Precalculus

Plotting polar points

A sonar screen shows each contact as a distance and a direction from the ship at its center. How do you name a point that way, and why does the same point end up with many names?

Where you are 3 in Polar coordinates

When you reach for it

Yuina watches a sonar screen on a ship. The ship sits at the center, rings mark distances 1 km apart, and a contact shows up as a blip: so many kilometers out, in some direction. To report it as (x, y) she would need how far east and how far north it is, which the screen does not show. What it shows directly is a distance and a direction.

Polar coordinates name points exactly that way: a direction, given as an angle, and a distance along it. Reach for them whenever a picture is built around one center: a sonar screen, an air traffic control display, a wheel, and later whole curves whose polar equations are short. This lesson names and plots single points; turning a polar name into x and y is the next step, Converting between polar and rectangular.

The procedure

The pole, the polar axis and the grid

Fix a point O, called the pole, and a ray starting at O, called the polar axis. On paper the pole is placed where the origin would be and the polar axis along the positive x-axis, pointing right. Every direction out of the pole is then an angle in standard position: the turn from the polar axis, counterclockwise when the angle is positive and clockwise when it is negative. Angles here are in radians unless a degree sign is written (Angles and their measure).

A polar grid draws the two kinds of line this system needs: circles centered at the pole, one unit apart, and rays out of the pole at evenly spaced angles. Distance from the pole is read off the circles; direction is read off the rays.

0
\displaystyle \frac{\pi}{6}
\displaystyle \frac{\pi}{3}
\displaystyle \frac{\pi}{2}
\displaystyle \frac{2\pi}{3}
\displaystyle \frac{5\pi}{6}
\pi
\displaystyle \frac{7\pi}{6}
\displaystyle \frac{4\pi}{3}
\displaystyle \frac{3\pi}{2}
\displaystyle \frac{5\pi}{3}
\displaystyle \frac{11\pi}{6}
O
1 2 3 4
Figure 1A polar grid. Distance from the pole O is read off the circles, direction off the rays. The thick ray is the polar axis.

Polar coordinates: turn, then walk

A point P has polar coordinates (r, \theta) when you reach it from the pole by turning to face the ray at angle \theta and then walking r units along that ray. The angle \theta is the angular coordinate and r is the radial coordinate. The pair is written with r first, even though the turn happens first.

To plot \displaystyle \left(3, \frac{2\pi}{3}\right): the angle \displaystyle \frac{2\pi}{3} is two thirds of a half turn, that is 120^\circ, so the ray points up and to the left into quadrant II. Walk out along it to the third circle and mark the point.

\displaystyle \left(3, \frac{2\pi}{3}\right)
Figure 2Plotting \displaystyle \left(3, \frac{2\pi}{3}\right). Turn through \displaystyle \frac{2\pi}{3}, then walk out to the third ring.

Only the direction you end up facing matters. A negative \theta turns clockwise, a \theta above 2\pi turns more than once around, and coterminal angles end on the same ray: \displaystyle \frac{5\pi}{2} = 2\pi + \frac{\pi}{2} faces straight up, as \displaystyle \frac{\pi}{2} does.

A negative r: walk backward

The radial coordinate is a signed distance, like a number on a number line, where a negative number means “the other way”. When r < 0, face the ray at \theta exactly as before, then walk |r| units backward: through the pole and out the far side.

Take \displaystyle \left(-3, \frac{\pi}{3}\right). Face \displaystyle \frac{\pi}{3}, up and to the right, then walk 3 units backward. You go down and to the left, into quadrant III, and stop on the third circle. The ray directly opposite the one at \theta is a half turn further round, at \theta + \pi, so the point lies on the ray \displaystyle \frac{\pi}{3} + \pi = \frac{4\pi}{3}, three units out.

\displaystyle \theta = \frac{\pi}{3}
\displaystyle \left(-3, \frac{\pi}{3}\right)
Figure 3Plotting \displaystyle \left(-3, \frac{\pi}{3}\right). Face \displaystyle \frac{\pi}{3} (dashed), then walk 3 units backward. The point is the same as \displaystyle \left(3, \frac{4\pi}{3}\right).

