JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

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Application +4 | -1

The area of the triangle whose vertices are represented by the complex number

0,z,z{e^{i\alpha }},(0 < \mathrm{\alpha } < \pi )

equals

View answer and explanation

Correct option: B

\frac{1}{2}{\left| z \right|^2}\sin \mathrm{\alpha }

Option B (

\frac{1}{2}{\left| z \right|^2}\sin \mathrm{\alpha }

) is correct because the area of the triangle can be found using the determinant formula with the given vertices, which simplifies to the expression involving the sine of the angle.

More Questions from COMPLEX NUMBERS

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The number of non-zero integer solutions of the equation

{\left| {1 - i} \right|^x} = {2^x}

is If

|z|=1

and

z \neq \pm 1

, then for any complex number

z, \frac{z}{1-z^2}

Let

\left|z_r-r\right| \leq r, \forall r=1,2,3, \ldots, n

, then

\left|\sum\limits_{r=1}^n z_r\right|

is less than The roots of the cubic equation

(z+a b)^3=a^3

, where

a \neq 0

represents the vertices of a triangle. If

\alpha, \beta

and

\gamma

are the roots, then find the value of

|\alpha-\beta|

. Number of solutions of the equation

{z^2} + {\left| z \right|^2} = 0,

where

z \in C

is If

2x = - 1 + \sqrt 3 i

, then the value of

{\left( {1 - {x^2} + x} \right)^6} - {\left( {1 - x + {x^2}} \right)^6}

is equal to