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
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.
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
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.
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.