Z is a complex number , a is a constant . Find the maximum and minimum value for Z notice that :

- TrojanPoem

Z is a complex number , a is a constant . Find the maximum and minimum value for Z notice that :

- chestercat

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- TrojanPoem

\[\left| \frac{ 1 }{ Z } + Z \right| = a\]

- TrojanPoem

\[a \neq 0\]

- dan815

write out a+bi form how will that simplify

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## More answers

- Empty

Nah write it out in \(r e^{i\theta}\) form ;)

- dan815

ya thats faster

- dan815

|dw:1434894017464:dw|

- TrojanPoem

= a a constant

- Empty

Not quite dan, unless R=1 that won't work.

- TrojanPoem

What about triangular form ?

- dan815

the 2 lines mean we are looking at the modulus right

- dan815

ok i got no time right now g2g

- dan815

cya later

- TrojanPoem

Ok dan. Cya

- ganeshie8

we may (ab)use AM-GM inequality
\[z+\frac{1}{z}\ge 2\]

- Empty

This is wrong since z is complex we have no order. If a=0 that could happen since there's nothing restricting us from choosing z=i.
\[|i+\frac{1}{i}| = 0\]

- ganeshie8

Ahh right, thats so silly

- Empty

So for the case a=0 it's easy to see that a _possible_ max and min for z is i and -i.... But I don't know since the phrase "Find the maximum and minimum value for Z " is ambiguous, I'm just assuming this means "maximum and minimum length given the constant a".
I'm interpretting this as maximizing/minimizing the length of z as a function of a... but I don't really know how to do that lol.

- dan815

|dw:1434901923700:dw|

- dan815

@Trojenpoem i think there is a nice simplification with that triangular form u mentioned

- dan815

- dan815

|dw:1434902070085:dw|

- Empty

Let's let \(z=re^{i\theta}\). Now consider this picture: |dw:1434900774783:dw| See how \(a\) is just the resultant vector sum of these two z's? So in order to maximize the length of z (which is really just r) for a given \(a\) notice that as we increase \(\theta\) closer to \(\pi\) we end up having to pick larger and larger r in order to make our \(a\) vector remain constant length.
So let's look at the case when \(\theta = \pi\) so that we have:
\[|r-\frac{1}{r}|=a\]
We know that if we're trying to maximize r that r>1/r so we drop the absolute value signs and solve the quadratic for r in terms of a and pick the positive root:
\[r_{max}=\frac{a+\sqrt{a^2+4}}{2}\]
So this maximizes r since it's impossible to rotate two vectors further apart than 180 degrees for the same reason you can't run more than halfway into a forest- you will begin to run out of the forest.
Now you may think, hey, this only shows the maximum, what about the minimum? Well in fact we could have just walked through this entire argument with 1/r. So the min as you will easily find by inverting the last formula is just:
\[r_{min}=\frac{2}{a+\sqrt{a^2+4}}\]
So to clarify,
\[z_{min}=\frac{2}{a+\sqrt{a^2+4}}e^{i \pi /2}\]
\[z_{max}=\frac{a+\sqrt{a^2+4}}{2}e^{i \pi /2}\]

- dan815

howd u get that quadratic for rmax

- Empty

it's solving r-1/r=a

- Empty

Like really the main thing is inverses mean the angles are exactly lined up so the possible values will be like: |dw:1434903781229:dw|

- ParthKohli

\[\left| z + \frac{1}{z}\right|^2 = a^2\]\[\Rightarrow \frac{z^2 + 1}{z}\cdot \frac{\overline z ^2 + 1}{\overline z }= a^2\]\[\Rightarrow \frac{|z|^4 + z^2 + \overline z^2 + 1}{|z|^2}=a^2\]\[\Rightarrow |z|^2 + \frac{1}{|z|^2} + 2 \cos 2\theta = a^2 \]I was thinking along the lines of this

- TrojanPoem

@Empty , " we drop the absolute value signs and solve the quadratic" hmm ?

- Empty

Ok I'll try reexplaining myself since I realize it's hard to explain the visual reasoning.
\[|z+\frac{1}{z}| = a\]
The problem as I understand it is to find the minimum or maximum value of the length of z for a given constant a. So what this means is that z and \(\frac{1}{z}\) are complex numbers that when added together produce a new complex number with length of a. Make sure you understand this is what the equation you have says.
So if we write \[z=r e^{i \theta}\] then the length we are minimizing/maximizing is really the value r. Let's plug it in:
\[|r e^{i \theta}+\frac{1}{r} e^{-i \theta}| = a\]
Right now let's focus on maximizing r and we'll think about minimizing it later.
So this equation specifically shows the angles being the same but in opposite directions! This is the idea behind maximizing r. Let's look at the extremes. Either \(\theta=0\) or \(\theta = \pi/2\) Why these? Think of the picture, when the angle is zero then both of the vectors are adding to each other and have a length of a after, so we would have r+1/r=a for the length of a. See how they're pointing in the same direction? That means to get the constant value, a, our lengths are on top of each other in a sense. If the vectors had opposite sign, then they'd end up having to both be longer to reach that same distance. Here's a picture to illustrate: |dw:1434907265303:dw| So this other case I've written at the bottom has two vectors adding to a, but their combined lengths is actually longer than a, since we're subtracting one from the other. This happens exactly at \(\theta = \pi/2\) since this is the point where we have rotated both vectors r and 1/r to be pointing in opposite directions.
Now we have
\[|ri-\frac{1}{r}i| = a\] which since it only has components in the i direction is just the coefficients added together, so we have:
\[r-\frac{1}{r}= a\]
multiply this on both sides by r to get:
\[r^2-1= ar\]
rearrange:
\[r^2-ar-1=0\]
Then solve this quadratic for r, the positive root of course will be the maximum but due to the symmetry of r, we can see that whatever maximizes r also minimizes 1/r and the inverse of the positive root of this quadratic is also the minimum length of z as well, which I have put the formula above.
This will require some thinking to understand, but if you try to understand I can help you to understand. Good luck, and think about the picture of a continuously varying pair of z's that sum in the complex plane to give a constant length resultant vector a.

- TrojanPoem

I really like how you think.

- ParthKohli

@TrojanPoem +1

- Empty

Hahaha thanks :D I might fall asleep soon but this is a fun problem definitely.

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