Show that $4\leq |z+3|+|z^2-z+4|\leq 11$ where $|z|=2$

Question

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, $4\leq f(x)\leq 11$

Answer 1

Geometric answer for maximum:

I interpret the expression $|z+3|+ |z^2 -z+4|$ as the "the distance from $-3$ to a point on a circle of radius $2$ (red) plus the length of line segment connecting the point on circle of radius $2$ centered at origin to a point on circle of radius $4$ but with twice the angle centered at (4,0) (blue)". Hopefully that made sense. If it did not, I have kept a Geogebra link at the end.

For the expression $|z+3|+|z^2-z+4|$, we can consider the set of geometric configuration parameterized to it as a function of $\theta$, one can easily check that both $\theta=0$ is a maximum by noticing that the distance function doesn't change to first order in $\theta$ [ the distance function evaluate at $d(\theta+ \Delta \theta)=d(\theta-\Delta \theta)$ for small enough $\Delta \theta$]. Taking higher values of $\Delta \theta$ , we notice the function is decreasing, therefore, $\theta=\pi$ is a maximum.

Here is a picture to specify what I mean:

The above is when $\theta=\pi$, increasing a bit:

Decreasing a bit:

A more detailed way one could convince themself could be through Taylor:

$$ d(\theta + \Delta \theta) = d(\theta) + \Delta \theta d'(\theta)$$

$$ d(\theta- \Delta \theta) = d(\theta) - \Delta \theta d'(\theta)$$

Using the fact that $d(\theta-\Delta \theta)= d(\theta+\Delta \theta)$ , we have that $d'(\theta)=0$

Geometric answer for minimum

We have by triangle inequality ,

$$ |z+3 + z^2 - z+4| \leq |z+3| + |z^2 - z+4|$$

Rewriting:

$$ | z^2 +4 + 3 | \leq |z+3| + |z^2 - z+4|$$

The left most term is the distance from the fixed point three to a varying point on the blue circle. Geometric observation tells us that the least value it can take is $4$ (See FA). Done!

GeoGebra link

Answer 2

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 $x \in [\tfrac14, 1].$