Note that we must have $|z-1|=2$ and $\left|z^{2}-1\right|=6$$, so$$|z+1|=\frac{\left|z^{2}-1\right|}{|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.