1.1 Complex Numbers

Before we explore complex numbers, let's first consider the question: "Is there a real number whose square is negative?" To answer this question, we can examine a few simple examples. Let's take a look at equations 1 and 2 for better understanding.

This is because squaring any real number never results in a negative value. Therefore, if equation 2 has a solution, we need to introduce a new kind of number, an imaginary number, defined as the square root of $- 1$. This imaginary unit is represented by the symbol $i$ (iota). The imaginary number $i$ tells us that $i^{2} = - 1$.

$i^{17}=i^{16}\times i=(i^2{)}^8\times i=(-1{)}^8\times i=i$$i^{17} = i^{16}\times i = (i^{2})^{8}\times i = (-1)^{8}\times i = i$

$i^{-25}=\frac{1}{i^{25}}=\frac{1}{i^{24}\times i}=\frac{1}{(i^2{)}^{12}\times i}=\frac{1}{(-1{)}^{12}\times i}=\frac{1}{i}=\frac{1}{i}\times \frac{i}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i$$i^{- 25} = \frac{1}{i^{25}} = \frac{1}{i^{24}\times i} = \frac{1}{(i^{2})^{12}\times i} = \frac{1}{(-1)^{12}\times i} = \frac{1}{i} = \frac{1}{i}\times \frac{i}{i} = \frac{i}{i^{2}} = \frac{i}{-1} = -i$

We have observed that the equation $x^{2} + 1 = 0$ has no solution within the real number system. To address equations like this, we expand the real number system to include new types of numbers, leading to the development of the complex number system in rectangular form.