JEE MAIN - Mathematics (2021 - 18th March Evening Shift - No. 6)
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 :
4
$${1 \over 4}$$
2
$${1 \over 2}$$
Explanation
_18th_March_Evening_Shift_en_6_2.png)
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 = $${1 \over 2} \times 2 \times {1 \over 2} = {1 \over 2}$$
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