# What are imaginary numbers?

1. Aug 19, 2015

### Sobhan

What are imaginary numbers?
Does any one know a good book for it?

2. Aug 19, 2015

### Staff: Mentor

Last edited by a moderator: May 7, 2017
3. Aug 19, 2015

### micromass

What level are you at? What level are you looking for?

4. Aug 19, 2015

### Sobhan

i need a book for basics of it.

5. Aug 19, 2015

### micromass

You're not answering my question. What level are you at?

6. Aug 19, 2015

### Sobhan

highschool

7. Aug 19, 2015

### micromass

High school can mean a lot of things. It can mean you know calculus, or it can mean you don't even know algebra yet.

8. Aug 19, 2015

### Sobhan

i know calculus but not completely

9. Aug 19, 2015

### BvU

Well, my respect for your mathematical curiosity !

As you may know, we can't find a real number for which the square is -1. Imaginary numbers appear if we pretend we can: we call $i$ the number for which $i^2 = -1$ (so we simply imagine things ). From that moment on we can (pretend to ) do a whole lot of things: we can also solve for $\sqrt {-1}$, although a small duality arises: because if $i^2 = -1$, then also $(-i)^2 = -1$ !

Imaginary numbers are an extension of the real numbers so that we can invert a mathematical operation: the operation of "raising to a power".

In the same way we extended natural numbers with integer numbers so that we could invert addition:
with natural numbers we can't find the natural number to solve ? + 5 = 2
So we invented negative numbers

And with integer numbers we couldn't solve ? * 3 = 2
So we invented fractions and had a set of rational numbers

With rational numbers we still couldn't solve ? 2 = 2
so we invented real numbers

And with real numbers we couldn't solve ? 2 = -1
And that's where imaginary numbers appear on stage !

10. Aug 25, 2015

### jk22

$$?^2=2$$ needs algebraic numbers.

There are transcendental numbers too for example sin(?)=1 even im not sure it is transcendent.

Most real numbers are transcendent. Complex number is a extension of real numbers you can represent the imaginary unit by a real 2x2 matrix $$i=\left(\begin{array}{cc} 0&-1\\1&0\end{array}\right)$$

Last edited: Aug 25, 2015
11. Aug 25, 2015

### HallsofIvy

Another way of defining complex numbers is this: the set of complex numbers is the set of ordered pairs of real numbers, (a, b), with addition defined by (a, b)+ (c, d)= (a+ c, b+ d) and multiplication defined by (a, b)*(c, d)= (ac- bd, ad+ bc).

One thing we can show immediately is that pairs with second number, 0, (a, 0), have addition (a, 0)+ (b, 0)= (a+ b, 0+ 0)= (a+ b, 0) and multiplication (a, 0)*(b, 0)= (a*b- 0*0, a*0+ b*0)= (ab, 0) so that we can "identify" the real number, a, with the pair (a, 0) and, in that sense, think of the real numbers as being a "subset" of the complex numbers.

But (0, 1)*(0, 1)= (0*0- 1*1, 0*1+ 1*0)= (-1, 0) so that, in this "number system", unlike the real number, there exist a "complex number" (pair) is -1.
We can write (a, b)= (a, 0)+ (0, b)= a(1, 0)+ b(0, 1). We have already identified (1, 0) with the real number, 1. If we use "i" to represent the pair (0, 1), in the same way that we are using "1" to represent the pair (1, 0), we can write (a, b)= a+ bi in the more usual notation.

12. Jul 6, 2016

### QuantumForumUser

When a negative number is under a square root radical sign (√), you can use the multiplication property of square roots (√(ab)=√(a)×√(b)) to change how the answer "looks". Let's say √(-3). The number -3 can factor out as -1 times 3 so √(-3)=√(-1)×√(3). Since i is equal to √(-1), one can write √(-3) equals i×√(3).