1. How many complex numbers $z$ satisfy

$$
z^{2019}=(z+1)^{2019}=1 \quad ?
$$
1. From the equality $z^{2019}=1$ it follows that $\left|z^{2019}\right|=1$, and then $|z|^{2019}=1$, from which we get $|z|^{2019}=1$. In the same way, it follows that $|z+1|=1$. Let's write $z=a+bi$, $a, b \in \mathbb{R}$. The equality $|z|=1$ reduces to $a^{2}+b^{2}=1$, while $|z+1|=1$ reduces to $(a+1)^{2}+b^{2}=1$. Subtracting the second-to-last equality from the last one, we get $2a+1=0$, i.e., $a=-\frac{1}{2}$. Now we have $\left(-\frac{1}{2}\right)^{2}+b^{2}=1$, i.e., $b^{2}=\frac{3}{4}$, from which it follows that $b= \pm \frac{\sqrt{3}}{2}$. Therefore, the only possible candidates for the numbers that satisfy the required condition are $z=-\frac{1}{2} \pm i \frac{\sqrt{3}}{2}$. In this case, we have $z+1=\frac{1}{2} \pm i \frac{\sqrt{3}}{2}$ and then $(z+1)^{3}=$ $\left(\frac{1}{2} \pm \frac{\sqrt{3}}{2} i\right)^{3}=\left(\frac{1}{2}\right)^{3} \pm 3 \cdot\left(\frac{1}{2}\right)^{2} \cdot \frac{\sqrt{3}}{2} i+3 \cdot \frac{1}{2} \cdot\left(\frac{\sqrt{3}}{2} i\right)^{2} \pm\left(\frac{\sqrt{3}}{2} i\right)^{3}=\frac{1}{8} \pm \frac{3 \sqrt{3}}{8} i-\frac{9}{8} \mp \frac{3 \sqrt{3}}{8} i=-1$, so from there $(z+1)^{2019}=\left((z+1)^{3}\right)^{673}=(-1)^{673}=-1$, which does not meet the condition.

Therefore, there is no complex number that satisfies the given condition.