The area of the triangle formed by the comple | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $z = x + iy$. Then $iz = -y + ix$ and $z + iz = (x - y) + i(x + y)$.The vertices of the triangle are $(x, y)$,$(-y, x)$,and $(x - y, x + y)$.T

Vedclass · Accepted Answer

$\frac{1}{2} |z|^2$

Vedclass · Answer

(A) Let $z = x + iy$. Then $iz = -y + ix$ and $z + iz = (x - y) + i(x + y)$.The vertices of the triangle are $(x, y)$,$(-y, x)$,and $(x - y, x + y)$.The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Substituting the coordinates:Area $= \frac{1}{2} |x(x - (x + y)) + (-y)((x + y) - y) + (x - y)(y - x)|$Area $= \frac{1}{2} |x(-y) - y(x) + (x - y)(-(x - y))|$Area $= \frac{1}{2} |-xy - xy - (x - y)^2|$Area $= \frac{1}{2} |-2xy - (x^2 - 2xy + y^2)|$Area $= \frac{1}{2} |-2xy - x^2 + 2xy - y^2|$Area $= \frac{1}{2} |-x^2 - y^2| = \frac{1}{2} (x^2 + y^2) = \frac{1}{2} |z|^2$.

The area of the triangle formed by the complex numbers $z$,$iz$,and $z+iz$ as vertices in the Argand diagram is:

Similar Questions

$A$ circle whose radius is $r$ and centre is $z_0$,then the equation of the circle is

If ${z_1}, {z_2}, {z_3}, {z_4}$ are the affixes of four points in the Argand plane and $z$ is the affix of a point such that $|z - z_1| = |z - z_2| = |z - z_3| = |z - z_4|$,then ${z_1}, {z_2}, {z_3}, {z_4}$ are

Let $A = \{ z \in \mathbb{C} : |\frac{z+1}{z-1}| < 1 \}$ and $B = \{ z \in \mathbb{C} : \arg(\frac{z-1}{z+1}) = \frac{2\pi}{3} \}$. Then $A \cap B$ is

For any real number $r$, let $A_r = \{e^{i \pi r n} : n \in \mathbb{N}\}$ be a set of complex numbers. Then,

Let $z = x + iy$ be a point in the Argand plane. If the amplitude of $\left(\frac{z - 3}{z + 2i}\right)$ is $\frac{\pi}{2}$,then the locus of $z$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module