## IrishBoy123 one year ago z = complex

1. IrishBoy123

$$|z| = 2$$ $$|z^2 - 4z - 3| \ge ????$$

2. ganeshie8

so $$z$$ is on a circle of radius $$2$$

3. IrishBoy123

yes

4. IrishBoy123

i have used the Triangle Inequality 1/ looking for tips on execution 2/ looking for other approaches, especially drawing it i get >= 1

5. ganeshie8

im getting 7 as the minimum value

6. IrishBoy123

shall i post what i did?

7. mathmate

if |z|=2, then z^2=4, so z^2-4z-3=1-4z. where z is the circle of radius 2.|dw:1441022800703:dw| So I got -4+1=-3 as the minimum.

8. ganeshie8

$$|z_1-z_2| \ne |z_2|-|z_2|$$ in general right

9. ganeshie8

$$|z^2-4z-3|\ne |z^2|-|4z|-|3|$$

10. IrishBoy123

following....

11. mathmate

but |z^2-4z-3|= |z^2-3-4z|, I hope I am not wrong here.

12. ganeshie8

wolfram gave 7 as the minimum i really don't know how to work it haha

13. IrishBoy123

i 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

14. ganeshie8

|z^2-4z-3| >= 4 - |4z+3| ?

15. IrishBoy123

so for the first bit i get $$\ge |4 - |4z + 3| |$$ then $$|4z + 3| \ge \ | |4z| - |3| |= 5$$ $$|4z + 3| \le |4z| + |3| = 11$$

16. IrishBoy123

so is final answer |4-5| or |4-11|??

17. ganeshie8

wow! thats really very clever!

18. IrishBoy123

if it's clever its because @Loser66 & @freckles taught me

19. ganeshie8

looks you ppl have messed with triangle inequality a lot before !

20. Loser66

$|z^2 |=|z^2-4z-3+4z+3|\leq |z^2 -4z-3|+|4z+3|\leq |z^2-4z-3|+|4z|+|3|$ Pick the far left and far right, we have $|z|^2 \leq |z^2-4z-3| +|4z|+|3|$

21. Loser66

I think you can handle it from here.

22. ganeshie8

indeed it looks pretty

23. ganeshie8

wait a sec, do we have problems with signs ? don't we end up with $$\color{red}{-}7 \leq |z^2-4z-3|$$ ?

24. ganeshie8

which is useless..

25. Empty

Here's my way: $z=2e^{i\theta}$ $z^2-4z-3 = (2e^{i \theta})^2 -4 *2e^{i \theta} -3$ $4e^{i2\theta } -8 e^{i \theta} -3$ 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!

26. Empty

I 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 xy-plane (-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|

27. Empty

What 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.

28. Empty

Here 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

29. IrishBoy123

thank 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 $$|z^2 - 4z - 3| \ge 4 - 5 = -1$$

30. ganeshie8

Awesome! somehow im stuck at proving $$||z|-|w||\le |z-w|$$ from loser's work

31. Loser66

@ganeshie8 These are my prof's lecture note for that parts

32. Loser66

33. Loser66

Sorry for attach the first page twice. :)

34. ganeshie8

Gotcha! $|z-w|^2\\~\\ = |z|^2-2\Re(z\overline{w})+|w|^2\\~\\\ \ge |z|^2-2|z\overline{w}|+|w|^2\\~\\\ = |z|^2-2|z||w|+|w|^2\\~\\\ =(|z|-|w|)^2$ therefore $$|z-w|\ge ||z|-|w||$$

35. IrishBoy123

i've plugged into @Empty 's eqn and the max/min are 7, 9, 10.63

36. ganeshie8

thanks @Loser66 that looks a lot simpler than my mess..

37. IrishBoy123
38. ganeshie8

that is so cool!

39. IrishBoy123

i'm quite speechless actually. thank you everyone for all the help, so much to think about ... life is good!

40. ganeshie8

by definition, life is always good and math is the most interesting part! yah i stole the first phrase from Preetha :)