jinny1's Specialist Questions Thread :D

[jinny1]

we did square it. i^2 = -1.

She is talking about |z|, because she is wondering why it's not sqrt( x^2 + (iy)^2 ), and I was saying that the i^2 is not supposed to be there.

oh ok.. but isn't it technically wrong to just get rid of the "i" when we are evaluating the absolute value of z

Nope, it's not wrong. You can think of |z| as simply the square root of zz* (where z* is the conjugate).

hmmm.yea i understand but with any other absolution value functions: we always square the all the coeffcients of x and y. what makes 'i' so special that we can ignore it now??

This may be hard to understand, but it is because i is linearly independent from the real numbers. That means no matter what real number you have, you cant make the number i. It is much easier if you think of the complex numbers as simply a means of writing coordinates in the plane (even though, of course, there is much more to complex numbers than just that). Sorry I can't really explain, maybe someone else can try.

I'm not sure if this is entirely correct but here is how I look at it.
When we have |z|, it is similar to say taking the magnitude of a vector (because this is when the magnitude of z is say 2 e.t.c)
So when we do that with vectors, we take the coefficients in front of i, j, k and square them, then root them. Here we take the coffecients of the "real numbers" and "imaginary numbers" and square them then root the whole thing.

but when we are multiplying two vectors, we still just take the coeffecients and multiply them only..

when it comes to mutliplying two complex numbers, we multiply the 'i's together 

[brightsky]

ahh i see. i agree with dc302, whenever you try to 'picture' a complex number, picture the argand diagram. as dc302 said, complex numbers are an entirely different entity from real numbers. (and for this reason, it is wrong to compare complex numbers with real numbers, e.g. saying 2i < 3, etc.).
and yeah complex numbers is pretty much exactly the same as vectors and can effectively be treated as such.

[b^3]

Yes but that would be because thats the dot product, its a different concept. When we take the magnitude of vectors or complex numbers then we use the coefficients. Think of it this way, for |z| we have a triangle, the adjacent side is the real component, the opposite side is the imaginary component. To get the magnitude we need to find the hypotenuse, so we use the pythagoras theorem and end up with root(x^2+y^2).

I hope that made sense, I think I just confused myself.

[brightsky]

but when we are multiplying two vectors, we still just take the coeffecients and multiply them only..

this is done only for the purpose of efficiency. when we multiply position vectors, i*j = j*k = i*k = 0 since they are perpendicular.

[b^3]

ahh i see. i agree with dc302, whenever you try to 'picture' a complex number, picture the argand diagram. as dc302 said, complex numbers are an entirely different entity from real numbers. (and for this reason, it is wrong to compare complex numbers with real numbers, e.g. saying 2i < 3, etc.).
and yeah complex numbers is pretty much exactly the same as vectors and can effectively be treated as such.

Yeh cause even our cis(theta), cis(theta)=cos(theta)+i*sin(theta)
which is the vector addition to give the triangle that results in the ray that represents the cis(theta).

[dc302]

If we're talking about coefficients, you can do the same for complex numbers.

In fact, any complex number is in the form:

a * 1 + b * i, where 1 is the 'unit vector' in the real axis, and i is the 'unit vector' in the imaginary axis.

So now, when you find the magnitude, you are essentially taking the coefficients of 1 and i.

[b^3]

This is what I'm talking about with vector addition.
http://www.youtube.com/watch?v=ZPGHuuk2bKw&feature=related
Go to 0:50 seconds. and remember that e^ix=cis(theta)

[dc302]

but when we are multiplying two vectors, we still just take the coeffecients and multiply them only..

this is done only for the purpose of efficiency. when we multiply position vectors, i*j = j*k = i*k = 0 since they are perpendicular.

Keep in mind that we are not 'multiplying' vectors in the same sense we are multiplying complex numbers; the two cannot be compared in that way.

[b^3]

but when we are multiplying two vectors, we still just take the coeffecients and multiply them only..

this is done only for the purpose of efficiency. when we multiply position vectors, i*j = j*k = i*k = 0 since they are perpendicular.

Keep in mind that we are not 'multiplying' vectors in the same sense we are multiplying complex numbers; the two cannot be compared in that way.

Thats what I was trying (and that is the key word here) to explain before, the dot product is different to just mulitplying. Geez I have to go over my theory for complex numbers and vectors.

[jinny1]

Wow thank you so muchguys. Very helpful