The same reasoning works for every r and \theta: walking backward along a ray is walking forward along the opposite ray. So

(-r, \theta) \text{ and } (r, \theta + \pi) \text{ name the same point.}

Every name of a point

In rectangular coordinates every point has exactly one name. In polar coordinates every point has infinitely many, because two changes never move the point. Adding a full turn 2\pi to \theta leaves you facing the same ray. Changing the sign of r and adding a half turn \pi to \theta swaps walking forward for walking backward along the opposite ray. Doing either any number of times, for any integer k:

\begin{aligned} (r, \theta) &= (r, \theta + 2\pi k) \\ (r, \theta) &= (-r, \theta + \pi + 2\pi k) \end{aligned}

For example, \displaystyle \left(2, \frac{\pi}{6}\right) is also \displaystyle \left(2, \frac{13\pi}{6}\right), \displaystyle \left(2, -\frac{11\pi}{6}\right), \displaystyle \left(-2, \frac{7\pi}{6}\right) and \displaystyle \left(-2, -\frac{5\pi}{6}\right), among others.

These two families are all the names, for any point other than the pole. Let P be at distance d > 0 from the pole. It lies on exactly one ray out of the pole; call that ray’s angle \varphi. Any name (r, \theta) of P walks |r| units from the pole, so |r| = d, which leaves two cases. If r = d, the walk is forward along the ray at \theta, and it reaches P only if that ray is the ray through P, so \theta is coterminal with \varphi: \theta = \varphi + 2\pi k. If r = -d, the walk is backward, so it is the ray opposite \theta, at \theta + \pi, that must pass through P: \theta + \pi = \varphi + 2\pi k, so \theta = \varphi - \pi + 2\pi k, which is the same list as \varphi + \pi + 2\pi k with k shifted by one. Every name is therefore (d, \varphi + 2\pi k) or (-d, \varphi + \pi + 2\pi k).

This gives a test. Two names (r_1, \theta_1) and (r_2, \theta_2) with r_1 \ne 0 are the same point exactly when r_2 = r_1 and \theta_2 - \theta_1 is an even multiple of \pi, or r_2 = -r_1 and \theta_2 - \theta_1 is an odd multiple of \pi. If |r_2| \ne |r_1|, never.

The pole. When r = 0 you walk nowhere, whichever way you face. So (0, \theta) is the pole for every \theta, not only for angles that differ by multiples of \pi. The pole is the one point with no direction, and the argument above, which started from the one ray through P, does not apply to it.

One preferred name. Among all the names of a point other than the pole, exactly one has r > 0 and 0 \le \theta < 2\pi: the r must be d, and exactly one angle coterminal with \varphi lies in a window 2\pi wide. That is the name to report when one name is wanted.

Worked examples

Example 1

Plotting with positive r

Plot \displaystyle A = \left(4, \frac{7\pi}{6}\right), \displaystyle B = \left(2, -\frac{\pi}{3}\right) and \displaystyle C = \left(3, \frac{7\pi}{3}\right) on a polar grid, and say which quadrant each lies in.

Solution

A: \displaystyle \frac{7\pi}{6} = \pi + \frac{\pi}{6}, a half turn and then \displaystyle \frac{\pi}{6} more. A half turn faces left, and the extra counterclockwise \displaystyle \frac{\pi}{6} tips the ray below the horizontal, so it points down and to the left. Walk 4 units out, to the fourth circle. A is in quadrant III.

B: the angle is negative, so turn clockwise \displaystyle \frac{\pi}{3}, which is 60^\circ below the polar axis. Walk 2 units. B is in quadrant IV, on the ray also named \displaystyle -\frac{\pi}{3} + 2\pi = \frac{5\pi}{3}.

C: \displaystyle \frac{7\pi}{3} = 2\pi + \frac{\pi}{3}. A full turn brings you back to facing the polar axis, and the remaining \displaystyle \frac{\pi}{3} leaves you on the ray at \displaystyle \frac{\pi}{3}. Walk 3 units. C is in quadrant I, and it is the same point as \displaystyle \left(3, \frac{\pi}{3}\right).

Example 2

Plotting with negative r

Plot \displaystyle D = \left(-2, \frac{3\pi}{4}\right) and \displaystyle E = \left(-4, -\frac{\pi}{2}\right), and give each a name with positive r.

Solution

D: face \displaystyle \frac{3\pi}{4}, which is in quadrant II, halfway between straight up and straight left. Walk 2 units backward, through the pole. You end on the opposite ray, \displaystyle \frac{3\pi}{4} + \pi = \frac{7\pi}{4}, in quadrant IV, halfway between straight down and the polar axis. So \displaystyle D = \left(2, \frac{7\pi}{4}\right).

E: \displaystyle -\frac{\pi}{2} is a clockwise quarter turn, so you face straight down. Walking 4 units backward takes you straight up 4 units. So \displaystyle E = \left(4, \frac{\pi}{2}\right), using \displaystyle -\frac{\pi}{2} + \pi = \frac{\pi}{2}. The point lies on the vertical line through the pole, above it, so it is in no quadrant.

