If-z-C-satisfies-z-3-z-3-2-then-maximum-possible-value-of-z-z-1-is-

Question Number 8721 by swapnil last updated on 24/Oct/16

If z \in C satisfies

∣ z^{3} + z^{- 3} ∣⩽ 2

t hen maximum possible value of ∣ z + z^{- 1} ∣ is ?

Commented by 123456 last updated on 24/Oct/16

∣ z^{3} + z^{- 3} ∣⩽ 2

z = e^{θ ι}

z^{3} + z^{- 3} = e^{3 θ ι} + e^{- 3 θ ι} = 2 \cos (3 θ) \Rightarrow∣ \cos (3 θ) ∣⩽ 1

z + z^{- 1} = e^{θ ι} + e^{- θ ι} = 2 \cos θ

\cosh t = \frac{e^{t} + e^{- t}}{2}

\sinh t = \frac{e^{t} - e^{- t}}{2}

e^{θ ι} = \cos θ + ι \sin θ

\cosh (t ι) = \cos t

\sinh (t ι) = ι \sin t

\cos^{2} θ + \sin^{2} θ = 1

\cosh^{2} t - \sinh^{2} t = 1

Commented by prakash jain last updated on 24/Oct/16

z = {r e}^{ι θ}

∣ z^{3} + z^{- 3} ∣= 2 r^{3} ∣ \cos 3 θ ∣

∣ z + z^{- 1} ∣= 2 r ∣ \cos θ ∣

θ = \frac{π}{6}, 3 θ = \frac{π}{2}

∣ z^{3} + z^{- 3} ∣= 0

∣ z + z^{- 1} ∣= 2 r \cos \frac{π}{6} = r \sqrt{3}

You can choose r .

∣ z^{3} + z^{- 3} ∣ will remain 0 .

So there is no limit on ∣ z + z^{- 1} ∣