8. In the complex plane, the points $0, z, \frac{1}{z}, z+\frac{1}{z}$ form a parallelogram with an area of $\frac{35}{37}$. If the real part of $z$ is greater than 0, then the minimum value of $\left|z+\frac{1}{z}\right|$ is $\qquad$
8. $\frac{5 \sqrt{74}}{37}$

Analysis: It is known that $\frac{1}{z}=\frac{\bar{z}}{|z|^{2}}$, so the arguments of $\frac{1}{z}$ and $z$ on the complex plane are opposite numbers. Suppose $z=r e^{i \theta}$, then $\frac{1}{z}=\frac{e^{-i \theta}}{r}$. The area of the parallelogram can be expressed as: $|z| \cdot\left|\frac{1}{z}\right| \cdot \sin 2 \theta$. Given that $\sin 2 \theta=\frac{35}{37}$, then $\left|z+\frac{1}{z}\right|^{2}=\left(r e^{i \theta}+\frac{e^{-i \theta}}{r}\right) \cdot\left(r e^{-i \theta}+\frac{e^{i \theta}}{r}\right)=r^{2}+e^{i \cdot 2 \theta}+e^{-i \cdot 2 \theta}+\frac{1}{r^{2}}=r^{2}+\frac{1}{r^{2}}+2 \cos 2 \theta \geq \frac{50}{37}$. Therefore, $\left|z+\frac{1}{z}\right| \geq \frac{5 \sqrt{74}}{37}$.