# Argand Plane?

• Apr 28th 2009, 11:10 PM
fardeen_gen
Argand Plane?
If $\displaystyle \sum_{r = 1}^{4} a_{r} = 0$ and $\displaystyle \sum_{r = 1}^{4} a_{r}Z_{r} = 0$ where $\displaystyle a_{1},a_{2},a_{3},a_{4}$ are non-zero real numbers and $\displaystyle P(Z_{1}),Q(Z_{2}),R(Z_{3}), S(Z_{4})$ are concyclic points on Argand Plane, then prove that:
$\displaystyle \sum_{r = 1}^{4} a_{r}|Z_{r}|^2 = 0.48$
• Apr 29th 2009, 02:42 AM
Opalg
Quote:

Originally Posted by fardeen_gen
If $\displaystyle \sum_{r = 1}^{4} a_{r} = 0$ and $\displaystyle \sum_{r = 1}^{4} a_{r}Z_{r} = 0$ where $\displaystyle a_{1},a_{2},a_{3},a_{4}$ are non-zero real numbers and $\displaystyle P(Z_{1}),Q(Z_{2}),R(Z_{3}), S(Z_{4})$ are concyclic points on Argand Plane, then prove that:
$\displaystyle \sum_{r = 1}^{4} a_{r}|Z_{r}|^2 = 0.48$

This cannot be correct. Suppose that $\displaystyle a_{1},a_{2},a_{3},a_{4}$ and $\displaystyle Z_{1},Z_{2},Z_{3},Z_{4}$ satisfy these conditions. If we replace $\displaystyle a_{1},a_{2},a_{3},a_{4}$ by $\displaystyle 2a_{1},2a_{2},2a_{3},2a_{4}$ then the conditions will still be satisfied, but $\displaystyle \sum_{r = 1}^{4} a_{r}|Z_{r}|^2$ will be doubled.
• Apr 29th 2009, 04:55 AM
fardeen_gen
The problem set of the complex numbers section of my text seems to be full of wrong problems. This makes it two wrong problems in a day (Worried)