question 63639 again prove: ∀z∈C: ∣z+1∣+∣z^2 +z+1∣+∣z^3 +1∣≥1


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Question Number 63784 by MJS last updated on 09/Jul/19

$question 63639 again$ $prove :$ $\forall z \in C : ∣ z + 1 ∣ + ∣ z^{2} + z + 1 ∣ + ∣ z^{3} + 1 ∣⩾ 1$
Commented by Prithwish sen last updated on 09/Jul/19

$1 =∣ 1 + z + 1 + z^{3} - (z^{3} + z^{2}) - (z^{2} + z + 1) ∣$ $⩽ ∣ 1 + z ∣ + ∣ 1 + z^{3} ∣ + ∣ z^{2} ∣∣ 1 + z ∣ + ∣ z^{2} + z + 1 ∣ (∵ ∣ z_{1} - z_{2} ∣ ⩽ ∣ z_{1} ∣ + ∣ z_{2} ∣$ $⩽ ∣ 1 + z ∣ + ∣ 1 + z^{3} ∣ + ∣ z^{2} + z + 1 ∣$ $= ∣ 1 + z ∣ + ∣ z^{2} + z + 1 ∣ + ∣ z^{3} + 1 ∣ (∵ addition$ $of complex numbers follow commutative law .$ $Hence proved .$
Commented by Prithwish sen last updated on 09/Jul/19

$please check it sir .$
Commented by Prithwish sen last updated on 09/Jul/19

$Is it ok sir ?$
Commented by mr W last updated on 09/Jul/19

$h o w c a n y o u s a y :$ $∣ 1 + z ∣ + ∣ 1 + z^{3} ∣ + ∣ z^{2} ∣∣ 1 + z ∣ + ∣ z^{2} + z + 1 ∣$ $⩽ ∣ 1 + z ∣ + ∣ 1 + z^{3} ∣ + ∣ z^{2} + z + 1 ∣$ $?$
Commented by MJS last updated on 10/Jul/19
$so how can we solve it? for z=−1 0+1+0≥1 for z=−(1/2)±((√3)/2)i 1+0+2≥1 for z=(1/2)±((√3)/2)i (√3)+2+0≥1 is it enough to show that for z close to −1 the sum of the absolutes is rising in every direction? we can put x=−1+re^(iθ) with −π≤θ<π and choose any small value for r>0 ⇒ ∣z+1∣+∣z^2 +z+1∣+∣z^3 +1∣=f(θ)= =r+(√(r^4 −2r^3 cos θ −(1−4cos^2 θ)r^2 −2rcos θ +1))+r(√(r^4 −6r^3 cos θ +3(1+4cos^2 θ)r^2 −18rcos θ +9)) (df/dθ)=rsin θ ×[some complicated term which has no zeros for small values of r] ⇒ our function has minima/maxima at θ=−π∨θ=0 ⇒ { ((f(−π)=r+∣r^2 +r+1∣+r∣r^2 +3r+3∣)),((f(0)=r+∣r^2 −r+1∣+r∣r^2 −3r+3∣)) :} in both cases the minimum is at r=0 because r cannot be negative ⇒ z=−1 seems to be the absolute minimum with ∣z+1∣+∣z^2 +z+1∣+∣z^3 +1∣=1$
$so how can we solve it ?$ $for z = - 1$ $0 + 1 + 0 ⩾ 1$ $for z = - \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$ $1 + 0 + 2 ⩾ 1$ $for z = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$ $\sqrt{3} + 2 + 0 ⩾ 1$ $is it enough to show that for z close to - 1$ $the sum of the absolutes is rising in every$ $direction ?$ $we can put x = - 1 + r e^{i θ} with - π ⩽ θ < π and$ $choose any small value for r > 0 \Rightarrow$ $∣ z + 1 ∣ + ∣ z^{2} + z + 1 ∣ + ∣ z^{3} + 1 ∣= f (θ) =$ $= r + \sqrt{r^{4} - 2 r^{3} \cos θ - (1 - {4 \cos}^{2} θ) r^{2} - 2 r \cos θ + 1} + r \sqrt{r^{4} - 6 r^{3} \cos θ + 3 (1 + {4 \cos}^{2} θ) r^{2} - 18 r \cos θ + 9}$ $\frac{d f}{d θ} = r \sin θ \times [s o m e c o m p l i c a t e d t e r m w h i c h h a s$ $n o z e r o s f o r s m a l l v a l u e s o f r]$ $\Rightarrow our function has minima / maxima at$ $θ = - π \lor θ = 0$ $\Rightarrow {\begin{cases} f (- π) = r + ∣ r^{2} + r + 1 ∣ + r ∣ r^{2} + 3 r + 3 ∣ \\ f (0) = r + ∣ r^{2} - r + 1 ∣ + r ∣ r^{2} - 3 r + 3 ∣ \end{cases}$ $in both cases the minimum is at r = 0 because$ $r cannot be negative$ $\Rightarrow z = - 1 seems to be the absolute minimum$ $with ∣ z + 1 ∣ + ∣ z^{2} + z + 1 ∣ + ∣ z^{3} + 1 ∣= 1$
Commented by Prithwish sen last updated on 10/Jul/19

$Sir {is}^{'} nt the ansolute value indicate the$ $distance between z and the origin . If so then$ $why it {can}^{'} t be positive all the time ?$
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