question-63639-again-prove-z-C-z-1-z-2-z-1-z-3-1-1-

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 = - \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 ?