Let a complex number be w = 1 - \[{\sqrt 3 }\]i. Let another complex number z be such that |zw| = 1 and arg(z) - arg(w) = \[{\pi \over 2}\]. Then the area of the triangle with vertices origin, z and w is equal to :
Solutions · 1
Official solution
Given, w = 1 - \[\sqrt 3 \] i
\[ \Rightarrow |w| = \sqrt {{{(1)}^2} + {{( - \sqrt 3 )}^2}} = 2\]
Also, given, zw, = 1
\[ \Rightarrow \], z, | w, = 1 (using property)
\[ \Rightarrow \], z, = \[{1 \over 2}\]
Also, arg(z) - arg(w) = \[{{\pi {} } \over 2}\]
\[ \therefore \] Angle between two complex number z and w is \[{{\pi {} } \over 2}\].
\[ \therefore \] \[\angle zow = {{\pi {} } \over 2}\]
\[ \therefore \] \[\Delta\]zow is a right angle triangle with base ow=2 and height \[oz = {1 \over 2}\]
\[ \therefore \] Area = \[{1 \over 2} \times 2 \times {1 \over 2} = {1 \over 2}\]