Let $R$ be the region $z \leq 1$. Compute the maximum value of $|z^2 + z + 2|$ in $R$ and find out the point where this function reaches this value in $R$.
I let $$z = \cos \theta + i \sin \theta$$ Then $$|z^2 + z +2|^2 = (\cos 2 \theta + 2 \cos \theta)^2 + (\sin 2 \theta + 2 \sin \theta)^2$$ $$= \cos^2 2\theta + 4\cos^2 \theta+ 4 + 4\cos\theta \cos 2\theta+ 4\cos2\theta+8\cos\theta+\sin^2 2\theta+ 4\sin^2\theta+4\sin\theta\sin2\theta$$
I am having trouble simplifying this and going further to get $$8(\cos \theta + 3/4)^2 + 1/2$$ That's the answer(maximum value $5$, occurs at $\theta=0$) that is given. I would be grateful for any help to solve this question.
By the maximum modulus principle, the maximum must occur on the boundary. By the triangle inequality, $|z^2+z+2|\leq|z|^2+|z|+2\leq4$. We can check $1^2+1+2=4$, so the function attains its maximum when $z=1$.