Why are imaginary numbers defined the way they are?

Question

Any complex number (hopefully this is the right word) can be stated in the form $z = a + bi$ for reals $a$ and $b$ where $\text{Re}(z) = a$ and $\text{Im}(z) = b$, the "real" and "imaginary" parts respectfully. I'm assuming rationals, integers, etc are all considered "complex" (I don't know all the various "number categories" and what's defined as what and where, etc)

Anyway what is the reason for it being defined this way? How are complex/imaginary numbers derived? Are complex and imaginary numbers different things? Why is it $z = a + bi$?

Answer 1

First of all, it's good to understand that "imaginary" is a silly choice of name for a certain type of number. These numbers are neither more nor less "imaginary", in the common language sense, than any other mathematical object.

That said, any number of the form $a+bi$ is complex, where $i=\sqrt{-1}$. Those complex numbers with $b=0$ are called "real", and these include the familiar rational numbers, integers, etc. Those complex numbers with $a=0$ are called "imaginary", and they are precisely those numbers whose squares are non-positive "real" numbers.

Where do they come from? Complex numbers fill in solutions to equations that we could otherwise not solve, and then they turn out to have all kinds of applications. To illustrate what I mean about solving equations, consider that $x^2=a$ has two solutions whenever $a$ is a positive real number, and no real solutions when $a$ is a negative real number. Once we define imaginary numbers, then the equation has two solutions no matter what, with the only unusual case being $a=0$, in which the solution $0$ is repeated twice.

A nice thing about complex numbers it that they provide an "algebraically complete" structure: Any time we write down a polynomial equation, using complex numbers, it has solutions among the complex numbers - exactly as many solutions as its degree.

Answer 2

The algebraic numbers is the set of the roots of polynomial equations.

Some polynomials do not have real roots. i.e. $x^2 + 1 = 0$

The complex numbers generalizes the set of real numbers to include the algebraic numbers as a subset. So, yes, all real numbers are complex numbers.

It then turns out that these complex numbers have some interesting properties in their own right.

Answer 3

In a very short way, the are defined in this way to solve many problems that are not solvable in $\mathbb{R}$.

For example, since we can't find a real solution of quadratic equation in the form $x^2+1=0$ we set $i=\sqrt {-1}$ and then $x=\pm i$ is a solution.

Now we assume for the new object the same algebraic rules we use for all the numbers with the exception that $i^2=-1$, in this way we obtain the complex number $a+ib$. The incredible thing is that all the old algebra works also for them and now we can solve a lot of new problems.

Answer 4

When you want to solve a quadratic equation such as $$x^2+2x+2=0$$ you use the quadratic formula and get the solutions as $$x=-1\pm \sqrt {-1}$$

Now $ \sqrt {-1}$ is not a real number because the square of a real number is always non-negative.

At this point we have two choices.

Either say the equation has no real solution or call $ \sqrt {-1} =i$ which is an imaginary number and accept the complex solutions.

Complex numbers proved to be very useful in science and mathematics. They are now well situated and studied in detail.

College courses like complex analysis have been developed and many theorems have been proved regarding the properties of complex numbers.

Answer 5

It is helpful to see complex numbers as a 2 dimensional $\mathbb{R}-$Algebra,(as a ring which is also a 2 dimensional $\mathbb{R}-$Vector Space), where $\mathbb{R}$ with the usual addition and multiplication is embedded in it, meaning that you can identify it with the subspace and subring $\{a+0.i | a$$\in$$\mathbb{R}$$\}$. Complex numbers is a very natural extension of real numbers in that algebraic sense.