Argand Plane?

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April 28th 2009, 11:10 PM
fardeen_gen

Argand Plane?

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

Quote:

Originally Posted by fardeen_gen

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

This cannot be correct. Suppose that $a_{1},a_{2},a_{3},a_{4}$ and $Z_{1},Z_{2},Z_{3},Z_{4}$ satisfy these conditions. If we replace $a_{1},a_{2},a_{3},a_{4}$ by $2a_{1},2a_{2},2a_{3},2a_{4}$ then the conditions will still be satisfied, but $\sum_{r = 1}^{4} a_{r}|Z_{r}|^2$ will be doubled.
April 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)