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Question Number 140198 by EnterUsername last updated on 05/May/21

Let a > 0 and ∣ z + (1 / z) ∣= a (z \neq 0 is a complex number) .

Then the maximum and minimum values of ∣ z ∣ are

(A) \frac{a + \sqrt{a^{2} + 4}}{2} (B) \frac{2 a + \sqrt{a^{2} + 4}}{2}

(C) \frac{\sqrt{a^{2} + 4} - a}{2} (D) \frac{\sqrt{a^{2} + 4} - 2 a}{2}

Answered by Dwaipayan Shikari last updated on 05/May/21

∣ z ∣ + ∣ \frac{1}{z} ∣⩾∣ z + \frac{1}{z} ∣

F o r M a x i m u m ∣ z + \frac{1}{z} ∣=∣ z ∣ + ∣ \frac{1}{z} ∣

∣ z ∣ + \frac{1}{∣ z ∣} = a \Rightarrow∣ z ∣= \frac{a \pm \sqrt{a^{2} + 4}}{2}

M a x i m u m V a l u e ∣ z ∣_{m a x} = \frac{a + \sqrt{a^{2} + 4}}{2}

Commented by EnterUsername last updated on 05/May/21

Thanks!

Commented by mr W last updated on 05/May/21

∣ \frac{1}{z} ∣\neq \frac{1}{∣ z ∣}

Answered by mr W last updated on 05/May/21

l e t z = {r e}^{θ i}

∣ z ∣= r

\frac{1}{z} = \frac{1}{{r e}^{θ i}} = \frac{1}{r} e^{- θ i}

∣ z + \frac{1}{z} ∣=∣ {r e}^{θ i} + \frac{1}{r} e^{- θ i} ∣= \sqrt{r^{2} + \frac{1}{r^{2}} + 2 \cos 2 θ} = a

r^{2} + \frac{1}{r^{2}} = a^{2} - 2 \cos 2 θ = λ

m a x . o r m i n . f r o m r i s w h e n λ i s

a s l a r g e a s p o s s i b l e, i . e . λ = a^{2} + 2, a n d

θ = \frac{π}{2} .

r^{4} - (a^{2} + 2) r^{2} + 1 = 0

\Rightarrow r_{m a x / m i n}^{2} = \frac{a^{2} + 2 \pm \sqrt{{(a^{2} + 2)}^{2} - 4}}{2}

= \frac{a^{2} + 2 \pm a \sqrt{a^{2} + 4}}{2}

= \frac{a^{2} + 4 \pm 2 a \sqrt{a^{2} + 4} + a^{2}}{4}

= \frac{{(\sqrt{a^{2} + 4})}^{2} \pm 2 a \sqrt{a^{2} + 4} + a^{2}}{4}

= \frac{{(\sqrt{a^{2} + 4} \pm a)}^{2}}{4}

\Rightarrow r_{m a x / m i n} = \frac{\sqrt{a^{2} + 4} \pm a}{2}

i . e .

r_{m a x} = \frac{\sqrt{a^{2} + 4} + a}{2}

r_{m i n} = \frac{\sqrt{a^{2} + 4} - a}{2}

Commented by EnterUsername last updated on 06/May/21

Thank you Sir