College Algebra-Math 121

Dr. M. Hameed

Section 1.4 Complex Numbers

Lecture-4

Objectives

Introduce the concept of complex numbers &

imaginary unit

How to deal with minus in square roots.

Add & Subtract Complex Numbers

Multiply & Divide Complex Numbers

Imaginary Unit • Until now, you have always been told that you

can’t take the square root of a negative number. If you use imaginary units, you can,

• The imaginary unit is ¡.

• It is used to write the square root of a negative number. It is not a variable.

i 1

i2 1

In complex numbers, the imaginary unit 𝑖 and its higher powers will appear very frequently.

We will reduce all the higher powers down to a single power of 𝒊

To reduce higher powers, we will use following method.

We know 𝑖 = −1, then 𝑖2 = −1

Now if 𝑖3 is given, we will reduce it to 𝑖

𝑖3 = 𝑖2. 𝑖 = −1 . 𝑖 = −𝑖

𝑖4 = 𝑖2. 𝑖2 = −1 . (−1) = 1

𝑖5 = 𝑖4. 𝑖 = 1 . (𝑖) = 𝑖

𝑖6 = 𝑖4. 𝑖2 = 1 . −1 = −1

Few Considerations about imaginary unit 𝒊

How to deal with square root of negative numbers

• If b is a positive real number, then

bibbb 1))(1(

Examples:

3 3i 4 4i i2

• Before you do anything with negatives inside square-roots, you MUST convert the negative to 𝑖, then proceed.

Definition of Complex Numbers

If a and b are real numbers and i is the imaginary

unit, then the number 𝑎 + 𝑏𝑖 is called a complex

number in its standard form.

Example: (i) 2 + 6𝑖 is a complex number

As you can see, a complex number has two parts.

Real part: is the one without i = 2

Imaginary part is the one with i = 6

Examples Exp-1. Simplify and write complex numbers in standard form 1 + −8 Remember that, before you do anything, first get rid of negative inside the square-root By introducing the imaginary 𝑖

1 + −8 = 1 + 𝑖 8 = 1 + 𝑖2 2 = 1 + 2 2𝑖

Exp-2. Simplify and write complex numbers in standard form −4𝑖2 + 2𝑖 Get rid of negative inside the square-root, By introducing the imaginary 𝑖

−4𝑖2 + 2𝑖 = −4 −1 + 2𝑖 = 4 + 2𝑖

NOTE: By standard form we mean that we should be able to explicitly pick, a and b

Addition & Subtracting Complex Numbers

Addition If we two complex numbers, 𝑎 + 𝑏𝑖, and c + 𝑑𝑖, Then 𝑎 + 𝑏𝑖 + 𝑐 + 𝑑𝑖 = 𝑎 + 𝑐 + 𝑏 + 𝑑 𝑖 Remember it this way, when we adding two complex numbers, real adds to real part, and imaginary part adds to imaginary part. Then simplify to write it in standard form.

Subtraction If we two complex numbers, 𝑎 + 𝑏𝑖, and c + 𝑑𝑖, Then 𝑎 + 𝑏𝑖 − 𝑐 + 𝑑𝑖 = 𝑎 − 𝑐 + 𝑏 − 𝑑 𝑖 Again, when we subtracting two complex numbers, change signs of what is being subtracted. Then combine real with real, and imaginary with imaginary. Make sure to write it in standard form.

Examples Adding & Subtracting Complex Numbers

Exp-1. Perform addition or subtraction and write the result in standard form . 13 − 2𝑖 + −5 + 6𝑖 Solution: Remove parenthesis and combine like terms,

13 − 2𝑖 − 5 + 6𝑖 = 13 − 5 + 6𝑖 − 2𝑖 = 8 + 4𝑖

Exp-2. Perform addition or subtraction and write the result in standard form .

