Complex Number (Part 2) Flipbook PDF

Complex Number (Part 2)
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Complex Number Part 2 Learning Outcome: Understand the graphical representation of complex numbers through an Argand diagram.

Understand complex numbers in other forms.

~mys2020~

Complex Number (Graphical Representation) π‘°π’Ž(𝒛)

Argand Diagram

𝒁 = 𝒂 + π’ƒπ’Š

𝒃

Modulus 𝒁 =

𝒂 𝟐 + π’ƒπŸ Argument

𝜽 = π’‚π’“π’ˆ 𝒛 = tanβˆ’πŸ

𝒃 𝒂 𝑹𝒆(𝒛)

𝒂

** 𝜽 = angle between the positive 𝒙 βˆ’ π’‚π’™π’Šπ’” and the line Z. ~mys2020~

2

Complex Number (Graphical Representation) Examples: Represent 𝑧 = 4 + 3𝑖 in an Argand diagram and find its modulus and argument.

π‘°π’Ž(𝒛)

Modulus: 𝒁 =

𝒂 𝟐 + π’ƒπŸ

= πŸ’πŸ + πŸ‘πŸ = 25 =πŸ“ 𝒁 = πŸ’ + πŸ‘π’Š

πŸ‘

Argument:

πŸ“

𝒃 𝒂 πŸ‘ = tanβˆ’πŸ πŸ’ = πŸ‘πŸ”. πŸ–πŸ•Β°

π’‚π’“π’ˆ 𝒛 = tanβˆ’πŸ

𝜽 = πŸ‘πŸ”. πŸ–πŸ•Β° πŸ’

𝑹𝒆(𝒛)

~mys2020~

3

Complex Number (Graphical Representation) Exercise 1: Represent 𝑧 = βˆ’3 + 4𝑖 in an Argand diagram and find its modulus and argument. π‘°π’Ž(𝒛)

Modulus: 𝒁 =

=πŸ“

Argument: π’‚π’“π’ˆ 𝒛 =

𝑹𝒆(𝒛)

= πŸπŸπŸ”. πŸ–πŸ”πŸ—πŸ—Β° ~mys2020~

4

Complex Number (Graphical Representation) Exercise 2: Represent 𝑧 = βˆ’6 βˆ’ 5𝑖 in an Argand diagram and find its modulus and argument. Modulus: 𝒁 =

= πŸ”πŸ

Argument: π’‚π’“π’ˆ 𝒛 =

= πŸπŸπŸ—. πŸ–πŸŽπŸ“πŸ”Β° ~mys2020~

5

Complex Number (Graphical Representation) Exercise 3: Represent 𝑧 = 4 βˆ’ 7𝑖 in an Argand diagram and find its modulus and argument.

Modulus: 𝒁 =

= πŸ”πŸ“

Argument: π’‚π’“π’ˆ 𝒛 =

= πŸπŸ—πŸ—. πŸ•πŸ’πŸ’πŸ—Β° ~mys2020~

6

Complex Number (Graphical Representation) Exercise 4: Represent 𝑧 = 5 + 9𝑖 in an Argand diagram and find its modulus and argument.

Modulus: 𝒁 =

= πŸπŸŽπŸ”

Argument: π’‚π’“π’ˆ 𝒛 =

= πŸ”πŸŽ. πŸ—πŸ’πŸ“πŸ’Β° ~mys2020~

7

Complex Number (Graphical Representation) Exercise 5: Represent 𝑧 = βˆ’8 + 4𝑖 in an Argand diagram and find its modulus and argument. Modulus: 𝒁 =

= πŸ–πŸŽ

Argument: π’‚π’“π’ˆ 𝒛 =

= πŸπŸ“πŸ‘. πŸ’πŸ‘πŸ’πŸ—Β° ~mys2020~

8

Complex Number (Graphical Representation) Exercise 6: Given 𝑧1 = 5 + 3𝑖 and 𝑧2 = 4 βˆ’ 6𝑖, represent each of the following in an Argand diagram and find its modulus and argument. a.

𝑧1 + 𝑧2

b.

= πŸ—πŸŽ & πŸπŸπŸ—. πŸ–πŸŽπŸ“πŸ”Β° ~mys2020~

𝑧1 βˆ’ 𝑧2

= πŸ–πŸ & πŸ–πŸ‘. πŸ”πŸ“πŸ—πŸ–Β° 9

Complex Number (Graphical Representation)

c.

𝑧1 𝑧2

d.

