(A) Let $z = x + iy$. Substituting this into the equation: $|(x+1) + i(y-1)| = |(x-1) + i(y+1)|$. Squaring both sides: $(x+1)^2 + (y-1)^2 = (x-1)^2 +

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Straight line passing through the origin and the first and third quadrant.

(A) Let $z = x + iy$. Substituting this into the equation: $|(x+1) + i(y-1)| = |(x-1) + i(y+1)|$. Squaring both sides: $(x+1)^2 + (y-1)^2 = (x-1)^2 + (y+1)^2$. Expanding the terms: $x^2 + 2x + 1 + y^2 - 2y + 1 = x^2 - 2x + 1 + y^2 + 2y + 1$. Simplifying: $2x - 2y = -2x + 2y$. $4x = 4y$,which implies $y = x$. The equation $y = x$ represents a straight line passing through the origin and the first and third quadrants.

Class 11 Mathematics 4-1.Complex numbers Geometry of complex numbers

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The equation $|z+1-i|=|z-1+i|$ represents a (where $z$ is a complex number)

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Straight line passing through the origin and the first and third quadrant.
B
Straight line passing through the origin and the second and fourth quadrant.
C
Straight line passing through the point $(1, -1)$ and having slope $-1$.
D
Straight line passing through the point $(2, 1)$ and having slope $\frac{1}{2}$.

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The equation $|z+1-i|=|z-1+i|$ represents a (where $z$ is a complex number)

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Let a complex number be $w = 1 - \sqrt{3} i$. Let another complex number $z$ be such that $|zw| = 1$ and $\arg(z) - \arg(w) = \frac{\pi}{2}$. Then the area of the triangle with vertices at the origin,$z$,and $w$ is equal to ........ .

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