$|2z^2 + 3iz + 1| < 6, \quad |z| = 1$

Question

Show that $|2z^2 + 3iz + 1| < 6, \ |z| = 1$ holds for complex z. (note, its not supposed to be $\leq$ !)

Answer 1

hint $\vert a+b+c\vert\le \vert a\vert+\vert b\vert+\vert c\vert$

Edit:

1)This is a general method:

Let $f(z)=2z^2+3iz+1$, and $g(z)=f(z)\overline{f(z)}$, then $g(z)$ is a real valued function. Since $|z|=1$, you can write $z$ as $e^{i\theta}$, $\quad\theta\in[0,2\pi)$. Express $g(z)$ in terms of $\theta$ and find the maximum of $g(z)$, if the max is $36$ then your inequality is $\le$, if not that stays $<$

2) You can also consider $M_1,M_2$ and $I$ as respectives images of the complexes $2z^2, 3iz$ and $1$. Recall that the equality occurs if and only if these 3 points lie in the same line i.e $\widehat{M_1IM_2}=0\text{ or }\pi$

Answer 2

Hint:

For any three complex numbers $\;a,b,c\;$, $\;|a+b+c|\le |a|+|b|+|c|\;$ .

Answer 3

$$\begin{align} |z^2+3iz+1| &\leq |z^2|+3|iz|+1 \qquad (\textrm{by triangle inequality})\\ &= |z|^2+3|i||z|+1 \quad (|zw|=|z||w| \textrm{ for all } z,w\in \mathbb C)\\ &= 1^2+3\cdot 1^2+1 \qquad (|z|=1 \textrm{ and } |i|=1) \\ &= 5 \\ &< 6. \end{align}$$