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QLDSpecialist MathematicsQuick questions

Unit 3: Mathematical induction, and further vectors, matrices and complex numbers

Quick questions on Regions and curves in the complex plane in QCE Specialist Mathematics Unit 3

10short Q&A pairs drawn directly from our worked dot-point answer. For full context and worked exam questions, read the parent dot-point page.

What is modulus as distance?
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For z=x+iyz = x + iy, the expression ∣z−z0∣|z - z_0| is the distance from the point zz to the fixed point z0z_0 on the plane. So
What are perpendicular bisector from equal distances?
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The condition ∣z−z1∣=∣z−z2∣|z - z_1| = |z - z_2| says zz is equidistant from z1z_1 and z2z_2. The set of such points is the perpendicular bisector of the segment joining z1z_1 and z2z_2, a straight line. The inequality ∣z−z1∣<∣z−z2∣|z - z_1| < |z - z_2| is the half-plane closer to z1z_1.
What are argument conditions give rays?
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The condition arg⁡(z−z0)=α\arg(z - z_0) = \alpha describes all points whose direction from z0z_0 is the fixed angle α\alpha. This is a ray (half-line) starting at z0z_0 (open endpoint, since z=z0z = z_0 has no defined argument) at angle α\alpha to the positive real direction. A condition like 0≤arg⁡(z−z0)≤π30 \leq \arg(z - z_0) \leq \tfrac{\pi}{3} is the wedge-shaped region between two rays.
What are boundaries?
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Strict inequalities (<<, >>) exclude the boundary, drawn as a dashed curve. Inclusive inequalities (≤\leq, ≥\geq) include the boundary, drawn solid. The marker checks both the correct shape and the correct boundary style, and whether the interior or exterior is shaded.
What are combining conditions?
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Two or more conditions joined by "and" give the intersection of the regions, the overlap. Sketch each region, then shade only the common part. This is where careful boundary handling matters most, since the answer is the set satisfying every condition at once.
What is interpret the first condition?
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∣z−2∣≤2|z - 2| \leq 2 is the closed disc centred at z0=2z_0 = 2 (the point (2,0)(2, 0)) with radius 22. The boundary circle (x−2)2+y2=4(x - 2)^2 + y^2 = 4 is solid because the inequality is inclusive.
What is interpret the second condition?
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With z=x+iyz = x + iy, Im⁡(z)=y\operatorname{Im}(z) = y, so y≥1y \geq 1 is the closed half-plane on and above the horizontal line y=1y = 1, drawn solid.
What is find the overlap?
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Shade the part of the disc lying on or above y=1y = 1. The line y=1y = 1 cuts the circle where (x−2)2+1=4(x - 2)^2 + 1 = 4, so (x−2)2=3(x - 2)^2 = 3 and x=2±3x = 2 \pm \sqrt 3.
What is describe the result?
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The region is the circular segment of the disc above the chord from (2−3, 1)(2 - \sqrt3,\, 1) to (2+3, 1)(2 + \sqrt3,\, 1), up to the top of the circle at (2,2)(2, 2). Both boundaries are solid since both inequalities are inclusive.
What is check a point?
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Take z=2+1.5iz = 2 + 1.5i: ∣z−2∣=1.5≤2|z - 2| = 1.5 \leq 2 holds, and Im⁡(z)=1.5≥1\operatorname{Im}(z) = 1.5 \geq 1 holds, so this interior point is correctly in the shaded region.

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