Today we will learn about the

**complex numbers**. The crucial point about complex numbers is that we accept a very special number that we will denote it by $i$. This number $i$ is very special because it satisfies the following identity $$i^2 = -1.$$

So a complex number will have the form $$a + ib$$ where $a$ and $b$ are two

*real numbers*. When $b=0$ then $a + ib = a$ is just a normal

*real number*, and when $a=0$ then $a + ib = ib$ is called a

*pure imaginary number*. Here are some examples of complex numbers: $$1+ i, ~~ 2 - 3i, ~~ -\sqrt{3} + 4i, ~~5i - 4, ~~6, ~~i, ~~-3i, ~~4 + 2i, \dots$$

First, let us learn about some basic arithmetic like addition, subtraction, multiplication, division of complex numbers.

**Addition and Subtraction:**are quite simple and trivial as follows $$(3 + 2i) + (4 + i) = 7 + 3i, ~~(3 + 2i) - (4+i) = -1 + i,$$ $$(2 + i) + (3 - 4i) = 5 - 3i, ~~ (2 + i) - (3 - 4i) = -1 + 5i,$$ $$2i + (3i + 1) = 1 + 5 i, ~~ 2i - (3i + 1) = -1 - i,$$

In general, we have $$(a + i b) + (c + i d) = (a+c) + i (b+d) , ~~ (a + i b)- (c + i d) = (a-c) + i (b - d). $$

**Multiplication:**we need to use the identity $i^2 = -1$ in multiplication, for examples, $$(3 + 2i) (4 + i) = 12 + 3i + 8i + 2 i^2 = 12 + 11 i - 2 = 10 + 11 i,$$ $$(2 + i) ( 3 - 4i) = 6 - 8i + 3i - 4 i^2 = 6 - 5i + 4 = 10 - 5i,$$ $$2i (3i+1) = 6 i^2 + 2i = -6 + 2i .$$

In general, we have $$(a + i b)(c + i d) = ac + i ad + i bc + i^2 bd = (ac - bd) + i (bc + ad ) .$$

**Division:**let us recall the division of the radicals, suppose that we want to calculate $$\frac{1 + 4 \sqrt{3}}{5 + 2 \sqrt{3}},$$ then we will use the following identity $$(5 + 2 \sqrt{3})(5 - 2 \sqrt{3}) = 5^2 - (2 \sqrt{3})^2 = 25 - 12 = 13.$$

We have $$\frac{1 + 4 \sqrt{3}}{5 + 2 \sqrt{3}} = \frac{(1 + 4 \sqrt{3})(5 - 2 \sqrt{3})}{(5 + 2 \sqrt{3})(5 - 2 \sqrt{3})}= \frac{5 - 2 \sqrt{3} + 20 \sqrt{3} - 24}{25 - 12} = \frac{-19 + 18 \sqrt{3}}{13} = -\frac{19}{13} + \frac{18}{13} \sqrt{3}.$$

Similarly, for complex numbers, we will use the identity $$(a + i b)(a - ib ) = a^2 - i^2 b^2 = a^2 + b^2 .$$ We have $$\frac{c + i d}{a + i b} = \frac{(c + i d)(a - ib)}{(a + ib)(a - ib)} = \frac{(ac + bd) + i(ad - bc)}{a^2 + b^2} = \frac{ac + bd}{a^2 + b^2} + i \frac{ad - bc}{a^2 + b^2}.$$

Here are some examples, $$\frac{3 + 2i}{4 + i} = \frac{(3 + 2i)(4 - i)}{(4+i)(4-i)} = \frac{12 - 3i + 8 i - 2 i^2}{16 - i^2} = \frac{14 + 5i}{17} = \frac{14}{17} + \frac{5}{17} i,$$ $$\frac{2+i}{3 - 4i} = \frac{(2+i)(3 + 4i)}{(3 - 4i)(3+4i)} = \frac{6 + 8i + 3 i + 4 i^2}{9 - 16 i^2} = \frac{2 + 11 i}{25} = \frac{2}{25} + \frac{11}{25} i,$$ $$\frac{2i}{3i + 1} = \frac{2i(1 - 3i)}{(1+3i)(1-3i)} = \frac{2i - 6 i^2}{1 - 9 i^2} = \frac{6 + 2i}{10} = \frac{3}{5} + \frac{1}{5} i,$$

**Conjugation:**as we have seen above, the two complex numbers $a + i b$ and $a - ib$ are used together in division because their product is a real number $$(a + i b)(a - ib ) = a^2 - i^2 b^2 = a^2 + b^2.$$

These two complex numbers are called

**conjugate**and they are very useful when they are used together. We will use the notation $\overline{z}$ to denote the conjugate of $z$, thus, $$\overline{a + i b} = a - ib, ~~~~\overline{a- ib} = a + ib.$$

In special cases, the conjugate of a real number $a$ is $a$, and the conjugate of $ib$ is $-ib$.

