Alzebra

Algebra Level 5

If $\left| z \right| \geq 3$ ,then the least value of $\left| z+\dfrac{1}{z} \right|$ is expressed as $\dfrac{a}{b}$ . Find the value of $a+b$ .

DETAILS : $z \in \mathbb C$

The answer is 11.

2 solutions

Sandeep Bhardwaj
Dec 1, 2014

Using the simple and very useful property of complex numbers : $\large \boxed{\left| z_1 +z_2 \right| \geq \left| |z_1|-|z_2| \right|}$ Therefore, $\left| z+\dfrac{1}{z} \right| \geq \left| |z| -\dfrac{1}{|z|} \right|$

$\implies \left| z+\dfrac{1}{z} \right| \geq |3-\dfrac{1}{3}|$

$\implies \left| z+\dfrac{1}{z} \right| \geq \dfrac{8}{3}$

$therefore,\ a=8,b=3$

Hence the value of a+b is $\boxed{11}$

Note: You should show that the minimum can indeed be achieved.
You have only found a lower bound, you need to show that it is the greatest lowest bound, in order for it to be the least value.

Calvin Lin Staff - 6 years, 6 months ago

This can be done by euler's notation $(re^{\iota\theta})$

Let $z=3e^{\iota\theta}$ .

Then $\dfrac{1}{z}=\dfrac{1}{3}e^{-\iota\theta}$

Now the difference of both amplitudes must be $\pi$ to get the minima.

So $\theta-(-\theta)=\pm\pi$ or $\theta=\pm\dfrac{\pi}{2}$

Hence $z=\pm 3\iota$

Pranjal Jain - 6 years, 6 months ago

Yeah Too easy question !!

Karan Shekhawat - 6 years, 6 months ago

John Ashley Capellan
Dec 9, 2014

Consider z = r(cos a + i sin a). Using De Moivre's Theorem: 1/z = (1/r)(cos a - i sin a) Hence, z + 1/z = (r + 1/r)(cos a) + (r - 1/r)(i sin a) abs(z + 1/z) = sqrt((r^2 + 1/r^2 + 2)(cos^2 a) + (r^2 + 1/r^2 - 2)(sin^2 a)) = sqrt((r^2 + 1/r^2) + 2(cos^2 a - sin^2 a)) = sqrt((r^2 + 1/r^2) + 2 cos (2a)).

Since abs(z) >= 3, it implies that abs(r) >= 3. Since the minimum of cos(2a) = -1, it implies that min(|z + 1/z|) = sqrt((r^2 + 1/r^2) - 2) = sqrt((r - 1/r)^2) = |r - 1/r|. Since r >=3, it implies also that -1/r >= -1/3. Hence, min(|z + 1/z|) = |3 - 1/3| = 8/3.

Alzebra

The answer is 11.

2 solutions

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