Example 3

Three more names for one point

Give a name for \displaystyle P = \left(4, \frac{\pi}{6}\right) that has (a) r > 0 and -2\pi \le \theta < 0; (b) r < 0 and 0 \le \theta < 2\pi; (c) r < 0 and -2\pi \le \theta < 0.

Solution

(a) Keep r = 4 and take away a full turn: \displaystyle \frac{\pi}{6} - 2\pi = -\frac{11\pi}{6}, which lies in the window. The name is \displaystyle \left(4, -\frac{11\pi}{6}\right).

(b) Change the sign of r and add a half turn: \displaystyle \frac{\pi}{6} + \pi = \frac{7\pi}{6}. The name is \displaystyle \left(-4, \frac{7\pi}{6}\right): face down and to the left, walk backward up and to the right.

(c) Take a full turn away from the angle in (b): \displaystyle \frac{7\pi}{6} - 2\pi = -\frac{5\pi}{6}. The name is \displaystyle \left(-4, -\frac{5\pi}{6}\right).

Example 4

Same point or not

Which of these name the same point as \displaystyle \left(5, \frac{2\pi}{3}\right)?

(a) \displaystyle \left(5, -\frac{4\pi}{3}\right) \qquad (b) \displaystyle \left(-5, \frac{5\pi}{3}\right) \qquad (c) \displaystyle \left(-5, -\frac{\pi}{3}\right) \qquad (d) \displaystyle \left(5, -\frac{2\pi}{3}\right)

Solution

Use the test. With the same r the angles must differ by an even multiple of \pi; with the opposite r, by an odd multiple of \pi.

(a) Same r. \displaystyle -\frac{4\pi}{3} - \frac{2\pi}{3} = -2\pi, an even multiple of \pi. Same point.

(b) Opposite r. \displaystyle \frac{5\pi}{3} - \frac{2\pi}{3} = \pi, an odd multiple. Same point.

(c) Opposite r. \displaystyle -\frac{\pi}{3} - \frac{2\pi}{3} = -\pi, an odd multiple. Same point.

(d) Same r. \displaystyle -\frac{2\pi}{3} - \frac{2\pi}{3} = -\frac{4\pi}{3}, not a multiple of \pi at all. A different point: it is in quadrant III, while \displaystyle \left(5, \frac{2\pi}{3}\right) is in quadrant II.

Example 5

Two contacts on the sonar screen

On Yuina’s sonar screen the ship is at the pole, the rings are 1 km apart, and angles are measured counterclockwise from the polar axis. Contact A is at \displaystyle \left(3, \frac{5\pi}{4}\right). Riko logs contact B as \displaystyle \left(-2, \frac{5\pi}{4}\right). Where is B? Give it a name with positive r, and find how far apart the two contacts are.

Solution

A is 3 km out on the ray at \displaystyle \frac{5\pi}{4}, down and to the left. For B, face the same way and walk 2 km backward: B is on the opposite ray, \displaystyle \frac{5\pi}{4} - \pi = \frac{\pi}{4}, up and to the right. So \displaystyle B = \left(2, \frac{\pi}{4}\right). (Adding \pi gives \displaystyle \frac{9\pi}{4}, coterminal with \displaystyle \frac{\pi}{4}.)

A
B
Figure 4Contacts A and B in Example 5. They lie on opposite rays, so the segment between them passes through the ship.

The rays at \displaystyle \frac{5\pi}{4} and \displaystyle \frac{\pi}{4} point in opposite directions, so together they form one straight line through the ship. A and B sit on that line on opposite sides of the pole, so the distance between them is the sum of their distances from the pole:

AB = 3 + 2 = 5 \text{ km}

Had Riko logged B as \displaystyle \left(2, \frac{5\pi}{4}\right) instead, both contacts would sit on the same ray, and they would be 3 - 2 = 1 km apart.

Example 6

An air traffic control display

Lan reads an air traffic control display. The radar station is at the center, and each aircraft is shown by its range in miles and its bearing: the angle in degrees measured clockwise from north An aircraft is at range 40 miles, bearing 150^\circ. Taking the polar axis to point east, give its polar coordinates with \theta in degrees and 0^\circ \le \theta < 360^\circ. Then give a name with negative r.

Solution

A bearing starts from north and turns clockwise; a polar angle starts from east and turns counterclockwise. As in Bearings and navigation, north is the polar direction 90^\circ. A clockwise turn of 150^\circ from there is a polar turn of -150^\circ, so

\theta = 90^\circ - 150^\circ = -60^\circ

which is coterminal with -60^\circ + 360^\circ = 300^\circ. The aircraft is at (40, 300^\circ). Check: a bearing of 150^\circ lies between east and south, and the ray at 300^\circ points down and to the right.

