(B) Let $z = x + iy$. Then the equation becomes $|x + iy - i| = |x + iy - 1|$.This simplifies to $|x + i(y - 1)| = |(x - 1) + iy|$.Squaring both sides

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the line through the origin with slope $1$

(B) Let $z = x + iy$. Then the equation becomes $|x + iy - i| = |x + iy - 1|$.This simplifies to $|x + i(y - 1)| = |(x - 1) + iy|$.Squaring both sides,we get $x^2 + (y - 1)^2 = (x - 1)^2 + y^2$.Expanding the terms: $x^2 + y^2 - 2y + 1 = x^2 - 2x + 1 + y^2$.Canceling $x^2, y^2,$ and $1$ from both sides,we get $-2y = -2x$,which simplifies to $y = x$.This is the equation of a line passing through the origin with slope $1$.

Class 11 Mathematics 1 Complex Numbers and Quadratic Equations: Complex Numbers Geometry of complex numbers

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The equation $|z - i| = |z - 1|$,where $i = \sqrt{-1}$,represents:

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A
a circle of radius $\frac{1}{2}$
B
the line through the origin with slope $1$
C
a circle of radius $1$
D
the line through the origin with slope $-1$

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1 Complex Numbers and Quadratic Equations: Complex Numbers

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Geometry of complex numbers

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The equation $|z - i| = |z - 1|$,where $i = \sqrt{-1}$,represents:

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Let $S = \{z \in \mathbb{C} - \{i, 2i\} : \frac{z^2 + 8iz - 15}{z^2 - 3iz - 2} \in \mathbb{R} \}$. If $\alpha - \frac{13}{11}i \in S$ and $\alpha \in \mathbb{R} - \{0\}$,then $242\alpha^2$ is equal to

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