complex no

• Nov 9th 2011, 07:09 AM
prasum
complex no
if z1,z2,z3...zn lie on circle mod(z)=2 then the value of mod(z1+z2+...+zn)-4*mod(1/z1+1/z2+1/z3...+1/zn) is
• Nov 9th 2011, 07:26 AM
Amer
Re: complex no
is the question

$\mid z_1+z_2+...+z_n -4 \mid \cdot \mid \frac{1}{z_1} + \frac{1}{z_2} + ...+\frac{1}{z_n}\mid$

or

$\left(\mid z_1+z_2+...+z_n \mid -4 \right) \cdot \mid \frac{1}{z_1} + \frac{1}{z_2} + ...+\frac{1}{z_n}\mid$
• Nov 9th 2011, 08:16 AM
prasum
Re: complex no
• Nov 9th 2011, 09:02 AM
Plato
Re: complex no
Quote:

Originally Posted by prasum
if z1,z2,z3...zn lie on circle mod(z)=2 then the value of
|(z1+z2+...+zn)|-4|(1/z1+1/z2+1/z3...+1/zn)| is

$\left| {\sum\limits_{k = 1}^n {z_k } } \right| - 4\left| {\sum\limits_{k = 1}^n {\frac{1}{{z_k }}} } \right| = 0$
• Nov 9th 2011, 09:19 AM
Amer
Re: complex no
Quote:

Originally Posted by Plato
$\left| {\sum\limits_{k = 1}^n {z_k } } \right| - 4\left| {\sum\limits_{k = 1}^n {\frac{1}{{z_k }}} } \right| = 0$

I know that

$\mid z_1 + z_2 + ... +z_n \mid\;\; \leq\; \mid z_1 \mid\; + \mid z_2 \mid\; + ...+\mid z_n\mid = 2n$

$\left| \frac{1}{z_1} +\frac{1}{z_2} + ...+ \frac{1}{z_n} \right| \;\; \leq\; \left| \frac{1}{z_1}\right|\; + \;\left| \frac{1}{z_2} \right| \; + ...+\;\left| \frac{1}{z_n}\right| = \frac{n}{2}$

you consider the equally without the bigger why ?
• Nov 9th 2011, 09:26 AM
veileen
Re: complex no
Let $S=\sum_{k=1}^{n} z_k$.

$|z_k|=2\Rightarrow z_k\cdot \overline{z_k}=4\Leftrightarrow \frac{1}{z_k}=\frac{\overline{z_k}}{4}$

$4\left | \sum_{k=1}^{n} \frac{1}{z_k} \right |=\left | \sum_{k=1}^{n} \frac{4}{z_k} \right |$ $=\left | \sum_{k=1}^{n} \overline{z_k} \right |=\left | \overline{S}\right |=\left | S\right |$

* That inequalities aren't really useful, if you multiply by -4 the second one it becomes $-4\left |\sum_{k=1}^{n} \frac{1}{z_k} \right| \geq -2n$ (so you can't add them).