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Question Number 63784 by MJS last updated on 09/Jul/19
question 63639 again prove: ∀z∈C: ∣z+1∣+∣z^2 +z+1∣+∣z^3 +1∣≥1
$$\mathrm{question}\:\mathrm{63639}\:\mathrm{again} \\ $$$$\mathrm{prove}: \\ $$$$\forall{z}\in\mathbb{C}:\:\mid{z}+\mathrm{1}\mid+\mid{z}^{\mathrm{2}} +{z}+\mathrm{1}\mid+\mid{z}^{\mathrm{3}} +\mathrm{1}\mid\geqslant\mathrm{1} \\ $$
Commented by Prithwish sen last updated on 09/Jul/19
1=∣1+z +1 + z^3 − (z^3 + z^2 ) − (z^2 + z + 1)∣ ≤ ∣1+z∣ + ∣1+z^3 ∣ + ∣z^2 ∣∣1+z∣ +∣z^2 +z + 1∣ ( ∵ ∣z_1 −z_2 ∣ ≤ ∣z_1 ∣ +∣z_2 ∣ ≤ ∣1+z∣ +∣1+z^3 ∣ + ∣ z^2 +z+1∣ = ∣1+z∣ + ∣z^2 +z+1∣ + ∣z^3 +1∣ (∵ addition of complex numbers follow commutative law. Hence proved.
$$\mathrm{1}=\mid\mathrm{1}+\mathrm{z}\:+\mathrm{1}\:+\:\mathrm{z}^{\mathrm{3}} −\:\left(\mathrm{z}^{\mathrm{3}} \:+\:\mathrm{z}^{\mathrm{2}} \right)\:−\:\left(\mathrm{z}^{\mathrm{2}} \:+\:\mathrm{z}\:+\:\mathrm{1}\right)\mid \\ $$$$\leqslant\:\mid\mathrm{1}+\mathrm{z}\mid\:+\:\mid\mathrm{1}+\mathrm{z}^{\mathrm{3}} \mid\:+\:\mid\mathrm{z}^{\mathrm{2}} \mid\mid\mathrm{1}+\mathrm{z}\mid\:+\mid\mathrm{z}^{\mathrm{2}} +\mathrm{z}\:+\:\mathrm{1}\mid\:\left(\:\because\:\mid\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} \mid\:\leqslant\:\mid\mathrm{z}_{\mathrm{1}} \mid\:+\mid\mathrm{z}_{\mathrm{2}} \mid\right. \\ $$$$\leqslant\:\mid\mathrm{1}+\mathrm{z}\mid\:+\mid\mathrm{1}+\mathrm{z}^{\mathrm{3}} \mid\:+\:\mid\:\mathrm{z}^{\mathrm{2}} +\mathrm{z}+\mathrm{1}\mid\: \\ $$$$=\:\mid\mathrm{1}+\mathrm{z}\mid\:+\:\mid\mathrm{z}^{\mathrm{2}} +\mathrm{z}+\mathrm{1}\mid\:+\:\mid\mathrm{z}^{\mathrm{3}} +\mathrm{1}\mid\:\left(\because\:\mathrm{addition}\right. \\ $$$$\mathrm{of}\:\mathrm{complex}\:\mathrm{numbers}\:\mathrm{follow}\:\mathrm{commutative}\:\mathrm{law}. \\ $$$$\boldsymbol{\mathrm{Hence}}\:\boldsymbol{\mathrm{proved}}. \\ $$
Commented by Prithwish sen last updated on 09/Jul/19
please check it sir.
$$\mathrm{please}\:\mathrm{check}\:\mathrm{it}\:\mathrm{sir}. \\ $$
Commented by Prithwish sen last updated on 09/Jul/19
Is it ok sir ?
