# Imaginary number's definition misunderstanding

1. Aug 5, 2013

### alingy1

1. The problem statement, all variables and given/known data

I'm in this self-learning course. I came on this problem I thought of.

So, i^2=-1.
But, isn't i=sqroot of -1?
If so, the product of the two minus -1 and the square root of that should give 1.

Am I not getting something?
I searched the web with the keywords of my question, but I couldn't find any result...

2. Aug 5, 2013

### SteamKing

Staff Emeritus
Is i positive or negative?

The relation i^2 = -1 is the converse of saying i = SQRT(-1). If a = SQRT(b), then a^2 = b.

Then, there are these relations:

i = SQRT(-1)
i^2 = -1
i^3 = -i
i^4 = +1

3. Aug 5, 2013

### HallsofIvy

Staff Emeritus
I do not understand what you mean by "The product of two minus -1 and the square of that". Literally interpreted, "two minus -1" is 2- (-1)= 3 but surely that is not what you mean. Did you mean "the product of -1, twice"? Does "that" refer to i? Yes, $i^2= -1$. The product of "-1, twice" is $(-1)^2= 1$. What does that have to do with $i^2= -1$?

If there is something you are "not getting", perhaps it is the difference between the "square" and the "square root". The square of -2 is (-1)(-1)= 1. But the square root of -1 is i.

(There is a "technical" point here: while every number has a single square, every complex number has two square roots. The two square roots of -1 are i and -i.)

4. Aug 5, 2013

### alingy1

Oh boy, sorry. I'm a second language speaker so speaking English messes my train of thoughts. :P
Let me repair the situation:
So, I have been taught that:''the product of two radicals is the same as the radical of the product, and vice versa.''-purplemath
But, now, dealing with the imaginary number:
We know that i=sqroot(-1).
So, i^2 should equal 1 since:
i^2=sqroot(-1)xsqroot(-1)=sqroot(-1x-1)=sqroot(1)=1

But, that makes no sense because it is said that i^2=-1

Is the property that I have been taught wrong? If so, when can I not use it?

5. Aug 5, 2013

### Staff: Mentor

The square root has an ambiguity in the complex numbers. You cannot use all the regular operations possible with positive real numbers.

6. Aug 5, 2013

### LCKurtz

@alingly1: You may find this post in the FAQ section by micromass interesting: