# Imaginary numbers

1. Mar 30, 2005

### minase

I don't get the use of imaginary numbers. To find the square root of negative numbers :grumpy: but it does not exist and it is not a real number. Can u please explain it to me.

2. Mar 30, 2005

### Gale

they certainly do exist... they just aren't real. they're just a special sort of number. Like, what's the deal with negative numbers??? i mean, we don't have negative apples, or negative space... they make no sense... (matter of fact, the greeks refused to accept negative numbers for just that reason.) just like many things in math, imaginary numbers are just a concept, that when manipulated correctly, help to explain physical phenomena.

3. Mar 30, 2005

### Delta

One of these concepts being electronic engineering especially in tuned circuits, etc, where the principles of working with imaginery numbers are very evident.

4. Mar 30, 2005

### tutor69

The "problem" that leads to complex numbers concerns solutions of equations. All quadratic equations and Polynomials will have defined roots if we introduce complex numbers. for example

a) x^2 = 1 , has two roots: x= 1 and x = -1, both are real and you can visualise them on xy plot in a real plane.

B) x^2 = -1, has two roots again $$x = \sqrt{-1}$$
and $$x = - \sqrt{-1}$$
but how will you visualise this concept in real plane. For simplifying this because there are many such problems in science which evokes such roots and numbers. For this we introduce the complex plane, C, made of points each of which represents complex number. Then the roots of above equation are x = -i and x = i. Now you can clearly see these two points on complex plane.

Complex numbers can be represented in cartesian coordinates and their conversion to polar coordinates is quite easy.

Thus complex numbers made the concepts of Physics and Math more transparent.

5. Mar 30, 2005

### Galileo

The square root of a negative number isn't anymore real or less real than any other number, since numbers only exist in our heads. (The adjectives 'real' and 'imaginary' are ill-chosen).
We can invent, or define a new entity which we shall call i, to which we assign the property that i^2=-1.
Ofcourse, you can define anything you want, but the question is: Is it allowed? logically consistent? at all useful?
The answer is yes. Complex numbers (which have a real and an imaginary part) have many nice properties and can be viewed as an extension of the set of real numbers.

6. Mar 30, 2005

### HallsofIvy

Staff Emeritus
We might also mention the "cubic formula" that lead to the use of complex numbers. Cardano's formula requires that you take the square root of a number and then do various other things. There are some cubic equations where it is easy to show that they HAVE three real roots- yet Cardano's formula requires that you take the square root of negative numbers. It turns out that the imaginary parts then cancel in the final answer- but you need to use them in order to get the real number solutions!

By the way, what part of this was homework?

7. Mar 30, 2005

### roger

If the property of i is i^2=-1 why is it that we can't apply the rule, root -1 *root -1 = root ( -1*-1) = root 1 ?

8. Mar 30, 2005

### Data

Why must square roots of negative real numbers obey the same rules as square roots of non-negative real numbers?

9. Mar 31, 2005

### Galileo

Note that I didn't say that i is the square root of -1. In fact, THE square root of a number doesn't exist, since there are always 2 roots (expect for 0).
So when I define i through the property $i^2=-1$, I cannot distinguish between i and -i, since $(-i)^2=-1$ also.
Therefore care must be taken when dealing with roots of numbers (or multi-valued functions in general).
In the real case there would be no problem, since there we can make a choice to take the positive root (but that's just a convention).
In the complex case, we can no longer do so, since the set of complex numbers is essentially a plane, they have no ordering property. (there's no such thing as a positive or negative complex number, unless perhaps it happens to be real).

So you have to make a choice. If $\sqrt{-1}=i$, then
$$\sqrt{-1}\sqrt{-1}=i^2=-1$$, if $\sqrt{-1}=-i$ then:
$$\sqrt{-1}\sqrt{-1}=(-i)^2=-1$$

Last edited: Mar 31, 2005