, 2018

Solution:
Answer: $\frac{5}{4}$
Note that we must have $|z-1|=2$ and $|z^{2}-1|=6$, so $|z+1|=\frac{|z^{2}-1|}{|z-1|}=3$. Thus, the distance from $z$ to $1$ in the complex plane is $2$ and the distance from $z$ to $-1$ in the complex plane is $3$. Thus, $z$, $1$, $-1$ form a triangle with side lengths $2$, $3$, $3$. The area of a triangle with sides $2$, $2$, $3$ can be computed to be $\frac{3 \sqrt{7}}{4}$ by standard techniques, so the length of the altitude from $z$ to the real axis is $\frac{3 \sqrt{7}}{4} \cdot \frac{2}{2}=\frac{3 \sqrt{7}}{4}$. The distance between $1$ and the foot from $z$ to the real axis is $\sqrt{2^{2}-\left(\frac{3 \sqrt{7}}{4}\right)^{2}}=\frac{1}{4}$ by the Pythagorean Theorem. It is clear that $z$ has positive imaginary part as the distance from $z$ to $-1$ is greater than the distance from $z$ to $1$, so the distance from $0$ to the foot from $z$ to the real axis is $1+\frac{1}{4}=\frac{5}{4}$. This is exactly the real part of $z$ that we are trying to compute.