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

$$\mathrm{If}\mathrm{z}\in \mathrm{C}\mathrm{satisfies}$$$$\mid {\mathrm{z}}^3+{\mathrm{z}}^{-3}\mid ⩽2$$$$t\mathrm{hen}\mathrm{maximum}\mathrm{possible}\mathrm{value}\mathrm{of}\mid \mathrm{z}+{\mathrm{z}}^{-1}\mid \mathrm{is}?$$

Commented by 123456 last updated on 24/Oct/16

$$\mid z^3+z^{-3}\mid ⩽2$$$$z=e^{\theta \iota }$$$$z^3+z^{-3}=e^{3\theta \iota }+e^{-3\theta \iota }=2\cos \left(3\theta \right)\Rightarrow \mid \cos \left(3\theta \right)\mid ⩽1$$$$z+z^{-1}=e^{\theta \iota }+e^{-\theta \iota }=2\cos \theta$$$$\cosh t=\frac{e^t+e^{-t}}{2}$$$$\sinh t=\frac{e^t-e^{-t}}{2}$$$$e^{\theta \iota }=\cos \theta +\iota \sin \theta$$$$\cosh \left(t\iota \right)=\cos t$$$$\sinh \left(t\iota \right)=\iota \sin t$$$${\cos }^2\theta +{\sin }^2\theta =1$$$${\cosh }^{2}t-{\sinh }^{2}t=1$$

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

$$z={re}^{\iota \theta }$$$$\mid z^3+z^{-3}\mid =2r^3\mid \cos 3\theta \mid$$$$\mid z+z^{-1}\mid =2r\mid \cos \theta \mid$$$$\theta =\frac{\pi }{6},3\theta =\frac{\pi }{2}$$$$\mid z^3+z^{-3}\mid =0$$$$\mid z+z^{-1}\mid =2r\cos \frac{\pi }{6}=r\sqrt{3}$$$$\mathrm{You}\mathrm{can}\mathrm{choose}r.$$$$\mid z^3+z^{-3}\mid \mathrm{will}\mathrm{remain}0.$$$$\mathrm{So}\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{limit}\mathrm{on}\mid z+z^{-1}\mid$$

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