= πŸπŸ•πŸ”πŸ– & πŸ‘πŸ‘πŸ’. πŸ”πŸ“πŸ‘πŸ–Β° ~mys2020~

𝑧1 𝑧2

= 𝟎. πŸ–πŸŽπŸ–πŸ” & πŸ–πŸ•. πŸπŸ•πŸ‘πŸ•Β° 10

Complex Number (Polar form, Exponential form and Trigonometric form)

Polar form

Cartesian form

𝑧 = 𝑧 βˆ πœƒΒ° @ 𝑧 = π‘Ÿβˆ πœƒΒ°

𝑧 = 3 + 5𝑖

Trigonometric form

Exponential form

𝑧 = |𝑧| cos πœƒ + 𝑖 sin πœƒ @ 𝑧 = π‘Ÿ cos πœƒ + 𝑖 sin πœƒ

𝑧 = 𝑧 𝑒 πœƒπ‘– πœƒ 𝑖𝑛 π‘Ÿπ‘Žπ‘‘π‘–π‘Žπ‘›π‘  @ 𝑧 = π‘Ÿπ‘’ πœƒπ‘–

~mys2020~

11

Complex Number (Polar form, Exponential form and Trigonometric form) Example: Convert z = 3 + 5𝑖 into trigonometric form, polar form and exponential form. Step 1: Find the values of modulus and argument 5 3 = πŸ“πŸ—. πŸŽπŸ‘πŸ”πŸΒ°

π‘Ÿ = 32 + 52 = πŸ‘πŸ’

πœƒ = tanβˆ’1

Step 2: Plot the complex number in an Argand diagram π‘°π’Ž(𝒛) πŸ“

𝒓 = πŸ‘ + πŸ“π’Š

𝜽 = πŸ“πŸ—. πŸŽπŸ‘πŸ”πŸΒ° πŸ‘

𝑹𝒆(𝒛) ~mys2020~

12

Complex Number (Polar form, Exponential form and Trigonometric form) Step 3: Substitute the values of π‘Ÿ and πœƒ into the formulas. Trigonometric form

𝑧 = π‘Ÿ cos πœƒ + 𝑖 sin πœƒ = 34 cos 59.0362 + 𝑖 sin 59.0362 Polar form 𝑧 = π‘Ÿβˆ πœƒΒ° = 34∠59.0362Β° Exponential form πœƒ0 = 59.0362 Γ— = 1.0304 π‘Ÿπ‘Žπ‘‘

πœ‹ 180Β°

𝑧 = π‘Ÿπ‘’ πœƒπ‘– = 34𝑒1.0304𝑖 ~mys2020~

13

Complex Number (Polar form, Exponential form and Trigonometric form) Exercise 1: Convert z = 4 + 6𝑖 into trigonometric form, polar form and exponential form.

~mys2020~

= πŸ“πŸ cos πŸ“πŸ”. πŸ‘πŸŽπŸ—πŸ—Β° + π’Š sin πŸ“πŸ”. πŸ‘πŸŽπŸ—πŸ—Β° = πŸ“πŸβˆ πŸ“πŸ”. πŸ‘πŸŽπŸ—πŸ—Β° = πŸ“πŸπ’†πŸŽ.πŸ—πŸ–πŸπŸ–π’Š 14

Complex Number (Polar form, Exponential form and Trigonometric form) Exercise 2: Convert 𝑧 = 25∠48Β° into trigonometric form, cartesian form and exponential form.

= πŸπŸ“ cos πŸ’πŸ–Β° + π’Š sin πŸ’πŸ–Β° = πŸπŸ”. πŸ•πŸπŸ•πŸ“ + πŸπŸ–. πŸ“πŸ•πŸ•πŸ“π’Š = πŸπŸ“π’†πŸŽ.πŸ–πŸ‘πŸ•πŸ–π’Š ~mys2020~

15

Complex Number (Polar form, Exponential form and Trigonometric form) Exercise 3: Given 𝑧1 = 12∠45Β° , 𝑧2 = 24 cos 10Β° + 𝑖 sin 10Β° and 𝑧3 = 14𝑒 0.342𝑖 . Calculate: a.

𝑧1 𝑧2

b.

(in polar form)

= πŸπŸ–πŸ–βˆ πŸ“πŸ“Β°

𝑧1 𝑧3

(in polar form)

= πŸπŸ”πŸ–βˆ πŸ”πŸ’. πŸ“πŸ—πŸ“πŸΒ° ~mys2020~

16

Complex Number (Polar form, Exponential form and Trigonometric form) c.

𝑧2 𝑧1

d.

(in trigonometric form)

= 𝟐 cos βˆ’πŸ‘πŸ“ + π’Š sin βˆ’πŸ‘πŸ“ ~mys2020~

𝑧3 𝑧2

(in exponential form)

= 𝟎. πŸ“πŸ–πŸ‘πŸ‘π’†πŸŽ.πŸπŸ”πŸ•πŸ“π’Š 17

Complex Number (Polar form, Exponential form and Trigonometric form) Exercise 4: Solve the following expression in an exponential form.

10 cos 200Β° + 𝑖 sin 200Β° Γ— 6 cos 10Β° + 𝑖 sin 10Β° 20 cos 70Β° + 𝑖 sin 70

= πŸ‘π’†πŸ.πŸ’πŸ’πŸ‘πŸ“π’Š ~mys2020~

18

To be continued ~mys2020~

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