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Complex number


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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