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Complex Numbers

ENGIN 211, Engineering Math

Imaginary Number and the Symbol 𝐽

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Consider the solutions for this quadratic equation: 𝑥2 + 1 = 0

𝑥 = ± −1

−1 is called the imaginary number, and we use the symbol 𝑗 to

represent it: 𝑗 = −1.

Thus, the solutions 𝑥 = ±𝑗

Obviously, 𝑗2 = −1

This notation allowed us to deal with a large number of quadratic equations of

the form: 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0

with the solution 𝑥 =−𝑏± 𝑏2−4𝑎𝑐

2𝑎

In case 𝑏2 − 4𝑎𝑐 < 0, so we can rewrite

𝑥 =−𝑏 ± 𝑗 4𝑎𝑐 − 𝑏2

2𝑎

Complex Numbers

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Example: 𝑥2 + 2𝑥 + 3 = 0

We have two roots

𝑥1 = −1 + 𝑗 2 and 𝑥2 = −1 − 𝑗 2

Both roots consisting of a real number and an imaginary number are

complex number.

Re(𝑥1) = −1 – real part of the complex number 𝑥1

Im(𝑥1) = 2 – imaginary part of the complex number 𝑥1

Re(𝑥2) = −1 – real part of the complex number 𝑥2

Im 𝑥2 = − 2 – imaginary part of the complex number 𝑥2

Note: Both Re(𝑥) and Im(𝑥) are real numbers themselves.

Operations of Complex Numbers

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Addition: −2 + 𝑗6 + 3 − 𝑗4 = 1 + 𝑗2

Subtraction: −2 + 𝑗6 − 3 − 𝑗4 = −5 + 𝑗10

Powers of 𝑗,

𝑗2 = −1, 𝑗3 = 𝑗2𝑗 = −𝑗, 𝑗4 = 𝑗2𝑗2 = −1 −1 = 1, …

Multiplication:

−2 + 𝑗6 3 − 𝑗4 = −2 3 + −2 −𝑗4 + 𝑗6 3 + (𝑗6)(−𝑗4)

= −6 + 𝑗8 + 𝑗18 − 𝑗224 = −6 + 𝑗26 − −1 24= 18 + 𝑗26

Complex Conjugates

Complex conjugate:

𝑎 + 𝑗𝑏 and 𝑎 − 𝑗𝑏 are complex conjugate pairs

𝑎 + 𝑗𝑏 𝑎 − 𝑗𝑏 = 𝑎2 − 𝑗𝑏 2 = 𝑎2 + 𝑏2

– always a real number

Example:

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3 + 𝑗4 3 − 𝑗4 = 32 + 42 = 25

Division of Complex Numbers

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Division by a real number 3+𝑗4

2= 1.5 + 𝑗2

Division by another complex number

3 + 𝑗4

2 − 𝑗3=

3 + 𝑗4 2 + 𝑗3

2 − 𝑗3 2 + 𝑗3

=6 − 12 + 𝑗(8 + 9)

4 + 9=

−6 + 𝑗17

13= −0.46 + 𝑗1.31

Division by itself

3 + 𝑗4

3 + 𝑗4=

3 + 𝑗4 3 − 𝑗4

3 + 𝑗4 3 − 𝑗4=

25

25= 1

Equal Complex Numbers

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If two complex number are equal, their corresponding real and imaginary

parts must be equal:

𝑎 + 𝑗𝑏 = 𝑐 + 𝑗𝑑

Implies:

𝑎 = 𝑐, and 𝑏 = 𝑑

Because we can write

𝑎 − 𝑐 = 𝑗(𝑑 − 𝑏)

And a real number never equals an imaginary number.

Example:

𝑥 = 𝑥1 + 𝑗𝑥2 = 6 + 𝑗8

Then 𝑥1 = 6, and 𝑥2 = 8.

Complex Plane (Argand Diagram)

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Real axis

Imaginary axis 𝑧 = 𝑎 + 𝑗𝑏

(𝑎, 𝑏)

𝑎

𝑏

Graphical addition of two complex numbers

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2

3

1

3

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Polar Form

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Real axis

Imaginary axis

𝑧 = 𝑎 + 𝑗𝑏

𝑎

𝑏 θ

𝑟

Conversion between polar and rectangular forms

𝑟 = 𝑎2 + 𝑏2, 𝜃 = tan−1𝑏

𝑎

𝑎 = 𝑟 cos 𝜃 , 𝑏 = 𝑟 sin 𝜃 , −𝜋 ≤ 𝜃 ≤ 𝜋

Example: 𝑧 = 6 − 𝑗8 can be converted into polar form,

𝑟 = 62 + 82 = 10, 𝜃 = tan−1 −8

6= −36𝑜52′

𝑟: modulus of the complex number 𝑧

𝜃: argument of the complex number 𝑧

Exponential Form

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Maclaurin Series

𝑒𝑥 = 1 + 𝑥 +𝑥2

2!+

𝑥3

3!+

𝑥4

4!+

𝑥5

5!+ ⋯

Let 𝑥 = 𝑗𝜃,

𝑒𝑗𝜃 = 1 + 𝑗𝜃 +𝑗𝜃 2

2!+

𝑗𝜃 3

3!+

𝑗𝜃 4

4!+

𝑗𝜃 5

5!+ ⋯

= 1 −

𝜃2

2!+

𝜃4

4!− ⋯ + 𝑗 𝜃 −

𝜃3

3!+

𝜃5

5!− ⋯

= cos 𝜃 + 𝑗 sin 𝜃 − 𝐄𝐮𝐥𝐞𝐫′𝐬 𝐢𝐝𝐞𝐧𝐭𝐢𝐭𝐲

Thus, 𝑟 cos 𝜃 + 𝑗 sin 𝜃 = 𝑟𝑒𝑗𝜃 – exponential form

Three ways: 𝑧 = 𝑎 + 𝑗𝑏 = 𝑟 cos 𝜃 + 𝑗 sin 𝜃 = 𝑟𝑒𝑗𝜃

Which Form Is Better?

