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Find the area (in sq. units) of the region satisfying both arg(z - i + 2) = pi/6 and arg(z + 4 - 3i) = -pi/4, where z = x + iy is a complex number.

1. 0
2. pi/4
3. 1
4. The two rays do not intersect

Correct answer: 0

Solution

arg(z - i + 2) = pi/6 means the ray from point (-2, 1) at angle pi/6 (30 deg) above horizontal: parameterically z = -2 + r*cos(pi/6) + i*(1 + r*sin(pi/6)) for r > 0, giving y - 1 = (1/sqrt(3))(x + 2) for x > -2. arg(z + 4 - 3i) = -pi/4 means the ray from point (-4, 3) at angle -pi/4 (45 deg below horizontal): y - 3 = -(x + 4) for x > -4, i.e., y = -x - 1. The question asks for the region satisfying BOTH conditions simultaneously. The intersection of two rays (each with zero area) is at most a single point. Area of a point = 0.

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