Question

z = complex

Answer 1

\(|z| = 2\)

\(|z^2 - 4z - 3| \ge ????\)

Answer 2

so \(z\) is on a circle of radius \(2\)

Answer 3

yes

Answer 4

i have used the Triangle Inequality

1/ looking for tips on execution
2/ looking for other approaches, especially drawing it

i get >= 1

Answer 5

im getting 7 as the minimum value

Answer 6

shall i post what i did?

Answer 7

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

Answer 8

\(|z_1-z_2| \ne |z_2|-|z_2|\)

in general right

Answer 9

\(|z^2-4z-3|\ne |z^2|-|4z|-|3|\)

Answer 10

following....

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

?

Answer 15

so for the first bit i get \(\ge |4 - |4z + 3| |\)

then \(|4z + 3| \ge \ | |4z| - |3| |= 5\)

\(|4z + 3| \le |4z| + |3| = 11\)

Answer 16

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

Answer 17

wow! thats really very clever!

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

I think you can handle it from here.

Answer 22

indeed it looks pretty

Answer 23

wait a sec, do we have problems with signs ?

don't we end up with \(\color{red}{-}7 \leq |z^2-4z-3|\) ?

Answer 24

which is useless..

Answer 25

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!

Answer 26

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: