1. Convert the circle equation into Cartesian form

Let $z=x+iy.$ Then:

• $\operatorname{Re}(z)=x$
• $\operatorname{Im}(z)=y$
• $z^2=(x+iy)^2=x^2-y^2+2ixy$ So, $\operatorname{Re}(z^2)=x^2-y^2.$

Given: $\operatorname{Re}(z^2)+2(\operatorname{Im}(z))^2+2\operatorname{Re}(z)=0$ Substitute: $x^2-y^2+2y^2+2x=0$ $x^2+y^2+2x=0$ Complete the square: $x^2+2x+y^2=0$ $(x+1)^2+y^2=1$ So the circle has center $(-1,0).$

2. Find the vertex of the parabola

Given parabola: $x^2-6x-y+13=0$ Rearrange: $y=x^2-6x+13$ Complete the square: $y=(x-3)^2+4$ Hence the vertex is $(3,4).$

3. Equation of the line through the center and the vertex

The line passes through points $(-1,0)$ and $(3,4).$ Its slope is $m=\frac{4-0}{3-(-1)}=\frac{4}{4}=1.$

Using point-slope form through $(-1,0)$: $y-0=1(x+1)$ $y=x+1.$

4. Find the y-intercept

For the line $y=x+1,$ when $x=0$, $y=1.$ So the y-intercept is $\boxed{1}.$

5. Comparison with stored correct answer

Stored correct answer = $1$. Our derived answer also equals $1$, so they agree.