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

Maverick
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    What is your question? – lhf Apr 04 '22 at 23:48
  • Is there a way to do this without using calculus – Maverick Apr 04 '22 at 23:51
  • Hi @Maverick, I've edited the answer once again (for the last time) with more correct steps and detailed arguement. Please go through it and tell me if everything is clear – Clemens Bartholdy Apr 06 '22 at 07:47
  • A tag edit isn't worth bumping the question, but if there's occasion to edit could you please delete the complex-geometry tag? (The geometry tag might be appropriate to add, but this isn't complex geometry.) – Andrew D. Hwang Apr 07 '22 at 02:10
  • So geometry+ complex numbers $\neq$ complex geometry..? O_o @AndrewD.Hwang – Clemens Bartholdy Apr 07 '22 at 02:49
  • @Buraian Yes that's right. Same for analytic-geometry, algebraic-geometry, and others that don't come to mind offhand. <> Hovering over a tag pops up a description in a tooltip. The descriptions generally state explicitly when they should not be used (e.g., "For elementary questions about geometry in the complex plane, use the tags (complex-numbers) and (geometry) instead."), but that's usually at the bottom and isn't always visible. – Andrew D. Hwang Apr 07 '22 at 12:29

2 Answers2

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Geometric answer for maximum:

enter image description here

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:

enter image description here The above is when $\theta=\pi$, increasing a bit:

enter image description here

Decreasing a bit:

enter image description here

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

Clemens Bartholdy
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  • If you thought this was 'entertaining' , you can check out this post – Clemens Bartholdy Apr 05 '22 at 07:02
  • According to first diagram $d_2=|z^2+3|$ – Maverick Apr 05 '22 at 15:22
  • What should $\theta$ be if $d_1+d_2$ were to be minimum . – Maverick Apr 05 '22 at 15:41
  • Please tell for minimum – Maverick Apr 05 '22 at 15:53
  • @Buraian $;d_1+d_2 = 2 \sqrt{5} + \sqrt{13} \gt 8,$ when $,z=2i,$ so the maximum cannot be $,7,$. – dxiv Apr 05 '22 at 17:43
  • Totally redid the answer @dxiv – Clemens Bartholdy Apr 05 '22 at 21:54
  • @Buraian This latest edit looks a lot better now, however... 1) The claim that the differentials in question are zero is mostly handwaved. 2) Even accepting those claims, what they prove is that the respective points correspond to local extrema, but the question is about the global maximum and minimum. – dxiv Apr 06 '22 at 21:39
  • One can easily see that extremum occurs only at intersection between circle ,x=-2 and x=2. The secondo rder behaviour (seeing distance increases or decreases) as we vary the angle tells us if it is max or min. Now, that we know the set of all max and min and also knowing that the function is bounded, we can identify the globals @dxiv – Clemens Bartholdy Apr 06 '22 at 21:43
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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 $x \in [\tfrac14, 1].$

Calum Gilhooley
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    To get the expression for $f(x),$ we can use: $$ |z^2 - z + 4| = |z^2 - z + z\bar{z}| = |z|\cdot|z + \bar{z} - 1| = 2|2\Re{z} - 1|. $$ – Calum Gilhooley Apr 06 '22 at 01:27