Tag: triangle inequality

given  $|Z _1|=|Z _2|=|Z _3|=1$     and     $Z _1+Z _2+Z _3=0$
$\Rightarrow |Z _1 -Z _2|=2 (Cos 30)=\sqrt{3}$
$\Rightarrow  area =\frac{\sqrt{3}}{4}a^2$    &     $ a=\sqrt{3}$
So, area $=\frac{\sqrt{3}}{4}\cdot 3 =\frac{3\sqrt{3}}{4}$

$|z _1+z _2+z _3| = |z _1-a+z _2-b+z _3-c+(a+b+c)|$
$\leq |z _1-a|+|z _2-b|+|z _3-c|+|a+b+c|$
$\leq 2|a+b+c| $
Hence, $|z _1+z _2+z _3|$ is less than $2|a+b+c|$.
Statement 1 is false and Statement 2 is true

$z^n cos\theta _0+z^{n-1} cos\theta _1+.....+z cos\theta _{n-1}+cos\theta _n=2$

We have for any two complex numbers $\displaystyle \alpha$ and $\displaystyle \beta$
$\displaystyle ||\alpha|| \leq |\alpha - \beta|$
Now $\displaystyle ||Z|-|\frac {4}{|Z|}||\leq|Z-\frac {4}{Z}|$
$\displaystyle \Rightarrow |Z| - \frac {4}{|Z|}|\leq 2$
Set $\displaystyle |Z| = r > 0$, then $\displaystyle |r-\frac {4}{r}|\leq 2$
$\displaystyle \Rightarrow -2 \leq r - \frac {4}{r} \leq 2$
The left inequality gives
$\displaystyle r^2 + 2r - 4 \geq 0$
The corresponding roots are
$\displaystyle r = \frac {-2\pm \sqrt {20}}{2} = -1 \pm \sqrt 5$
Thus $\displaystyle r \geq \sqrt 5 - 1$ or $\displaystyle r \leq -1 - \sqrt 5$
implies that $\displaystyle r \geq \sqrt 5 - 1$ (As r > 0) ... (i)
Again consider the right inequality
$\displaystyle r - \frac {4}{r} \leq 2 \Rightarrow r^2 - 2 r - 4 \leq 0$
The corresponding roots are
$\displaystyle r = \frac {2 \pm \sqrt {20}}{2} = 1 \pm \sqrt 5$
Thus $\displaystyle 1 - \sqrt 5 \leq r \leq 1 + \sqrt 5$
But r > 0, hence $\displaystyle r \leq 1 + \sqrt 5$ .... (ii)
(i) and (ii) gives
$\displaystyle \sqrt 5 - 1 \leq r \leq \sqrt 5 + 1$
So, the greatest value is $\displaystyle \sqrt 5 + 1$.

We have $\displaystyle \left| z \right| =\left| z+\frac { 2 }{ z } -\frac { 2 }{ z }  \right| \le \left| z+\frac { 2 }{ z }  \right| +\frac { 2 }{ \left| z \right|  } $

$\left| { z } _{ 1 }-1 \right| \le 1,\quad \left| { z } _{ 2 }-2 \right| \le 2,\quad \left| { z } _{ 3 }-3 \right| \le 3...\left| { z } _{ n }-n \right| \le n$
Adding these and using triangle inequality:
$\left| { z } _{ 1 }+{ z } _{ 2 }+...{ z } _{ n }-(1+2+...n) \right| \le 1+2+...n\quad =>\quad \left| { z } _{ 1 }+{ z } _{ 2 }+...{ z } _{ n }-\left(\dfrac { n(n+1) }{ 2 } \right) \right| \le \dfrac { n(n+1) }{ 2 } $
Thus, $\left| { z } _{ 1 }+{ z } _{ 2 }+...{ z } _{ n } \right| -\left(\dfrac { n(n+1) }{ 2 } \right)\le \dfrac { n(n+1) }{ 2 } \quad =>\quad \left| { z } _{ 1 }+{ z } _{ 2 }+...{ z } _{ n } \right| \le n(n+1)$
Hence, (c) is correct.

$|1+z+z^{2}+..z^{n}|\leq 1+|z|+|z|^{2}+..|z|^{n}$
$\leq \dfrac{|z|^{n+1}-1}{|z|-1}$

$\leq\dfrac{(|z|.|z|^{n}-1)}{|z|-1}$

$\leq\dfrac{|z|}{|z|-1}.[|z|^{n}-\dfrac{1}{|z|}]$
Hence 
It cannot be less than $[|z|^{n}-\dfrac{1}{|z|}]$.

Given: