The locus of the point z satisfying the equat | vedclass

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(A) Given equation: $|iz - 1| + |z - i| = 2$ $|i(z - 1/i)| + |z - i| = 2$ $|i(z + i)| + |z - i| = 2$ $|i| |z + i| + |z - i| = 2$ Since $|i| = 1$,we ha

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$A$ straight line

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(A) Given equation: $|iz - 1| + |z - i| = 2$ $|i(z - 1/i)| + |z - i| = 2$ $|i(z + i)| + |z - i| = 2$ $|i| |z + i| + |z - i| = 2$ Since $|i| = 1$,we have $|z - (-i)| + |z - i| = 2$ This is of the form $|z - z_1| + |z - z_2| = 2a$,where $z_1 = -i$ and $z_2 = i$. The distance between $z_1$ and $z_2$ is $|z_2 - z_1| = |i - (-i)| = |2i| = 2$. Since the sum of the distances from $z$ to two fixed points $z_1$ and $z_2$ is equal to the distance between the points themselves $(|z - z_1| + |z - z_2| = |z_1 - z_2|)$,the locus of $z$ is the line segment joining $z_1$ and $z_2$. Thus,the locus is a straight line.

The locus of the point $z$ satisfying the equation $|iz - 1| + |z - i| = 2$ is

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