Question

If `abs(z+1/z) = a` then what is the possible range of values of `abs(z)` ?

Answer 1

As the result of clarification for this question, I have deleted my original response and will flag it as needing help.

Explanation:

Rather than completely delete this answer, I've left it in order to preserve the conversation between myself and Raghav Yadav.
(note: I have/will make slight variation to the question to make `zinCC` obvious).

Basically, as I see it, we are looking for a way to express `abs(z)` in terms of `a`.

Answer 2

While the following does not prove whether they are truly the max and min values, it does show how to arrive at the bounds from the discussion in the other answer

Explanation:

We will make use of the reverse triangle inequality:
`|a+b| >= |(|a|-|b|)|`
(parentheses added due to formatting issues)

`a = |z+1/z| = |(z^2+1)/z| = |z^2+1|/|z|`

`=> a|z| = |z^2+1| >= |(|z^2|-|1|)| >= (|z|)^2-1`

`=> (|z|)^2-a|z|<= 1`

`=> (|z|)^2-a|z| + a^2/4 <= 1+a^2/4 = (a^2+4)/4`

`=> (|z|-a/2)^2 <= (a^2 +4)/4`

`=> sqrt((|z|-a/2)^2) <= sqrt((a^2+4)/4)`

`=> ||z|-a/2| <= sqrt(a^2+4)/2`

`=> -sqrt(a^2+4)/2 <= |z|-a/2 <= sqrt(a^2+4)/2`

`=> -sqrt(a^2+4)/2 +a/2 <= |z| <= sqrt(a^2+4)/2+a/2`

`:.(a-sqrt(a^2+4))/2 <= |z| <= (a+sqrt(a^2+4))/2`

Answer 3

The minimum and maximum values of `abs(z)` occur for `z` on the imaginary axis and the exact range is:

`2/(a+sqrt(a^2+4)) <= abs(z) <= (a+sqrt(a^2+4))/2`

Explanation:

First part

Show that maximum and minimum values of `abs(z)` occur for `z` on the imaginary axis.

Let `z = r(cos theta + i sin theta)` with `r > 0` and `theta in (-pi, pi]`

Then we find:

`z + 1/z = r(cos theta + i sin theta) + 1/r(cos theta - i sin theta)`

`=(r+1/r)cos theta + (r - 1/r) i sin theta`

So:

`a^2 = abs(z+1/z)^2 = (r+1/r)^2 cos^2 theta + (r-1/r)^2 sin^2 theta`

`=(r^2+1/r^2) +2(cos^2theta - sin^2theta)`

`=(r^2+1/r^2) + 2cos 2theta`

Hence, for a given value of `a^2`, the maximum possible value of `r^2+1/r^2` occurs when `cos 2theta = -1`. That is when `theta = +-pi/2`, i.e. when `z` is on the imaginary axis.

Note that if `r > 1` then increasing the value of `r` increases the value of `r^2+1/r^2`, and if `r < 1` then decreasing the value of `r` (while keeping `r > 0`) also increases the value of `r^2 + 1/r^2`.

So the maximum possible value of `r^2+1/r^2` corresponds to both the maximum and minimum possible values of `r^2`, that is of `abs(z)`.

`color(white)()`
Second part

Now look at values of `z` on the imaginary axis.

Suppose `z=qi` with `q > 0`.

Then:

`abs(z+1/z) = abs(qi+1/(qi)) = abs(i(q-1/q)) = abs(q-1/q)`

So if `q >= 1` then `q-1/q = abs(q-1/q) = a`.

Multiplying both ends by `q` then subtracting `aq` from both sides this becomes:

`q^2-aq-1 = 0`

Then the quadratic formula gives us:

`q = (a+-sqrt(a^2+4))/2`

We can discard one of these roots since it is negative and our assumption was `q >= 1`, hence:

`q = (a+sqrt(a^2+4))/2`

Then since `abs(q-1/q) = abs(1/q-q)` is symmetric in `q` and `1/q`, the other positive solution is:

`q = 2/(a+sqrt(a^2+4))`

These then are the two points at which the curve described by `abs(z+1/z) = a` intersects the positive imaginary axis.

If these are indeed the maximum and minimum values then the solution to the original question is:

`2/(a+sqrt(a^2+4)) <= abs(z) <= (a+sqrt(a^2+4))/2`