Show that $4\leq |z+3|+|z^2-z+4|\leq 11$ where $|z|=2$ Let $z\in \mathbb{C}$ and$|z|=2$.Show that $$4\leq |z+3|+|z^2-z+4|\leq 11$$
Is there a non-calculus way to do this
My Attempt:
I used method which involved calculus.
Let $z=2(\cos t+i\sin t)$
$\Rightarrow|z+3|=\sqrt{13+12\cos t}$
$\Rightarrow|z^2-z+4|=2|4\cos t-1|$
Let $f(x)=\sqrt{13+12x}+2|4x-1|$ where $x\in [-1,1]$
Now here $f(-1)=f(1)=11$
$f'(x)=\frac{6}{\sqrt{13+12x}}-8<0$ for $x\in [-1,\frac{1}{4})$
and $f'(x)=\frac{6}{\sqrt{13+12x}}+8>0$ for $x\in (\frac{1}{4},1]$
So, clearly $x=\frac{1}{4}$ is a critical point as $f(x)$ is not differentiable at $x=\frac{1}{4}$.
$f(\frac{1}{4})=4$
So, $4\leq f(x)\leq 11$

From the OP's proof by calculus:

Let $f(x )= \sqrt{13 + 12x} + 2|4x - 1|$ where $x \in [-1, 1].$

The expression $u = \sqrt{13 + 12x}$ is an increasing function of
$x$ in $[-1, 1],$ and $12x = u^2 - 13,$ therefore:
$$
f(x) = \begin{cases}
u + 2 - \tfrac23(u^2 - 13) = 11 - \tfrac13(u - 1)(2u - 1) &
(-1 \leqslant x \leqslant \tfrac14, \ 1 \leqslant u \leqslant 4), \\
u - 2 + \tfrac23(u^2 - 13) = \tfrac13(u + 1)(2u + 1) - 11 &
(\tfrac14 \leqslant x \leqslant 1, \ 4 \leqslant u \leqslant 5).
\end{cases}
$$
No calculus is needed to deduce that $f(x)$
decreases from $11$ to $4$ for $x \in [-1, \tfrac14]$ and
increases from $4$ to $11$ for $x \in [\tfrac14, 1].$
