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IrishBoy123
 one year ago
z = complex
IrishBoy123
 one year ago
z = complex

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\(z = 2\) \(z^2  4z  3 \ge ????\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3so \(z\) is on a circle of radius \(2\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1i have used the Triangle Inequality 1/ looking for tips on execution 2/ looking for other approaches, especially drawing it i get >= 1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3im getting 7 as the minimum value

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1shall i post what i did?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1if z=2, then z^2=4, so z^24z3=14z. where z is the circle of radius 2.dw:1441022800703:dw So I got 4+1=3 as the minimum.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\(z_1z_2 \ne z_2z_2\) in general right

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\(z^24z3\ne z^24z3\)

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1but z^24z3= z^234z, I hope I am not wrong here.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3wolfram gave 7 as the minimum i really don't know how to work it haha

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1i used this because its backwards \[z_1 + z_2 \ge  z_2  z_2 \] then i said \(z_1 = z^2\), and \(z_2 = (1)(4z+3) \) and then i did that again for \(z_2\) i got \( \ge 1\) or \(\ge 7\) i may just have chosen the wrong one

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3z^24z3 >= 4  4z+3 ?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1so for the first bit i get \(\ge 4  4z + 3 \) then \(4z + 3 \ge \  4z  3 = 5\) \(4z + 3 \le 4z + 3 = 11\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1so is final answer 45 or 411??

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3wow! thats really very clever!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1if it's clever its because @Loser66 & @freckles taught me

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3looks you ppl have messed with triangle inequality a lot before !

Loser66
 one year ago
Best ResponseYou've already chosen the best response.3\[z^2 =z^24z3+4z+3\leq z^2 4z3+4z+3\leq z^24z3+4z+3\] Pick the far left and far right, we have \[z^2 \leq z^24z3 +4z+3\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.3I think you can handle it from here.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3indeed it looks pretty

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3wait a sec, do we have problems with signs ? don't we end up with \(\color{red}{}7 \leq z^24z3\) ?

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Here's my way: \[z=2e^{i\theta}\] \[z^24z3 = (2e^{i \theta})^2 4 *2e^{i \theta} 3\] \[4e^{i2\theta } 8 e^{i \theta} 3\] We want to minimize \[\sqrt{(4e^{i2\theta } 8 e^{i \theta} 3)(4e^{i2\theta } 8 e^{i \theta} 3)^*}\] Which has a minimum at the same point the square does, so let's just look at the minimum of the expression and its complex conjugate: \[f(\theta) = 24 \cos (2 \theta ) 16 \cos (\theta) +89\] \[f'(\theta)= 0 = 24 \sin(2 \theta)+ 16 \sin (\theta)\] \[\sin \theta (1+3 \cos \theta) = 0\] So here we find all the local mins, 0, \(\pi\), \(\cos^{1}(\frac{1}{3})\), \(\cos^{1}(\frac{1}{3})\). Then we can just plug them in and see which is smallest in magnitude. Not the prettiest!

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I think these 4 points are simplest to see on a picture of \(4e^{i 2 \theta} 8e^{i \theta} 3\) Reading right to left, we center ourselves at the point in the xyplane (3,0) and then we create a circle of radius 8 here, and noting the negative sign we rewrite for clarity: \(4e^{i 2 \theta} +8e^{i (\theta+ \pi)} 3\) So now in half of rotation we extend a little "moon" orbit at radius 4 that goes twice as fast: dw:1441024362202:dw

Empty
 one year ago
Best ResponseYou've already chosen the best response.3What I mean is: \(4e^{i 2 \theta} 8e^{i \theta} 3\) Think of a point at 3 as the sun which has a little earth orbiting 8 away from it counter clockwise starting at the left. Then this earth has a little moon orbiting 4 away from it which starts to the right of the earth and completes a single orbit around the earth in the time it takes the earth to go halfway around the sun.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Here enjoy this is exactly what I'm doing (and if the name Fourier means anything to you, that's what this is) https://www.youtube.com/watch?v=QVuU2YCwHjw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1thank you very much @Empty i am somewhat familiar with Fourier albeit v rusty but that is extremely interesting and will be read by me for sure in detail in due course looking again at the inequality \(z_1 + z_2  \le z_1 + z_2\) so \(z_1 \ge z_1 + z_2   z_2\) \(z_1 = z^2  4z  3\) and \(z_1 + z_2 = z^2 \) makes \(z_2 = 4z+3\) \(z^2  4z  3 \ge z^2   4z+3 \) \( 4z  3 \le 4z+3 \le 4z + 3\) \( 5 \le 4z+3 \le 11\) \(z^2  4z  3 \ge 4 11 = 7 \) or \(z^2  4z  3 \ge 4  5 = 1\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Awesome! somehow im stuck at proving \(zw\le zw\) from loser's work

Loser66
 one year ago
Best ResponseYou've already chosen the best response.3@ganeshie8 These are my prof's lecture note for that parts

Loser66
 one year ago
Best ResponseYou've already chosen the best response.3Sorry for attach the first page twice. :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Gotcha! \[zw^2\\~\\ = z^22\Re(z\overline{w})+w^2\\~\\\ \ge z^22z\overline{w}+w^2\\~\\\ = z^22zw+w^2\\~\\\ =(zw)^2 \] therefore \(zw\ge zw\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1i've plugged into @Empty 's eqn and the max/min are 7, 9, 10.63

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3thanks @Loser66 that looks a lot simpler than my mess..

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1i'm quite speechless actually. thank you everyone for all the help, so much to think about ... life is good!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3by definition, life is always good and math is the most interesting part! yah i stole the first phrase from Preetha :)
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