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# Test of Mathematics Solution Subjective 88 - Complex Numbers with a Property  This is a Test of Mathematics Solution Subjective 88 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

## Problem:

A pair of complex numbers $z_1, z_2$ is said to have the property $P$ if for every complex number $z$ we find real numbers $r$ and $s$ such that $z=rz_1 + sz_2$.Show that a pair of complex numbers has property $P$ if and only if the points $z_1,z_2$ and $0$ on the complex plane are not collinear.

## Solution:

Let the complex numbers $z_1,z_2,0$ be collinear, and the line joining them make an angle $\theta$ with the x-axis. This means that:

$arg(z_1) =arg(z_2) = \theta$

$=> z_1 = |z_1| (cos\,\theta + i sin\, \theta)$

Similarly,

$=> z_2 = |z_2| (cos\,\theta + i sin\, \theta)$

Therefore, $z=rz_1 + sz_2$

$=> z =r |z_1| (cos\,\theta + i sin\, \theta) + s|z_2| (cos\,\theta + i sin\, \theta)$

$=> z =(r |z_1| + s|z_2|) (cos\,\theta + i sin\, \theta)$

Which implies that $z$ lies on the same line that joins $z_1$ and $z_2$. But that is not true, as $z$ can be any complex number.

Thus the assumption that $z_1, z_2, 0$ are collinear is false.

Hence Proved. This is a Test of Mathematics Solution Subjective 88 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

## Problem:

A pair of complex numbers $z_1, z_2$ is said to have the property $P$ if for every complex number $z$ we find real numbers $r$ and $s$ such that $z=rz_1 + sz_2$.Show that a pair of complex numbers has property $P$ if and only if the points $z_1,z_2$ and $0$ on the complex plane are not collinear.

## Solution:

Let the complex numbers $z_1,z_2,0$ be collinear, and the line joining them make an angle $\theta$ with the x-axis. This means that:

$arg(z_1) =arg(z_2) = \theta$

$=> z_1 = |z_1| (cos\,\theta + i sin\, \theta)$

Similarly,

$=> z_2 = |z_2| (cos\,\theta + i sin\, \theta)$

Therefore, $z=rz_1 + sz_2$

$=> z =r |z_1| (cos\,\theta + i sin\, \theta) + s|z_2| (cos\,\theta + i sin\, \theta)$

$=> z =(r |z_1| + s|z_2|) (cos\,\theta + i sin\, \theta)$

Which implies that $z$ lies on the same line that joins $z_1$ and $z_2$. But that is not true, as $z$ can be any complex number.

Thus the assumption that $z_1, z_2, 0$ are collinear is false.

Hence Proved.

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