Specialist Math Questions

[#1procrastinator]

If you're asked to evaluate the argument of a complex number in the interval [-2π, 0], we go clockwise right? But why are we going from 'down' rather than 'up'?

. e.g. if it's -i, then apparently, it's -π/2, but you're starting from -2pi so to me it's like going from -360 degrees to -90 degrees.

Or are you supposed to look at -π/2 as simply meaning going clockwise 90 degrees from -2π? Why isn't just π/2?

I think what's confusing me is the interval [-2π, 0]…go clockwise UP to zero, but the negative values are getting larger rather than smaller..


And for the Argument, does it being in the interval (-pi, pi] just mean that if it's larger than greater than pi, you go clockwise and if less, than anticlockwise?

----

Let z = cisθ
a) Show |z| = 1
b) 1/z = cis(-θ)

For a), is it enough to just say that since r = 1, and |z| is 1?  I'm not sure how to approach b)…not sure if it requires any trig knowledge from the preceding chapters


----

EDIT:

For the complex numbers u = a+bi, v=c+di and w=p+qi prove that:

a) u+v=v+u
b) (u+v)+w = u(v+w)
c) uv=vu
d) (uv)w=u(vw)

http://www.smbc-comics.com/comics/20110404.gif
 

[Somye]

for your 1/z question, that's the same as
Applying DeMoivre's theorem you get which equals

For the other questions, split it up into LHS and RHS and simplify them till both answers equal the same thing
eg. LHS = a+c + (b+d)i
RHS = c+a + (d+b)i = LHS

[kamil9876]

EDIT:

For the complex numbers u = a+bi, v=c+di and w=p+qi prove that:

a) u+v=v+u
b) (u+v)+w = u(v+w)
c) uv=vu
d) (uv)w=u(vw)

I think it's more appropriate to have this as a continuation of this discussion. Formal definition of complex numbers

In general it's very difficult to ask for help on proving something if the other person doesn't know what definitions you are using/what you already know (this is one way in which you get that graph from your picture). Secondly, specialist math probably wouldn't deal with this sort of stuff.

edit: actually this probably applies to your whole post.

[#1procrastinator]

@SomyeThanks Somye, hope the first part isn't worded too badly lol

@kamil: Ok, but they're actually questions from an old specialist textbook ('Core Specialist Math' or something). The properties of real numbers are given...I'll try have another go at them.

Also, does that mean that I have to somehow prove |z| = r = 1?

[#1procrastinator]

In example 27, chapter 4 (complex numbers) of the Essential book, they show how to find the locus defined by |z+3| = 2|z-i|

When they square both sides and root it, I'm guessing that's to get rid of the absolute value signs, but why does the i with the y disappear? Shouldn't it become negative y since i^2 = -1?



Then they square both sides which I think is to get rid of square roots but then complete the square to get the equation for a circle. How did they know to do that? Is this what you should always do if it's not immediately obvious what the locus is?

[Somye]

okay, so when you put a number inside the absolute value signs, what do you do?
for z = x +yi, your =
so in this, we do the opposite, with the i not actually in the absolute value.

So, there's two ways to do this question. One way is to think about it geometrically, ie. what set of points denotes (-3,0) being twice the distance away from (0,i). This relationship is denoted by a circle, but that's a little harder to think about.

An easier way is that once you have your and terms, if the pronumerals before the two numbers is equal, and they are added together, you'll have a circle. If the pronumerals are different and added together, you have an ellipse. And finally if theres subtraction between the two terms, you have a hyperbola.

An easy way to think about it though is that if you the modulus of a set of point = the modulus of a second point with a pro numeral in front, it's a circle.

[#1procrastinator]

Why is a linear combination called a linear combination and not a combination?

If in R^2, a set of 3 vectors will have one redundant vector, is that only true if the set if linearly dependent?

If we have a set of three 2D vectors, will one always be redundant? If so, could we choose any two vectors and solve for the scalar factors (not sure if that's the right term, just the constants in front of them) rather than solve a 3-variable, 2 equation system, choosing some number for one of the scalars?

[kamil9876]

it's just a definition. I guess it's good to distinguish it from other possible "combinations". I remember a lecturer once called something like a "polynomial combination of and ". Or said something like and are "polynomialy independent" (probably slang just to add some humour and contrast with the linear case)

If in R^2, a set of 3 vectors will have one redundant vector, is that only true if the set if linearly dependent?

Having a "redundant vector" is actually slang for being linearly dependent, so trivially yes. The more interesting point is that in any set of three vectors is indeed linearly independent.

If we have a set of three 2D vectors, will one always be redundant? 

Yes by what I said before.

If so, could we choose any two vectors and solve for the scalar factors (not sure if that's the right term, just the constants in front of them) rather than solve a 3-variable, 2 equation system, choosing some number for one of the scalars?

I'm not sure what your point is. Maybe you are suggesting that if you want to represent some arbitrary vector in
and you want to represent is as a linear combination of then can you delete one and just represent as a linear combination of the other two? Yes you can though you have to be careful which one you delete (for example if a,b are parralel and c is not, then by deleting c you lose a "dimension", so then there are some vectors you can't represent in terms of just a and b)

[#1procrastinator]

^ Thanks for the helpful reply

it's just a definition. I guess it's good to distinguish it from other possible "combinations". I remember a lecturer once called something like a "polynomial combination of and ". Or said something like and are "polynomialy independent" (probably slang just to add some humour and contrast with the linear case)

Ah ok. I just asked cause in one of the Khan Academy videos on Linear Algebra he mentioned why it was called a linear combination but I didn't understand it...guess it's not too important though

If in R^2, a set of 3 vectors will have one redundant vector, is that only true if the set if linearly dependent?

Having a "redundant vector" is actually slang for being linearly dependent, so trivially yes. The more interesting point is that in any set of three vectors is indeed linearly independent.

If we have a set of three 2D vectors, will one always be redundant? 

Yes by what I said before.

If so, could we choose any two vectors and solve for the scalar factors (not sure if that's the right term, just the constants in front of them) rather than solve a 3-variable, 2 equation system, choosing some number for one of the scalars?

I'm not sure what your point is. Maybe you are suggesting that if you want to represent some arbitrary vector in
and you want to represent is as a linear combination of then can you delete one and just represent as a linear combination of the other two? Yes you can though you have to be careful which one you delete (for example if a,b are parralel and c is not, then by deleting c you lose a "dimension", so then there are some vectors you can't represent in terms of just a and b)

Well there was an example where he had a set of 3 vectors and wanted to find the scale factors to (I think, don't have the notes I wrote down with me!) determine whether it was linearly dependent or not. 3 variables and 2 equations and he chose -1 for one of the vectors (not sure how he chose it). But that does answer my question :>