The number of solutions for the equations |z | vedclass

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(A) Let $z = x + iy$. From $|z - 1| = |z - 2|$,we have $|x + iy - 1| = |x + iy - 2|$. Squaring both sides: $(x - 1)^2 + y^2 = (x - 2)^2 + y^2$. $x^2 -

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(A) Let $z = x + iy$. From $|z - 1| = |z - 2|$,we have $|x + iy - 1| = |x + iy - 2|$. Squaring both sides: $(x - 1)^2 + y^2 = (x - 2)^2 + y^2$. $x^2 - 2x + 1 = x^2 - 4x + 4$,which simplifies to $2x = 3$,so $x = \frac{3}{2}$. From $|z - 1| = |z - i|$,we have $|x + iy - 1| = |x + iy - i|$. Squaring both sides: $(x - 1)^2 + y^2 = x^2 + (y - 1)^2$. $x^2 - 2x + 1 + y^2 = x^2 + y^2 - 2y + 1$,which simplifies to $-2x = -2y$,so $x = y$. Substituting $x = \frac{3}{2}$ into $x = y$,we get $y = \frac{3}{2}$. Thus,there is only one solution $z = \frac{3}{2} + i\frac{3}{2}$.

The number of solutions for the equations $|z - 1| = |z - 2| = |z - i|$ is

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