Rays and Regions on the Argand Diagram

A direction from a

z − a is the arrow from a to z, so arg(z − a) is the direction from a to z. Fixing it fixes a direction but leaves the distance free: arg$\left(z-a\right)=\theta$(z − a) = θ is the half-line that starts at a and heads off at angle $\theta$θ.

For arg(z − 1 − i) = 45°, a = 1 + i. The point 2 + 2i is on the half-line, because 2 + 2i − (1 + i) = 1 + i has argument 45°; so is 3 + 3i.

The point a itself is not on it. At z = 1 + i, z − a = 0, and 0 has no argument, because an arrow of length 0 points nowhere. The half-line is drawn with a hollow circle at its start to show that a is left out.

arg(z − 1 − i) = 45°: the half-line from 1 + i at 45°, through 2 + 2i and 3 + 3i. The hollow circle at 1 + i marks the start, which is not part of the set.

Only half of the line

The whole line through 1 + i at 45° also runs down and to the left, but those points fail. For z = 0, z − (1 + i) = −1 − i, which points down and to the left, at −135°. Every point on that half has argument −135° measured from 1 + i.

So arg(z − 1 − i) = 45° and arg(z − 1 − i) = −135° are two different half-lines from the same start, pointing opposite ways, and neither includes 1 + i.

With a = 0 the half-line starts at the origin: arg z = 90° is the positive imaginary axis, without 0. The point 3i is on it, and −3i, at −90°, is not.

Two half-lines from 1 + i. The one marked 45° is arg(z − 1 − i) = 45°. The one through 0 is arg(z − 1 − i) = −135°, a different set. Both start hollow at 1 + i.

From the rim to the disc

|z| = 3 is the circle of radius 3. Loosen it to $|z|\le 3$|z| ≤ 3, at most 3 from the origin, and every point inside counts as well as the rim: the locus is the whole disc.

Test points: 2 + 2i is about 2.83 from 0, so it is inside; 3i is exactly 3 from 0, on the rim, and $\le$≤ allows it; 3 + i is $\sqrt{10}\approx 3.16$√10 ≈ 3.16 from 0, so it is outside. The rim is drawn solid because it belongs to the region.

The region $|z|\le 3$|z| ≤ 3, shaded, with its rim solid. 2 + 2i is inside, 3i is on the rim and included, and 3 + i is outside.

A strict inequality leaves the rim out

|z − 2 − i| < 2 is every point less than 2 from 2 + i. The rim, exactly 2 away, is not included, so it is drawn dashed. The center 2 + i is in, and so is 3 + 2i, about 1.41 away. The point 4 + i is exactly 2 away, on the dashed rim, and is left out.

Reversing the sign shades the other side: |z − 2 − i| > 2 is everything outside the circle, again with a dashed rim, and $|z-2-i|\ge 2$|z − 2 − i| ≥ 2 is the outside with the rim included.

The region |z − 2 − i| < 2: the disc about 2 + i with its rim dashed. 2 + i and 3 + 2i are inside. 4 + i sits on the rim, drawn hollow, because it is exactly 2 away and so not in the region.

One side of a bisector

|z − 2i| = |z + 2i| is the perpendicular bisector of 2i and −2i, which is the real axis. $|z-2i|\le |z+2i|$|z − 2i| ≤ |z + 2i| asks for points at least as near 2i as −2i: the half of the plane on 2i’s side, and the bisector itself.

In Cartesian form, $x^2+{\left(y-2\right)}^2\le x^2+{\left(y+2\right)}^2$x² + (y − 2)² ≤ x² + (y + 2)². The $x^2$x² cancels, and so do the $y^2$y² and the 4, leaving $-4y\le 4y$−4y ≤ 4y, so $y\ge 0$y ≥ 0: everything on or above the real axis. The boundary is solid, since $\le$≤ includes it. Test 1 + i: it is $\sqrt{2}$√2 from 2i and $\sqrt{10}$√10 from −2i, so it is in the region.

The region $|z-2i|\le |z+2i|$|z − 2i| ≤ |z + 2i|: everything on or above the real axis, the solid boundary included. 1 + i is in it, nearer 2i than −2i.

A sector: a distance and a direction together

Two conditions together shade the points that pass both. $|z|\le 3$|z| ≤ 3 with $0\le$0 ≤ arg $z\le 60^{\circ }$z ≤ 60° is a slice of the disc: within 3 of the origin, and between the direction 0° and the direction 60°. The origin itself has no argument, so it is left out.

Test 2 + i: $|2+i|=\sqrt{5}\approx 2.24\le 3$|2 + i| = √5 ≈ 2.24 ≤ 3, and arg$\left(2+i\right)\approx 26.57^{\circ }$(2 + i) ≈ 26.57°, between 0° and 60°, so it is in the sector. Test 1 + 2i: $|1+2i|=\sqrt{5}$|1 + 2i| = √5 as well, but arg$\left(1+2i\right)\approx 63.43^{\circ }$(1 + 2i) ≈ 63.43°, past 60°, so it is outside. The area is $\frac{1}{2}{r}^2\theta$½r²θ with $\theta$θ in radians: $\frac{1}{2}\times 9\times \frac{\pi }{3}=\frac{3\pi }{2}\approx 4.71$½ × 9 × π/3 = 3π/2 ≈ 4.71.

The sector $|z|\le 3$|z| ≤ 3, $0\le$0 ≤ arg $z\le 60^{\circ }$z ≤ 60°. 2 + i, at about 26.57°, is inside. 1 + 2i is the same distance out but at about 63.43°, outside the slice. The origin is drawn hollow: it has no argument.

The usual mistakes

Drawing the whole line. arg(z − 2) = 60° is the half-line from 2; the points behind 2 are at −120° from it. For example, $3+\sqrt{3}i-2=1+\sqrt{3}i$3 + √3 i − 2 = 1 + √3 i is at 60°, but $1-\sqrt{3}i-2=-1-\sqrt{3}i$1 − √3 i − 2 = −1 − √3 i is at −120°.

Including the start. arg(z − 2) = 60° does not contain 2: there z − 2 = 0, which has no argument.

Measuring from the origin. arg(z − a) is the direction from a, so subtract a before finding the angle.

Shading only the rim, or the outside. $|z-i|\le 2$|z − i| ≤ 2 is the whole disc of radius 2 about i, rim included.

Drawing a strict boundary solid. |z| < 4 leaves out the points exactly 4 away, so its edge is dashed.

A searchlight on a pier

The application below lights the region $|z-1|\le 4$|z − 1| ≤ 4, $0\le$0 ≤ arg$\left(z-1\right)\le \frac{\pi }{4}$(z − 1) ≤ π/4: a sector like the one above, with its vertex moved to the searchlight at 1, radius 4, and angle $\frac{\pi }{4}$π/4, which is 45°.