let $z$ be a complex number (not lying on $x$-axis) of maximum modulus such that $\Big|z+\dfrac{1}{z}\Big|=1$. Then prove that $\mathcal{Re}(z)=0$
My Attempt $$ |re^{ia}+\frac{1}{r}e^{-ia}|=|\cos a(r+\frac{1}{r})+i\sin a(r-\frac{1}{r})|=1\\ r^2+\frac{1}{r^2}+2\cos 2a=1\implies 2r\frac{dr}{da}-\frac{2}{r^3}\frac{dr}{da}-4\sin 2a=0\\ \frac{dr}{da}.(r-\frac{1}{r^3})=2\sin 2a=0\implies\sin 2a=0\\ a=0\quad\text{or}\quad a=\pi/2\implies a=\pi/2\implies\cos a=0\\ Re(z)=0 $$ Can I prove it without using differentiation ?
The way I would do it is like this: if $ab \neq 0$, $|a+b| = |a|-|b|$ if and only if $a,b$ make a 180 degree angle when represented geometrically as vectors in the complex plane by the cosine theorem; but now $1 = |z+ \frac{1}{z}| \geq |z|-\frac{1}{|z|}$, so $|z|$ is at most the highest root of $x^2-x-1 = 0$ (which is the golden ratio) and we have equality precisely when $z$ and $\frac{1}{z}$ make a 180 degree angle which is easily seen to be equivalent to $z$ purely imaginary