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Complex Numbers: (lesson 1 of 2)

Complex number arithmetic

Definitions:

1. i=1i = \sqrt { - 1}, i2=1{i^2} = - 1

2. A complex number is any number of the form a + bi where a and b are real numbers.

Addition and Subtraction of complex numbers

To add or subtract two complex numbers, you add or subtract the real parts and the imaginary parts.

(a + bi) + (c + id) = (a + c) + (b + d)i.

(a + bi) - (c + id) = (a - c) + (b - d)i.

Example 1:

(3 - 5i) + (6 + 7i) = (3 + 6) + (-5 + 7)i = 9 + 2i.

(3 - 5i) - (6 + 7i) = (3 - 6) + (-5 - 7)i = -3 - 12i.

Exercise 1: Addition and Subtraction

Level 1

(2+3i)+(45i)= \color{blue}{(2 + 3i) + (4 - 5i) = } 6+2i 6 + 2i
2+6i 2 + 6i
26i 2 - 6i
62i 6 - 2i

Level 2

(26i)(3+3i)= \color{blue}{(2 - 6i) - ( - 3 + 3i) = } 59i 5 - 9i
5+9i 5 + 9i
95i 9 - 5i
9+5i 9 + 5i

Multiplying complex numbers

Example 2:

Let's take specific complex numbers to multiply, say 2 + 3i and 2 - 5i.

(2 + 3i)(2 - 5i) = 4 - 10i + 6i - 15i2 = 4 - 4i - 15i2

The definition of i tells us that i2 = -1 . Therefore,

(2 + 3i)(2 - 5i) = 4 - 4i -15(-1) = 19 - 4i.

If you generalize this example, you'll get the general rule for multiplication

(x + yi)(u + vi) = (xu - yv) + (xv + yu)i

Exercise 2: Multiplying complex numbers

Level 1

(3+2i)(23i)= \color{blue}{(3 + 2i) \cdot (2 - 3i) = } 12+5i 12+5i
125i 12-5i
12+5i -12+5i
125i -12-5i

Level 2

(25i)(2+5i)= \color{blue}{( - 2 - 5i) \cdot ( - 2 + 5i) = } 29i 29i
29 29
29i -29i
29 -29

Conjugate complex numbers

We define the conjugate of a + bi as a+bi=abi\overline {a + bi} = a - bi

Example 3: 2+3i=23i\overline {2 + 3i} = 2 - 3i

Conjugates are important because a complex number times its conjugate is a real number.

Example 4: (3+4i)(34i)=912i+12i16i2=916(1)=25(3 + 4i) \cdot (3 - 4i) = 9 - 12i + 12i - 16{i^2} = 9 - 16 \cdot ( - 1) = 25

Modulus of a complex number

We define modulus or absolute value of complex number a + bi as a2+b2\sqrt {{a^2} + {b^2}}. We write modulus of a + bi as |a + bi|.

Example 4:

3+4i=32+42=5{\rm{|3 + 4i|}} = \sqrt {{3^2} + {4^2}} = 5

Exercise 3: Conjugate and modulus

Level 1

(32i)(3+2i)= \color{blue}{\overline {(3 - 2i)} \cdot (3 + 2i) = } 13 -13
13i 13i
13 13
13i -13i

Level 2

43i= \color{blue}{\left| {4 - 3i} \right| = } 5 5
5i 5i
25 25
25i 25i

Division of complex numbers

The process of division of complex numbers:

step 1: Find the conjugate of a denominator.

step 2: Multiply the complex fraction, both top and bottom complex number.

Here is the complete division problem:

2+3i45i=2+3i45i4+5i4+5i==8+10i+12i+15i216+20i20i25i2==8+22i+15(1)1625(1)==7+22i41==741+2241i \begin{align} \frac{{2 + 3i}}{{4 - 5i}} &= \frac{{2 + 3i}}{{4 - 5i}} \cdot \frac{{4 + 5i}}{{4 + 5i}} = \\ &= \frac{{8 + 10i + 12i + 15{i^2}}}{{16 + 20i - 20i - 25{i^2}}} = \\ &= \frac{{8 + 22i + 15 \cdot ( - 1)}}{{16 - 25 \cdot ( - 1)}} = \\ &= \frac{{ - 7 + 22i}}{{41}} = \\ &= - \frac{7}{{41}} + \frac{{22}}{{41}}i \end{align}

Now, we can write down a general formula for division of complex numbers

a+bicdi=ac+bdc2+d2+bcadc2+d2\frac{{a + bi}}{{c - di}} = \frac{{ac + bd}}{{{c^2} + {d^2}}} + \frac{{bc - ad}}{{{c^2} + {d^2}}}

Exercise 4: Divide complex numbers

Level 1

1i1+i= \color{blue}{\frac{{1 - i}}{{1 + i}} = } 1i -1-i
1+i -1+i
i i
i -i

Level 2

2+3i2i= \color{blue}{\frac{{2 + 3i}}{{2 - i}} = } 125+825i \frac{1}{{25}} + \frac{8}{{25}}i
825+125i \frac{8}{{25}} + \frac{1}{{25}}i
15+85i \frac{1}{5} + \frac{8}{5}i
85+15i \frac{8}{5} + \frac{1}{5}i

Fundamental Theorem of Algebra

Every nth - order polynomial possess exactly n complex roots.