سوال ۲۷ کنکور ارشد علوم کامپیوتر ۱۴۰۲

The equation $|z - 1 - 3i| = |z - 1 + i|$ says that $z$ is equidistant from the two fixed points

$A = 1 + 3i = (1,\, 3), \qquad B = 1 - i = (1,\, -1).$

Geometrically the set of all such points is the perpendicular bisector of segment $AB$. Since $A$ and $B$ share the same real part and differ only in their imaginary part, the bisector is the horizontal line through the midpoint.

Algebraic check. Write $z = x + iy$. Squaring both moduli:

$(x-1)^2 + (y-3)^2 = (x-1)^2 + (y+1)^2.$

The $(x-1)^2$ terms cancel, leaving

$(y-3)^2 = (y+1)^2 \;\Longrightarrow\; y^2 - 6y + 9 = y^2 + 2y + 1 \;\Longrightarrow\; -8y + 8 = 0 \;\Longrightarrow\; y = 1.$

So the locus is the horizontal line $y = 1$, i.e.

$\{\, x + i \mid x \in \mathbb{R} \,\}.$

This matches the midpoint $\tfrac{A+B}{2} = 1 + i$ lying on it, confirming option 2.

Exam trick: $|z - a| = |z - b|$ is always the perpendicular bisector of segment $ab$. When $a$ and $b$ differ only in the imaginary part (same real part), the answer is instantly a horizontal line $y = \tfrac{\operatorname{Im}a + \operatorname{Im}b}{2}$ — here $\tfrac{3 + (-1)}{2} = 1$.