$$\mathrm{Is}\:\mathrm{it}\:\mathrm{ok}\:\mathrm{sir}\:? \\ $$
Commented by mr W last updated on 09/Jul/19
how can you say: ∣1+z∣ + ∣1+z^3 ∣ + ∣z^2 ∣∣1+z∣ +∣z^2 +z + 1∣ ≤ ∣1+z∣ +∣1+z^3 ∣ + ∣ z^2 +z+1∣ ?
$${how}\:{can}\:{you}\:{say}: \\ $$$$\mid\mathrm{1}+\mathrm{z}\mid\:+\:\mid\mathrm{1}+\mathrm{z}^{\mathrm{3}} \mid\:+\:\mid\mathrm{z}^{\mathrm{2}} \mid\mid\mathrm{1}+\mathrm{z}\mid\:+\mid\mathrm{z}^{\mathrm{2}} +\mathrm{z}\:+\:\mathrm{1}\mid\: \\ $$$$\leqslant\:\mid\mathrm{1}+\mathrm{z}\mid\:+\mid\mathrm{1}+\mathrm{z}^{\mathrm{3}} \mid\:+\:\mid\:\mathrm{z}^{\mathrm{2}} +\mathrm{z}+\mathrm{1}\mid\: \\ $$$$? \\ $$
Commented by MJS last updated on 10/Jul/19
so how can we solve it? for z=−1 0+1+0≥1 for z=−(1/2)±((√3)/2)i 1+0+2≥1 for z=(1/2)±((√3)/2)i (√3)+2+0≥1 is it enough to show that for z close to −1 the sum of the absolutes is rising in every direction? we can put x=−1+re^(iθ) with −π≤θ<π and choose any small value for r>0 ⇒ ∣z+1∣+∣z^2 +z+1∣+∣z^3 +1∣=f(θ)= =r+(√(r^4 −2r^3 cos θ −(1−4cos^2 θ)r^2 −2rcos θ +1))+r(√(r^4 −6r^3 cos θ +3(1+4cos^2 θ)r^2 −18rcos θ +9)) (df/dθ)=rsin θ ×[some complicated term which has no zeros for small values of r] ⇒ our function has minima/maxima at θ=−π∨θ=0 ⇒ { ((f(−π)=r+∣r^2 +r+1∣+r∣r^2 +3r+3∣)),((f(0)=r+∣r^2 −r+1∣+r∣r^2 −3r+3∣)) :} in both cases the minimum is at r=0 because r cannot be negative ⇒ z=−1 seems to be the absolute minimum with ∣z+1∣+∣z^2 +z+1∣+∣z^3 +1∣=1
$$\mathrm{so}\:\mathrm{how}\:\mathrm{can}\:\mathrm{we}\:\mathrm{solve}\:\mathrm{it}? \\ $$$$\mathrm{for}\:{z}=−\mathrm{1} \\ $$$$\mathrm{0}+\mathrm{1}+\mathrm{0}\geqslant\mathrm{1} \\ $$$$\mathrm{for}\:{z}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$$\mathrm{1}+\mathrm{0}+\mathrm{2}\geqslant\mathrm{1} \\ $$$$\mathrm{for}\:{z}=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$$\sqrt{\mathrm{3}}+\mathrm{2}+\mathrm{0}\geqslant\mathrm{1} \\ $$$$\mathrm{is}\:\mathrm{it}\:\mathrm{enough}\:\mathrm{to}\:\mathrm{show}\:\mathrm{that}\:\mathrm{for}\:{z}\:\mathrm{close}\:\mathrm{to}\:−\mathrm{1} \\ $$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{absolutes}\:\mathrm{is}\:\mathrm{rising}\:\mathrm{in}\:\mathrm{every} \\ $$$$\mathrm{direction}? \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{put}\:{x}=−\mathrm{1}+{r}\mathrm{e}^{\mathrm{i}\theta} \:\mathrm{with}\:−\pi\leqslant\theta<\pi\:\mathrm{and} \\ $$$$\mathrm{choose}\:\mathrm{any}\:\mathrm{small}\:\mathrm{value}\:\mathrm{for}\:{r}>\mathrm{0}\:\Rightarrow \\ $$$$\mid{z}+\mathrm{1}\mid+\mid{z}^{\mathrm{2}} +{z}+\mathrm{1}\mid+\mid{z}^{\mathrm{3}} +\mathrm{1}\mid={f}\left(\theta\right)= \\ $$$$={r}+\sqrt{{r}^{\mathrm{4}} −\mathrm{2}{r}^{\mathrm{3}} \mathrm{cos}\:\theta\:−\left(\mathrm{1}−\mathrm{4cos}^{\mathrm{2}} \:\theta\right){r}^{\mathrm{2}} −\mathrm{2}{r}\mathrm{cos}\:\theta\:+\mathrm{1}}+{r}\sqrt{{r}^{\mathrm{4}} −\mathrm{6}{r}^{\mathrm{3}} \mathrm{cos}\:\theta\:+\mathrm{3}\left(\mathrm{1}+\mathrm{4cos}^{\mathrm{2}} \:\theta\right){r}^{\mathrm{2}} −\mathrm{18}{r}\mathrm{cos}\:\theta\:+\mathrm{9}} \\ $$$$\frac{{df}}{{d}\theta}={r}\mathrm{sin}\:\theta\:×\left[{some}\:{complicated}\:{term}\:{which}\:{has}\right. \\ $$$$\left.{no}\:{zeros}\:{for}\:{small}\:{values}\:{of}\:{r}\right] \\ $$$$\Rightarrow\:\mathrm{our}\:\mathrm{function}\:\mathrm{has}\:\mathrm{minima}/\mathrm{maxima}\:\mathrm{at} \\ $$$$\theta=−\pi\vee\theta=\mathrm{0} \\ $$$$\Rightarrow\:\begin{cases}{{f}\left(−\pi\right)={r}+\mid{r}^{\mathrm{2}} +{r}+\mathrm{1}\mid+{r}\mid{r}^{\mathrm{2}} +\mathrm{3}{r}+\mathrm{3}\mid}\\{{f}\left(\mathrm{0}\right)={r}+\mid{r}^{\mathrm{2}} −{r}+\mathrm{1}\mid+{r}\mid{r}^{\mathrm{2}} −\mathrm{3}{r}+\mathrm{3}\mid}\end{cases} \\ $$$$\mathrm{in}\:\mathrm{both}\:\mathrm{cases}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{is}\:\mathrm{at}\:{r}=\mathrm{0}\:\mathrm{because} \\ $$$${r}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{negative} \\ $$$$\Rightarrow\:{z}=−\mathrm{1}\:\mathrm{seems}\:\mathrm{to}\:\mathrm{be}\:\mathrm{the}\:\mathrm{absolute}\:\mathrm{minimum} \\ $$$$\mathrm{with}\:\mid{z}+\mathrm{1}\mid+\mid{z}^{\mathrm{2}} +{z}+\mathrm{1}\mid+\mid{z}^{\mathrm{3}} +\mathrm{1}\mid=\mathrm{1} \\ $$
Commented by Prithwish sen last updated on 10/Jul/19
Sir is′nt the ansolute value indicate the distance between z and the origin. If so then why it can′t be positive all the time ?
$$\mathrm{Sir}\:\mathrm{is}'\mathrm{nt}\:\mathrm{the}\:\mathrm{ansolute}\:\mathrm{value}\:\mathrm{indicate}\:\mathrm{the} \\ $$$$\mathrm{distance}\:\mathrm{between}\:\mathrm{z}\:\mathrm{and}\:\mathrm{the}\:\mathrm{origin}.\:\mathrm{If}\:\mathrm{so}\:\mathrm{then} \\ $$$$\mathrm{why}\:\mathrm{it}\:\mathrm{can}'\mathrm{t}\:\mathrm{be}\:\mathrm{positive}\:\mathrm{all}\:\mathrm{the}\:\mathrm{time}\:?\: \\ $$

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