**Absolute value:**for real numbers, we have the concept of absolute value, $$|5| = 5, ~~~ |-5| = 5.$$

This concept can be extended for complex numbers. The absolute value of a complex number $z = a + ib$ is denoted by $|z|$ and defined as $$|z| = |a + ib| = \sqrt{a^2 + b^2}.$$

It follows that $$|z|^2 = a^2 + b^2 = (a + ib)(a - ib) = z ~ \overline{z}.$$

Here are some examples, $$|3 + 2i| = \sqrt{9 + 4} = \sqrt{13},$$ $$|2 - i| = \sqrt{4 + 1} = \sqrt{5},$$ $$|3i| = \sqrt{0+9} = 3,$$ $$|2| = \sqrt{4 + 0} = 2.$$

In special cases, we have $|5| = 5$, $|-5|=5$ and $|3i| = 3$, $|-3i| = 3$.

**Fibonacci's Identity**

Absolute value has the following important property $$|u|~|v| = |uv|.$$

This property gives rise to the following Fibonacci's identity $$(a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2 .$$

Indeed, if we take $u = a + ib$ and $v = c + id$ then $uv = (ac - bd) + i(ad + bc)$ and we have $$|u| = \sqrt{a^2 + b^2},$$ $$|v| = \sqrt{c^2 + d^2},$$ $$|uv| = \sqrt{(ac-bd)^2 + (ad + bc)^2}.$$

So the identity $|u| |v| = |uv|$ becomes $$(a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2 .$$

If you ever forget about this identity and would like to construct it back, you can follow the following steps

- First, write two arbitrary complex numbers and calculate the product $$(a + ib)(c + id) = (ac - bd) + i (ad + bc), $$

- Next, substitute the plus sign by the minus sign, $$(a - ib)(c - id) = (ac - bd) - i (ad + bc),$$

- Finally, multiply the above two equations together $$[(a + ib)(a - ib)] [(c + id)(c - id)] = [(ac - bd) + i (ad + bc)] [(ac - bd) - i (ad + bc)]$$ and we obtain the required identity $$(a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2 .$$

Today we have learned some basic facts about complex numbers. This may surprise you but complex numbers are actually used a lot in number theory. We will learn more about complex number in the next post. Hope to see you then.

*Homework.*

1. Perform addition, subtraction, multiplication and division for the following pairs of complex numbers:

- $3 + 4i$, $2 + 3i$;
- $4i$, $2-i$;
- $7 - i$, $2 + 4i$;
- $5 - 2i$, $3 + i$;
- $4 + 3i$, $-i$.

2. Find the complex conjugates for $5 + 3i$, $4-2i$, $1 + i$, $-4$, $-1 + 2i$, $5i$, $-4i$.

3. Calculate the absolute values for $2 + 3i$, $2-3i$, $4 + i$, $4-i$, $5 + 2i$, $5-2i$, $6$, $-6$, $2 - i$, $2 + i$, $4i$, $-4i$.

4. Calculate $$i^{3}, ~~~~i^{4}, ~~~~i^5, ~~~~i^6, \dots.$$

Find a general formula for $i^n$.

5. Prove that $$\left| \frac{u}{v} \right| = \frac{|u|}{|v|}.$$

6. Prove the following inequality $$|u + v| \leq |u| + |v|.$$

7. Show that for any complex number $z$, there exists $\phi$ such that $$z = |z| (\cos{\phi} + i ~ \sin{\phi}).$$

8. Use mathematical induction to prove the following identity $$(\cos{\phi} + i ~ \sin{\phi})^n = \cos{(n ~\phi)} + i ~ \sin{(n ~\phi)}.$$

9. Prove that $$(\cos{\alpha} + i ~\sin{\alpha})(\cos{\beta} + i ~\sin{\beta}) = \cos{(\alpha + \beta) + i ~\sin{(\alpha + \beta)}} .$$

10. Show that there exist two natural numbers $a$ and $b$ such that $$a^2 + b^2 = 5^{2013}.$$

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