4 Given the complex number $z=\cos \frac{2 \pi}{5}+i \sin \frac{2 \pi}{5}$, then $(1-z)\left(1-z^{2}\right) \cdot\left(1-z^{3}\right)\left(1-z^{4}\right)=$
4 5. Since $z=\cos \frac{2 \pi}{5}+i \sin \frac{2 \pi}{5}$ is a fifth root of 1, the five roots of the equation $x^{5}-1=0$ are $1, z, z^{2}, z^{4}$. Therefore, $x^{5}-1=(x-1)(x-z)\left(x-z^{2}\right)\left(x-z^{3}\right)\left(x-z^{4}\right)$. Also, $x^{5}-1=\left(1+x+x^{2}+x^{3}+x^{4}\right)(x-1)$, so $(x-z) \cdot\left(x-z^{2}\right)\left(x-z^{3}\right)$ $\left(x-z^{4}\right)=1+x+x^{2}+x^{3}+x^{4}$. By setting $x=1$, we obtain the result.