Question 102: Let the complex number $z$ satisfy $z+\frac{1}{z} \in[1,2]$, then the minimum value of the real part of $z$ is $\qquad$

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Question 102, Answer: According to the condition, $\operatorname{Im}(z)=-\operatorname{Im}\left(\frac{1}{z}\right)$. And $\frac{1}{z}=\frac{z}{|z|^{2}}$, so either $|z|=1$, or $z \in$ R.

If $z \in R$, since $1 \leq z+\frac{1}{z} \leq 2$, it can only be $z=1$, that is, $\operatorname{Re}(z)=z=1$; if $|z|=1$, then $z+\frac{1}{z}=2 \operatorname{Re}(z) \in[1,2]$, so $\operatorname{Re}(z) \geq \frac{1}{2}$. When $z=\frac{-1+\sqrt{3}}{2}$, it satisfies the condition, and $\operatorname{Re}(z)=\frac{1}{2}$.