The reason for the introduction of complex numbers is so that every quadratic equation will have a solution. For instance, an equation $x^2-1 = 0$ contains solutions in a set of real numbers, however $x^2+1=0$ does not contains solutions in a set of real numbers. Because of these and similar equations, we expand the set of real numbers ($\mathbb{R}$) to the set in which they will have the solution.
Let $i$ be the intended solution to the equation $x^2 + 1 =0$; therefore $i^2 = -1$. The number $i$ is called the unit imaginary number. The unit imaginary number has the main role in describing a set of complex numbers $\mathbb{C}$ which will be the extension of a set of real numbers $\mathbb{R}$.
The product of any real number $y$ and imaginary unit $i$ is a complex number. Numbers such as these are called imaginary numbers.
A complex number is the addition of a real and an imaginary number, that is, a complex number $z$ is the number of the shape $z= x + yi$, where $x$ and $y $ are real numbers. The number $x$ is called a real part, and $y$ is called an imaginary part of the complex number $z$. We write:
$$x = Re z, \quad \quad y= Im z.$$
A set of complex numbers is denoted as:
$$\mathbb{C} = \{x + yi : x, y \in \mathbb{R} \}.$$
Two complex numbers $z$ and $w$ are equal if
$$z=w \Leftrightarrow Re z = Re w, Im z = Im w.$$
Example 1. Determine $x$ and $y$ such that the following is valid:
$$(2 +3i) + (x+yi) = -7 +3i.$$
Solution:
$$(2 +3i) + (x+yi) = -7 +3i$$
$$(2+x) + (3+y)i = -7 +3i$$
Two complex numbers are equal iff their real and imaginary parts are equal. Therefore, we have:
$$2+x = -7 \Rightarrow x = -7 -2 = -9$$
and
$$3+y = 3 \Rightarrow y= 3-3 =0.$$
Powers of the imaginary unit
$$i^0 = 1$$
$$i^1 = i$$
$$i^2 = -1$$
$$i^3 = i^2 \cdot i = -1 \cdot i = -i$$
$$i^4 = i^3 \cdot i = -i \cdot i = -i^2 = 1.$$
If we count further
$$i^5 = i^4 \cdot i = 1 \cdot i = i$$
$$i^6 = i^5 \cdot i = i\cdot i = i^2 = -1$$
$$i^7 = i^6 \cdot i = -1 \cdot i = -i$$
$$i^8 = i^7 \cdot i = -i \cdot i = -i^2 = 1$$
we can observe that values of powers are repeated. Therefore,
$$i^n = i ^{4a + b} = i^{4a} \cdot i ^b = 1 \cdot i^b = i^b \quad b\in\{0,1,2,3\}, a \in \mathbb{Z}$$
is valid.
Example 2.
Calculate the following:
$$(-2i^{1023} – 3i^{343}) ( -7i^{234} + i^{456}).$$
Solution:
$1023$ by dividing with $4$ gives the rest $3$, that is
$$i^{1023} = i^{4 \cdot 255} \cdot i ^3 = 1 \cdot (-i) = -i.$$
Similarly, we obtain:
$$i^{343} = i^{4 \cdot 85 } \cdot i^3 = 1 \cdot (-i) = -i,$$
$$i^{234} = i^{4 \cdot 58 } \cdot i^2 = 1 \cdot (-1) =-1,$$
$$i^{456} = i^{4 \cdot 114} = 1.$$
Finally, we have:
$$(-2i^{1023} – 3i^{343}) ( -7i^{234} + i^{456}) =( -2 \cdot (-i) – 3 \cdot (-i)) (-7 \cdot(-1) + 1) $$
$$ = (2i + 3i)(7+1)$$
$$= 5i \cdot 8 $$
$$= 40i$$
Example 3. Calculate:
$$i + i ^2 + i ^3 + \ldots + i^{10}.$$
Solution:
Since
$$ i + i^2 + i^3 + i^4 = i + (-1) – i + 1 = 0$$
then we have
$$\underbrace{i + i^2 + i^3 + i^4 }_{=0} + \underbrace{i^5 + i^6 + i^7 + i^8 }_{=0} + i^9 +i^{10} = 0 + 0 + i^9 +i^{10} $$
$$= i^{4 \cdot 2} \cdot i + i^{4 \cdot 2} \cdot i^{2} $$
$$= 1 \cdot i + 1 \cdot (-1) $$
$$ = 1 -1 $$
$$ =0$$
