1. ## Imaginary unit

Why does i equal the square root of -1 and i^2 equal -1?

2. Originally Posted by HSS8
Why does i equal the square root of -1
that is simply how it is defined.

and i^2 equal -1?
well, squaring gets rid of the square root.

see here and here

3. Originally Posted by HSS8
Why does i equal the square root of -1 and i^2 equal -1?
As the previous reply correctly states, this is how "the imaginary" is defined. Math folks needed a number to do a certain thing, so they said "let's create that number, and call it $\displaystyle `i\mbox{'}$, since clearly it's 'imaginary' and can't have any possible 'real world' use or application." (This is, by the way, similar to how negative numbers got their start.)

There is a lot of math, science, and engineering that couldn't happen nowadays, had this number not been invented.

When you study complex numbers (numbers with the $\displaystyle i$ in them or involving taking the square root of negative values), make sure you remember the new rule: Always take care of the negative inside the square root first, before doing any other computations or simplifications!

4. Originally Posted by HSS8
Why does i equal the square root of -1 and i^2 equal -1?
The complex numbers are constructed so that $\displaystyle i^2=-1$, but there are 2 square roots of $\displaystyle -1$ and $\displaystyle i$ by construction is one of them.

(note I describe the complex numbers as constructed as they are built to have the properties needed so that certain algebraic process that we need make sense.)

CB

5. The "construction" CaptainBlack refers to is this: Define the complex numbers to be the set of pairs of real numbers (a, b) with addition defined by (a,b)+ (c,d)= (a+c,b+d), (a,b)*(c,d)= (ac-bd,ad+bc). Since (a,0)+ (c,0)= (a+c,0) and (a,0)*(c,0)= (ac-0(0),a(0)+0(c))= (ac,0), we can identify the complex numbers of the form (a,0) with the real numbers. Now, (0,1)*(0,1)= (0(0)-1(1),0(1)+1(0))= (-1, 0) which we associate with -1.

If we represent (0,1) by i, and, as I said, (1,0) by 1, we have (a,b)= a(1,0)+ b(0,1)= a+ bi. And, then, $\displaystyle i^2= (0,1)*(0,1)= (-1,0)= i$.