Recent content by d2j2003

the solution to 4|x|=12 is x=±3 BECAUSE if you plug x back in it works for both 3 and -3 ie. if you only list one of these as a solution then you have not completely solved the equation. Therefore if you have 4|det P|=12 then |det P| = 3 meaning that det P can be either 3 or -3 make sense?

right, but it would give an equation for a circle containing that point.. just not a specific equation because, as you said, you would need another point for that

if you know that the radius is 9 you can figure out the equation of the circle centered at the origin and then just translate it so that your point lies a distance of r from the center.

ok so \frac{d}{d}(e^{-3x})(y) = (by product rule) [\frac{d}{dx}(e^{-3x})]y + e^{-3x}[\frac{d}{dx}(y)] = (by chain rule) e^{-3x}[\frac{d}{dx}(-3x)]y+e^{-3x}[\frac{d}{dx}(y)] then go from there..

right but then if \alpha=\pi wouldn't it just be the entire plane? since if it is \pi/2 then it is half of the plane..

anyone??

d/dx [f(x)g(x)] = f'(x)g(x)+f(x)g'(x) (product rule) does this look familiar?

power rule is used when there are no variables in the raised part (ie. x^2 or 3x^6) but chain rule is used if there are variables in the raised part (ie. d/dx (e^(2x))=e^(2x) * d/dx (2x) = 2e^(2x) Chain rule is used in other situations but this is basically the situation where you would have to...

product rule for differentiation...

Homework Statement find linear fractional transformation from D={z:|Arg z| < \alpha}, \alpha≤\pi to the upper half plane Homework Equations The Attempt at a Solution The problem I am having here what exactly D is.. (visualizing it) D is just z such that |Arg z|≤\pi right? so...

I don't see anything in the chapter about symmetric points... so I'm assuming just 3 points to 3 points on the line.. but it seems like there would be more than one mapping... the answer the book gives is λ(1-i)\frac{z+1}{z-1} so I'm just trying to understand it

but I didn't think 0 and ∞ were on the circle |z|=1

I think I could map 1 to infinity and map -1 to 0 but after that I'm not sure..

For this problem since I am mapping from |z|=1 there is no inside and outside of the circle right? It is just the boundary that is being mapped.. So I would have to pick points from the boundary right?

If that's the case I would have to pick a point that maps to the origin and then pick another that maps to infinity? And then I'm not sure about the third...