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It depends on what you want to do:

For addition and subtraction, it is better to work with the

rectangular form 𝑧 = 𝑎 + 𝑗𝑏

For multiplication and division, it is better to work with

exponential form, 𝑧 = 𝑟𝑒𝑗𝜃, e.g.,

4𝑒𝑗60𝑜

2𝑒𝑗30𝑜 = 2𝑒𝑗 60𝑜−30𝑜= 2𝑒𝑗30𝑜

To convert between rectangular and exponential form, we work

thru polar form, 𝑧 = 𝑟 cos 𝜃 + 𝑗 sin 𝜃

3 + 𝑗4 = 5 cos 36𝑜52′ + 𝑗 sin 36𝑜52′ = 5𝑒𝑗36𝑜52′

Powers

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Consider: 𝑧 = 𝑟𝑒𝑗𝜃

𝑧2 = 𝑟𝑒𝑗𝜃 2= 𝑟2𝑒𝑗2𝜃 = 𝑟2 cos 2𝜃 + 𝑗 sin 2𝜃 , … 𝑧𝑛 = 𝑟𝑒𝑗𝜃 𝑛

=

𝑟𝑛𝑒𝑗𝑛𝜃 = 𝑟𝑛 cos 𝑛𝜃 + 𝑗 sin 𝑛𝜃 − DeMoivre’s theorem

If 𝑛𝜃 > 𝜋, we need to add or subtract multiples of 2𝜋 so that the equivalent

angle 𝜑 stays within 𝜑 ≤ 𝜋.

Example: 𝑧 = 2𝑒𝑗120𝑜= 2∠120𝑜, 𝑧5 = 25𝑒𝑗5×120𝑜

= 32∠600𝑜 = 32∠ −120𝑜

120𝑜

−120𝑜

𝑧

𝑧5

Roots

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Roots of a complex number is somewhat complicated and requires careful

attention in manipulation.

Example, 𝑤 = 32∠−120𝑜= 32𝑒−𝑗120𝑜, let’s find 𝑤1/5=?

1) We can take do the following

32𝑒−𝑗120𝑜 1/5= 32 1/5𝑒−𝑗120𝑜/5 = 2𝑒−𝑗24𝑜

= 𝟐∠ − 𝟐𝟒𝒐

But we certainly expect to recover 𝑧 = 2𝑒𝑗120𝑜= 2∠120𝑜 because

𝑧5 = 32∠ −120𝑜 from the previous example of powers, so what

happened?

2) Of course, we can say ∠ − 120𝑜 is also ∠360𝑜 − 120𝑜 = ∠240𝑜, so now

32𝑒−𝑗120𝑜 1/5= 32𝑒𝑗240𝑜 1/5

= 32 1/5𝑒𝑗240𝑜/5 = 2𝑒𝑗48𝑜= 𝟐∠𝟒𝟖𝒐

But the solution 𝑧 = 2𝑒𝑗120𝑜= 2∠120𝑜 is still not recovered.

Roots (Cont’d)

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3) Well, we can keep adding ∠720𝑜 − 120𝑜 = ∠600𝑜, so now

32𝑒−𝑗120𝑜 1/5= 32𝑒𝑗600𝑜 1/5

= 32 1/5𝑒𝑗600𝑜/5 = 2𝑒𝑗120𝑜= 𝟐∠𝟏𝟐𝟎𝒐 This is

it, but how do we know we have gotten them all?

4) We can also subtract 360𝑜, so ∠ − 360𝑜 − 120𝑜 = ∠ −480𝑜, then

32𝑒−𝑗120𝑜 1/5= 32𝑒−𝑗480𝑜 1/5

= 32 1/5𝑒−𝑗480𝑜/5 = 2𝑒−𝑗96𝑜= 𝟐∠ −𝟗𝟔𝒐

5) Of course, we can also obtain

32𝑒−𝑗120𝑜 1/5= 32𝑒−𝑗840𝑜 1/5

= 32 1/5𝑒−𝑗840𝑜/5 = 2𝑒−𝑗168𝑜= 𝟐∠ −𝟏𝟔𝟖𝒐

When do we stop?

Roots (Cont’d)

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Let’s plot these roots in the complex plane.

𝟐∠ − 𝟐𝟒𝒐

𝟐∠𝟒𝟖𝒐 𝟐∠𝟏𝟐𝟎𝒐

𝟐∠ −𝟗𝟔𝒐

𝟐∠ −𝟏𝟔𝟖𝒐

𝟑𝟐𝒆−𝒋𝟏𝟐𝟎𝒐

We can see that these five roots are

equally separated by 360𝑜

5= 72𝑜.

It is generally true that the n-th order roots have n number of values that are

equally separated by 𝟑𝟔𝟎𝒐

𝒏.

So in practice, we only need to find one as we did in Step 1), and the remaining 𝑛 − 1 roots will be generated similar to the illustration in the complex plane plot.

The one nearest to the positive real axis is called the principal root.

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Summary Key points:

Recognize 𝑗 = −1 for imaginary number

Three forms of the complex number: rectangular,

polar, and exponential

Conversions between the three forms

Complex conjugate pairs

Addition, subtraction, multiplication (powers), and

division

Roots of 𝑛-th order (equally spaced by 360𝑜/𝑛 on a

circle)

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