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If $|2z_1 + \bar z_2| = 2\sqrt2$ and $|1 + 2z_1z_2 | = 3 $ then minimum value of $(|z_1|^2 + 4|z_2|^2)$, is:-

Now

$|2z_1 + \bar z_2|^2 = 8$

$(2 z_1 + \bar z_2)(2\bar z_1 + z_2) = 8$

$4z_1\bar z_1 + 2 z_1z_2 + 2\bar z_1\bar z_2 + z_2\bar z_2 = 8$

$4|z_1|^2 + |z_2|^2 + 2 z_1z_2 + 2\bar z_1\bar z_2 = 8.......(1)$

Now

$|1 + 2z_1z_2 |^2 = 9 $

$(1 + 2z_1z_2)(1 + 2\bar z_1\bar z_2) = 9 $

$1 + 4|z_1|^2|z_2|^2 + 2z_1z_2 + 2\bar z_1\bar z_2= 9........(2)$

$(1) - (2)$

$4|z_1|^2 + |z_2|^2 = 4|z_1|^2|z_2|^2$

I am not able to make any progress after this step.

Deepak
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Mathking
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2 Answers2

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$$4|z_1|^2 + |z_2|^2 = 4|z_1|^2|z_2|^2$$

$$\dfrac{1}{|z_1|^2} + \dfrac{4}{|z_2|^2} = 4$$

By Cauchy-Schwarz,

$$\Bigl(\dfrac{1}{|z_1|^2} + \dfrac{4}{|z_2|^2} \Bigr)\Bigl(|z_1|^2 + 4|z_2|^2\Bigr) \ge (1 + 2\cdot2)^2 = 25 $$

$$|z_1|^2 + 4|z_2|^2 \ge \dfrac{25}{4} $$

with equality for

$$\dfrac{1}{|z_1|}\cdot2|z_2| - 2\dfrac{1}{|z_2|}\cdot|z_1|=0$$

$$\Rightarrow |z_1|=|z_2|=\dfrac{\sqrt5}{2}$$

cosmo5
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  • But at what points will we get our minimum? It remains to prove that there exist such numbers $z_1, z_2$ so $|z_1|=|z_2| = \frac{5}{2}$ and initial conditions are satisfied, Or am I missing something? – AO1992 Oct 19 '20 at 13:20
  • Yes, you're right. I added. – cosmo5 Oct 19 '20 at 13:30
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Let $|z_1|^2 = x$, $|z_2|^2 = y$. Then we have to find the minimum value of $x+4y$. Continuing from where you left off:

$$4x + y = 4xy \Rightarrow 4xy-y = 4x \Rightarrow y = \frac{4x}{4x-1}$$

and so:

$$x+4y=x + \frac{16x}{4x-1} = x + \frac{16x-4}{4x-1} + \frac{4}{4x-1} = x+4+\frac{4}{4x-1}$$

By AM-GM, $(x-\frac{1}{4})+\frac{4}{4x-1} ≥ 2 \sqrt{(x-\frac{1}{4}) \cdot \frac{4}{4x-1}} = 2$ since $x,y ≥ 0$.

Hence, the minimum value of $x + 4y = x - \frac{1}{4} + \frac{4}{4x-1} + \frac{17}{4}$ is $2 + \frac{17}{4} = \boxed{\frac{25}{4}}$.

Toby Mak
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