Find Least Value of \left|z+\frac{1}{z}\right|, |z| ≥ 3

[utkarshakash]

Homework Statement

Least value of $\left|z+\frac{1}{z}\right| if |z|\geq3 is$

Homework Equations

The Attempt at a Solution

$\left|z-\frac{-1}{z} \right| \geq |z|-\left| \frac{-1}{z} \right|$

So the minimum value will be

$|z|- \frac{1}{|z|}$

Now for minimum value
|z| = 3 as it is the minimum value of |z|

Substituting the value of |z| I get
$3- \frac{1}{3}$
=$\frac{8}{3}$

But the correct answer is $\frac{10}{3}$

 

[Ray Vickson]

Homework Statement

Least value of $\left|z+\frac{1}{z}\right| if |z|\geq3 is$

Homework Equations

The Attempt at a Solution

$\left|z-\frac{-1}{z} \right| \geq |z|-\left| \frac{-1}{z} \right|$

So the minimum value will be

$|z|- \frac{1}{|z|}$

Now for minimum value
|z| = 3 as it is the minimum value of |z|

Substituting the value of |z| I get
$3- \frac{1}{3}$
=$\frac{8}{3}$

But the correct answer is $\frac{10}{3}$

Your reasoning is wrong (or at least, incomplete---see below), but you have arrived at the correct answer. The minimum value is, indeed, 8/3, and the posted answer of 10/3 is incorrect. Hint: to show that 10/3 is incorrect, look for a value z = iy along the imaginary axis that gives z + 1/z = (8/3)i, so for that z we have |z + 1/z| = 8/3, which is less that the alleged minimum of 10/3.

So, what did you do wrong? Well, your inequality
$$\left|z-\frac{-1}{z} \right| \geq |z|-\left| \frac{-1}{z} \right|$$
is, indeed, true, but for some z it might be a strict inequality, so there is no guarantee that minimizing the right-hand-side will give an achievable minimum to the left-hand-side. I'll leave it up to you to see what else needs to be done to fix the argument.

RGV

 

[utkarshakash]

Your reasoning is wrong (or at least, incomplete---see below), but you have arrived at the correct answer. The minimum value is, indeed, 8/3, and the posted answer of 10/3 is incorrect. Hint: to show that 10/3 is incorrect, look for a value z = iy along the imaginary axis that gives z + 1/z = (8/3)i, so for that z we have |z + 1/z| = 8/3, which is less that the alleged minimum of 10/3.

So, what did you do wrong? Well, your inequality
$$\left|z-\frac{-1}{z} \right| \geq |z|-\left| \frac{-1}{z} \right|$$
is, indeed, true, but for some z it might be a strict inequality, so there is no guarantee that minimizing the right-hand-side will give an achievable minimum to the left-hand-side. I'll leave it up to you to see what else needs to be done to fix the argument.

RGV

I can't make out any fault in my reasoning. Can you please bring it out?

 

[Ray Vickson]

I can't make out any fault in my reasoning. Can you please bring it out?

I have already explained why your reasoning is incomplete: you have some expression A that you want to minimize, and you know that A ≥ B for some other expression B. Minimizing B does not necessarily minimize A.

For example, if f(x) = (x-1)²+ x⁴ and g(x) = (x-1)², we certainly have f(x) ≥ g(x) for all x (and f = g for some x). However, the minimum of g(x) is = 0, and it occurs at x = 1, while the minimum of f(x) is = 0.2892734 and it occurs at x = 0.5897545 (as obtained using numerical methods). So, in this case, minimizing f by minimizing the simpler function g would fail. However, if I change f to f(x) = x² + x⁴ and g(x) to g(x) = x², we would get the correct answer by minimizing g instead of f.

So, sometimes your method works and sometimes it fails. The issue is whether you can apply it in your particular problem.

RGV

 

[utkarshakash]

I have already explained why your reasoning is incomplete: you have some expression A that you want to minimize, and you know that A ≥ B for some other expression B. Minimizing B does not necessarily minimize A.

For example, if f(x) = (x-1)²+ x⁴ and g(x) = (x-1)², we certainly have f(x) ≥ g(x) for all x (and f = g for some x). However, the minimum of g(x) is = 0, and it occurs at x = 1, while the minimum of f(x) is = 0.2892734 and it occurs at x = 0.5897545 (as obtained using numerical methods). So, in this case, minimizing f by minimizing the simpler function g would fail. However, if I change f to f(x) = x² + x⁴ and g(x) to g(x) = x², we would get the correct answer by minimizing g instead of f.

So, sometimes your method works and sometimes it fails. The issue is whether you can apply it in your particular problem.

RGV

Hmmm... that seems quite correct. Thanks