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Complex Numbers: (lesson 1 of 2)

Complex number arithmetic

Definitions:

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

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

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

Level 2

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

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

Level 2

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

Conjugate complex numbers

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

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

Conjugates are important because of the fact that a complex number times its conjugate is real.

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

Modulus of a complex number

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

Example 4:

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

Exercise 3: Conjugate and modulus

Level 1

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

Level 2

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

$$ \begin{align} \frac{{2 + 3i}}{{4 - 5i}} &= \frac{{2 + 3i}}{{4 - 5i}} \cdot \frac{{4 + 5i}}{{4 + 5i}} = \\ &= \frac{{8 + 10i + 12i + 15{i^2}}}{{16 + 20i - 20i - 25{i^2}}} = \\ &= \frac{{8 + 22i + 15 \cdot ( - 1)}}{{16 - 25 \cdot ( - 1)}} = \\ &= \frac{{ - 7 + 22i}}{{41}} = \\ &= - \frac{7}{{41}} + \frac{{22}}{{41}}i \end{align} $$

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

$$\frac{{a + bi}}{{c - di}} = \frac{{ac + bd}}{{{c^2} + {d^2}}} + \frac{{bc - ad}}{{{c^2} + {d^2}}}$$

Exercise 4: Divide complex numbers

Level 1

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

Level 2

$$ \color{blue}{\frac{{2 + 3i}}{{2 - i}} = } $$ $ \frac{1}{{25}} + \frac{8}{{25}}i $
$ \frac{8}{{25}} + \frac{1}{{25}}i $
$ \frac{1}{5} + \frac{8}{5}i $
$ \frac{8}{5} + \frac{1}{5}i $

Fundamental Theorem of Algebra

Every nth - order polynomial possess exactly n complex roots.