSCalculator Skills Sheet

Set the MODE on the calculator to the screen shown.

Write down what you expected to happen when you type in and give a reason ____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Type in and write down the answer ___________________________________

Find the key and type in:

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 doexist.

Working with imaginary numbers.

From the laws of surds we can use to find other negative square roots

Examples

Try these.

−1

−1

i2nd

ix 2

−1

a) b) c) d)− − − −4 9 12 8

− = × − = × − =2 2 1 2 1 2.i

− = × − = × − =3 3 1 3 1 3.i


On the TI-83 it is possible to do all of the above, unfortunately it can not give answers in exact (surd form).

Try you should get

Try 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.

Powers of iPowers of i can be simplified as shown

Arithmetic with imaginary numbersCan 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.______________________________________________________________________________________________________________________________________________________________________________________________________________________________

3

− 3

a) b) c) d)− − − −4 9 12 8


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 = _________ b) 5i + 8i - 20i = _________c) 4i + 6i - 10i = ___________ d) 2i × (3i + 12i ) =__________e) (2i )2 =___________ f) (4i + 6i )3 = ______________

Try this on the TI-83: 2 + 3i = __________Comment on the result:______________________________________________________________________________________________________________________________________________________________________________________________________________


____

_______

__

Complex numbers on the TI-83When a real number and an imaginary number are added together (or subtracted) theexpression 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 + biwhere 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 numberIf 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


Conjugate complex numbersThe 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.

Generally: (a + bi) (a - bi) = a2 - abi + abi - b2i 2

= a2 + b2

If a complex number, (a + b ), is denoted by

then its conjugate, (a - b ), is denoted by or z*i

i

z

z

Conjugates are easily found on the TI-83.

Press and come across to CPX to see the complex numbers menu shown here

Press 1 (or ENTER

).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:

MATH

a i i i

d

)

)

2 + 5 b) - 7 - 4 c) 14

(7 + 3i) + (2 - 7i) e) (7 + 3i) (2 - 7i)

−

×

32


t.

Division

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.

This can be done on the TI-83 as shown

We can get fractional values by using MATH

menu to get the convert to fraction command

Press ENTER

to see

and ENTER

again to see the answer

Use the TI-83 to find the following simplifying your answer where possible

2 95 2

2 9 22 2

10 49 1825 4

2

2

+−

= + +− +

= + +−

ii

i ii i

i ii

( )(5 )(5 )(5 )

= − +

= − +

8 4929

829

4929

i

i

a) b) c)

d) e)

( +i) ( - i) i ( +i) (- + i) i

+ i-i

+ i( +i)

3 4 2 4 4 2 3

2 51

21 104 2

÷ ÷ ÷


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 !

Equating real and imaginary parts we get

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

(( )

15 815 8

2

2

2 2

+ = ++ = +

= + +

i a bii a bi

a b abi

a bab

2 2 152 8

− ==

a) (3 + i) b) 4i c) (-2 + 3i)

d) 2 + 5i1- i

e) 2 +16i(4 + 3i)


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.

The point A can be found in two ways:Go “a” along the real number axis and “b” up the i axisGo “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 this is usually refered to as the modulus of

and is denoted by

The angle �� with the positive direction of the real numbers axis can be found from

This is usually called the argument and denoted “arg”

So

and

A(a,b)

Real Numbers

i

θ

r

r a b= +( 2 2 a bi+

a bi+

tan−

1 ba

a bi r a b+ = = +( 2 2

arg( ) tana bi ba

+ = =

−α 1

� T3 Scotland Complex Numbers Page 10 of 11

Both can be easily found on the TI-83

Press MATH and come across to CPX to see the complexnumbers menu shown.

To find the angle use number 4 then ENTER

to paste angle( to the home screen.

Type in the complex number and press ENTER

(try it by hand to check the angle).

To find the modulus we use the abs (the American term)number 5 as above to give the screen shown.


Exercise

1. Simplify i 2, i -3, i 9, i -5, i 4n, i 4n+1

2. Add the following pairs of complex numbers:

3. Subtract the following pairs of complex numbers:

4. Express the following complex numbers in the form

5. Solve the following equations for x and y

a i i i i

c i i i i

)

)

3 + 2 and 4 + 6 b) 5 - and 3 +

2 - and - 4 - d) and 2 - 2

a i i i i

c i i i i

)

)

3 + 2 and 4 + 6 b) 5 - and 3 +

2 - and - 4 - d) and 2 - 2

a bi+a

iii

ii

d ii

ii

ii

)

)

23 + 2

b) 5 - c) 1+1-

3 + e) - 2 + 3 f) - 2 -

4 +

−

a x i i i x yi c x yi ii

) ( )( ) ) + y b) 3 += + − + = + =+

3 2 2 3 27

Teachers Teaching with Technology T · T3 Scotland Complex Numbers Page 6 of 11 Conjugate complex...

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