I would like to prove:
Let $p(z) = z^2 +az +b$, where a and b are positive, and $z \in \mathbb{C}$. Show that $|p(z)|$ attains minimum over the region $\Re(z)) \geq 0$ on the imaginary axis.
Here is what I have done:
By contradiction, suppose that |p(z)| does not attain its minimum on $\Re(z)) \geq 0$. So then, it must be that $|p(z)| = c$, some constant. Or we get that $|p(z)|^2 = c^2$. Taking partial derivatives, we get that $\frac{\partial |p(z)|^2}{\partial x}$ = $\frac{\partial |p(z)|^2}{\partial y}$ = $0$.
If $p(z)$ is as defined above, then we can of course separate into real and imaginary parts $u(x,y)$ and $v(x,y)$. Thus, we get: $\frac{\partial |p(z)|^2}{\partial x}$ = $2uu_x + 2vv_x = 0$.Similarly, we get $\frac{\partial |p(z)|^2}{\partial y}$ = $2uu_y + 2vv_y = 0$.Since $p(z)$ is entire and hence analytic, we can exploit the fact that Cauchy Riemman equations hold. So the equations become $uu_x -vu_y =0$ and $uu_y + vv_y = 0$. After multiplying the first by u and second by 2 and combining we get: $(u^2 + v^2)(u_x + u_x) = 0$. Since $c^2 = u^2 + v^2$, then $c^2(u_x + u_x) =0$. Similarly, multiplying first equation by v and second by u, and applying CR, then $(u^2 +v^2)v_x = c^2v_x = 0$, and so $v_x =0$.Again, using CR, we get that $v_x= -u_y= u_x=v_y=0$ by what we get above.So $p'(z) = 0$ and so $p(z) = const$, but that's a contradiction, since $p'(z) = 2z+a$.
I am not sure whether my logic here is right, especially at the beginning when stating that if |f| does not achieve a minimum, then |f| is constant.Can someone please verify?