The maximum distance from the origin of coord | vedclass

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(C) Let $z = r(\cos 	heta + i \sin 	heta)$.Then $\left| z + \frac{1}{z} ight| = a \implies \left| z + \frac{1}{z} ight|^2 = a^2$.Expanding this,

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$\frac{1}{2}(\sqrt{a^2 + 4} + a)$

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(C) Let $z = r(\cos 	heta + i \sin 	heta)$.Then $\left| z + \frac{1}{z} ight| = a \implies \left| z + \frac{1}{z} ight|^2 = a^2$.Expanding this,we get $\left| r(\cos 	heta + i \sin 	heta) + \frac{1}{r}(\cos 	heta - i \sin 	heta) ight|^2 = a^2$.This simplifies to $\left( r + \frac{1}{r} ight)^2 \cos^2 	heta + \left( r - \frac{1}{r} ight)^2 \sin^2 	heta = a^2$.Using $\cos^2 	heta = 1 - \sin^2 	heta$,we get $r^2 + \frac{1}{r^2} + 2 \cos 2	heta = a^2$.To maximize $r$,we need to minimize $\cos 2	heta$. The minimum value of $\cos 2	heta$ is $-1$ (at $	heta = \frac{\pi}{2}$).Substituting $\cos 2	heta = -1$,we get $r^2 + \frac{1}{r^2} - 2 = a^2$.This is $(r - \frac{1}{r})^2 = a^2$,so $r - \frac{1}{r} = a$ (taking the positive root for distance).$r^2 - ar - 1 = 0$.Solving for $r$ using the quadratic formula,$r = \frac{a + \sqrt{a^2 + 4}}{2}$.

The maximum distance from the origin of coordinates to the point $z$ satisfying the equation $\left| z + \frac{1}{z} \right| = a$ is

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