8 + −18 − 4 + 3 2𝑖

Solution: Start by converting negative square-roots by introducing i

8 + 𝑖 18 − 4 + 3 2𝑖 = 8 + 𝑖 (9)(2) − 4 + 3 2𝑖

= 8 + 𝑖3 2 − 4 + 3 2𝑖

Remove parenthesis, distribute minus and combine like terms,

= 8 − 4 + 3 2𝑖 − 3 2 = 4 It is in standard form, with a=4, b=0

Multiplication of complex Numbers

If we two complex numbers, 𝑎 + 𝑏𝑖, and c + 𝑑𝑖,

Then 𝑎 + 𝑏𝑖 𝑐 + 𝑑𝑖 = 𝑎𝑐 − 𝑏𝑑 + 𝑎𝑑 + 𝑏𝑐 𝑖

To multiply two complex numbers,

(i) Multiply like two factors in a normal way. (FOIL)

(ii) Convert 𝑖2 to −1,

(iii) Combine real and imaginary parts,

(iv) Make sure to put in Standard form

Examples Multiplying Complex Number Exp-1 . Perform multiplication and write the result in standard form . 6 − 2𝑖 2 − 3𝑖 Solution: Multiply in a usual way 6 − 2𝑖 2 − 3𝑖 = 12 − 18𝑖 − 4𝑖 + 6𝑖2 Put 𝑖2 = −1 = 12 − 18𝑖 − 4𝑖 − 6 = (12 − 6) − 18𝑖 − 4𝑖 = 6 − 22𝑖

EXP-2. Perform multiplication and write the result in standard form .

3 + 15𝑖 3 − 15𝑖

Solution: Multiply in a usual way

3 + 15𝑖 3 − 15𝑖 = 3 − 45𝑖 + 45𝑖 + 15𝑖2

Put 𝑖2 = −1 = 3 − 15 = −12

Conjugate of complex numbers

If we have a complex number 𝑎 + 𝑏𝑖, then its complex conjugate is 𝑎 − 𝑏𝑖 Similarly, the complex conjugate of 𝑎 − 𝑏𝑖 is 𝑎 + 𝑏𝑖

Basically, To get conjugate of a complex number, replace 𝑖 by −𝑖

EXAMPLE (1) Find complex conjugate of 𝟕 − 𝟏𝟐𝒊

Replace 𝑖 with – 𝑖, 7 − 12𝑖 = 7 − 12 −𝑖 = 7 + 12𝑖

(2) Find complex conjugate of 𝟏 + 𝟖

Replace 𝑖 with – 𝑖, But there is no 𝑖. So nothing to worry. Conjugate is same

𝟏 + 𝟖

Division of Complex numbers

If we have two complex number 𝑎 + 𝑏𝑖, and c + 𝑑𝑖

Then to divide complex numbers 𝑎+𝑏𝑖

𝑐+𝑑𝑖 , Follow these steps.

Steps, (i) Take a complex conjugate of denominator (𝑐 − 𝑑𝑖)

(ii) Multiply the conjugate of denominator 𝑎+𝑏𝑖

𝑐+𝑑𝑖.(

𝑐−𝑑𝑖

𝑐−𝑑𝑖)

(iii) Multiply out numerators and denominator (iv) Simplify and write result in standard form

Examples Divide Complex Numbers

Exp-1. Divide and write the result in standard form . 5

1−𝑖

Solution: Multiply top and both with conjugate

5

1−𝑖.

(1+i)

(1+i)

Multiply out numerators and denominators 5 + 5𝑖

1 − 𝑖 + 𝑖 − 𝑖2 =5 + 5𝑖

1 − 𝑖2

Put 𝑖2 = −1 =5+5𝑖

1+1=

5

2+

5

2𝑖

Exp-2. Divide and write the result in standard form 5𝑖

2+3𝑖 2

Solution: Expand denominator before you can take a conjugate 5𝑖

4+12𝑖+9𝑖2 =5𝑖

−5+12𝑖

Multiply top and both with conjugate

5𝑖

−5+12𝑖.

(−5−12i)

(−5−12i) =

−25𝑖−60𝑖2

25+60𝑖−60𝑖−144𝑖2

=60−25𝑖

169= −

60

169−

25

169𝑖

Thank you