z = complex

- IrishBoy123

z = complex

- chestercat

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- IrishBoy123

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

- ganeshie8

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

- IrishBoy123

yes

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## More answers

- IrishBoy123

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

- ganeshie8

im getting 7 as the minimum value

- IrishBoy123

shall i post what i did?

- 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.

- ganeshie8

\(|z_1-z_2| \ne |z_2|-|z_2|\)
in general right

- ganeshie8

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

- IrishBoy123

following....

- mathmate

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

- ganeshie8

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

- 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

- ganeshie8

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

- IrishBoy123

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

- IrishBoy123

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

- ganeshie8

wow! thats really very clever!

- ganeshie8

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

- 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
\[|z|^2 \leq |z^2-4z-3| +|4z|+|3|\]

- Loser66

I think you can handle it from here.

- ganeshie8

indeed it looks pretty

- ganeshie8

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

- ganeshie8

which is useless..

- 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!

- 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|

- 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.

- 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

- 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 \(|z^2 - 4z - 3| \ge 4 - 5 = -1\)

- ganeshie8

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

- Loser66

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

##### 2 Attachments

- Loser66

##### 1 Attachment

- Loser66

Sorry for attach the first page twice. :)

- 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 \(|z-w|\ge ||z|-|w||\)

- IrishBoy123

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

- ganeshie8

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

- IrishBoy123

https://www.desmos.com/calculator/gusfoxuqmh

- ganeshie8

that is so cool!

- IrishBoy123

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

- ganeshie8

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

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