Did I misunderstand it or my book is wrong?

Question

From my book, Prove that if $z \in \mathbb C $ where $|z|\leq 1$ then $|Im(1+\bar z+z^2)|\lt 3$

but, I have $|Im(1+\bar z+z^2)|\leq 3$

From $|Im(1+\bar z+z^2)|$ , I have

$$|Im(1+\bar z+z^2)|\leq|1+\bar z+z^2|\leq|1|+|\bar z|+|z|^2\leq1+|1|+|1|=3$$

Please check my solution, Thank you.

Answer 1

From the chain of inequalities you have written we see that equality can hold only when we have equality in each term, In particular we must have $|z|=1$. Let $z=e^{i\theta}$ with $\theta$ real. Then $Im (1+\overline {z}+z^{2})=-\sin \theta+\sin (2\theta)$. But $-\sin \theta+2\sin (\theta)\cos (\theta)=3$ is impossible because this forces both $|\sin \theta|$ and $|\cos \theta|=1$ contradicting the fact that $\sin^{2}(\theta)+\cos^{2}(\theta)=1$. Hence the final inequality is strict.

Answer 2

If $z=x+iy$ , then $Im(1+\overline{z}+z^2)=2xy-y.$ Now let $a=(x,y)$ and $b=(2y,-1).$ By Cauchy-Schwarz we get

$$|2xy-y|=| a \cdot b| \le ||a||_2 ||b||_2 =|z| \cdot \sqrt{4y^2+1} \le \sqrt{4y^2+1} \le \sqrt{5}<3.$$

Answer 3

Another way to see what you need to proof is to realize that the "$1$" at the beginning of the term is doing nothing to the imaginary part of that term, so

$$Im (1+\bar{z}+z^2) = Im (\bar{z}+z^2)$$

and hence using the same technique you used:

$$|Im (1+\bar{z}+z^2)| = |Im (\bar{z}+z^2)| \le |\bar{z} + z^2| \le |\bar{z}| + |z^2| \le |1| + |1| = 2 < 3.$$