Let z be a complex number (not lying on the X | vedclass

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(B) Let $z = r(\cos 	heta + i \sin 	heta)$.Given $\left| z + \frac{1}{z} ight| = 1$,we have $\left| z + \frac{1}{z} ight|^2 = 1$.Substituting $z

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$	ext{Re}(z) = 0$

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(B) Let $z = r(\cos 	heta + i \sin 	heta)$.Given $\left| z + \frac{1}{z} ight| = 1$,we have $\left| z + \frac{1}{z} ight|^2 = 1$.Substituting $z = r e^{i	heta}$,we get $\left| r e^{i	heta} + \frac{1}{r} e^{-i	heta} ight|^2 = 1$.This expands to $\left( r + \frac{1}{r} ight)^2 \cos^2 	heta + \left( r - \frac{1}{r} ight)^2 \sin^2 	heta = 1$.Simplifying,we get $r^2 + \frac{1}{r^2} + 2 \cos(2	heta) = 1$.To maximize $r$,we differentiate with respect to $	heta$ and set $\frac{dr}{d	heta} = 0$.This yields $-4 \sin(2	heta) = 0$,so $\sin(2	heta) = 0$,which means $	heta = 0, \frac{\pi}{2}, \pi, \dots$.Since $z$ does not lie on the $X$-axis,$	heta 
eq 0, \pi$. Thus,$	heta = \frac{\pi}{2}$ or $\frac{3\pi}{2}$.In these cases,$z$ is purely imaginary,so $	ext{Re}(z) = 0$.

Let $z$ be a complex number (not lying on the $X$-axis) of maximum modulus such that $\left| z + \frac{1}{z} \right| = 1$. Then:

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