For a name with negative r, add a half turn: 300^\circ + 180^\circ = 480^\circ, coterminal with 120^\circ. So the aircraft is also (-40, 120^\circ): face up and to the left, walk 40 miles backward.

Example 7

Every name in a window

List every name of \displaystyle Q = \left(-3, \frac{11\pi}{6}\right) with -2\pi \le \theta < 2\pi.

Solution

First find the ray Q lies on. It is 3 units out, on the ray opposite \displaystyle \frac{11\pi}{6}, which is \displaystyle \frac{11\pi}{6} - \pi = \frac{5\pi}{6}. So \displaystyle Q = \left(3, \frac{5\pi}{6}\right), in quadrant II.

Names with r = 3 have \displaystyle \theta = \frac{5\pi}{6} + 2\pi k. In the window that gives \displaystyle \frac{5\pi}{6} (k = 0) and \displaystyle \frac{5\pi}{6} - 2\pi = -\frac{7\pi}{6} (k = -1); every other k lands outside.

Names with r = -3 have \displaystyle \theta = \frac{11\pi}{6} + 2\pi k. In the window: \displaystyle \frac{11\pi}{6} and \displaystyle \frac{11\pi}{6} - 2\pi = -\frac{\pi}{6}.

The window is two full turns wide, so each family gives two names:

\left(3, \tfrac{5\pi}{6}\right),\ \left(3, -\tfrac{7\pi}{6}\right),\ \left(-3, \tfrac{11\pi}{6}\right),\ \left(-3, -\tfrac{\pi}{6}\right)

Boundaries

The pole has no direction. (0, \theta) is the pole for every \theta. The two families (r, \theta + 2\pi k) and (-r, \theta + \pi + 2\pi k) list every name only when r \ne 0, and the preferred name with r > 0 and 0 \le \theta < 2\pi exists for every point except the pole.

The coordinates mean nothing without the conventions. A polar name needs a pole, a polar axis, a positive direction of turning and a unit of angle. Bearings start at north and turn clockwise, so a bearing is not a polar angle until it is converted (Example 6). The angle unit must be stated: (3, 2) means a turn of 2 radians, about 115^\circ, not 2^\circ.

Negative r is a convention, not a measurement. A range on a sonar screen is never negative. Mathematics allows r < 0 because polar equations such as r = 2\sin\theta give negative r for some angles, and those points must mean something; walking backward is the meaning chosen.

Where this goes next

The right triangle under the walk links a polar name to a rectangular one: Converting between polar and rectangular gives x = r\cos\theta and y = r\sin\theta.

Once single points can be plotted, an equation in r and \theta draws a curve, one point per angle. That is how Polar circles and symmetry tests and Limacons and cardioids are drawn, and the many names of a point are why a polar curve can pass through it at several angles. A complex number is a point too, and its polar form names it by distance and direction, the angle again fixed only up to 2\pi k.

Exercises

Plotting

For each point, give the ray it lies on as an angle \theta with 0 \le \theta < 2\pi, its distance from the pole, and its quadrant, or where it lies if it is on a line through the pole rather than in a quadrant.

  1. \displaystyle \left(4, \frac{5\pi}{3}\right)
  2. \displaystyle \left(1, \frac{3\pi}{2}\right)
  3. \displaystyle \left(3, -\frac{5\pi}{6}\right)
  4. \displaystyle \left(2, \frac{11\pi}{4}\right)
  5. \left(2.5, \pi\right)

Negative r

Same instructions.

  1. \displaystyle \left(-1, \frac{\pi}{6}\right)
  2. \displaystyle \left(-4, \frac{2\pi}{3}\right)
  3. \displaystyle \left(-3, -\frac{3\pi}{4}\right)
  4. \left(-2, \pi\right)

Other names

  1. Give a name for \displaystyle \left(5, \frac{\pi}{4}\right) with (a) r > 0 and -2\pi \le \theta < 0; (b) r < 0 and 0 \le \theta < 2\pi.
  2. Give the name of \displaystyle \left(-2, \frac{4\pi}{3}\right) with r > 0 and 0 \le \theta < 2\pi.
  3. Give the name of \displaystyle \left(3, \frac{19\pi}{6}\right) with r > 0 and 0 \le \theta < 2\pi.
  4. Give the name of \displaystyle \left(-6, -\frac{\pi}{2}\right) with r > 0 and 0 \le \theta < 2\pi.
  5. List every name of \displaystyle \left(1, \frac{2\pi}{3}\right) with -2\pi \le \theta < 2\pi.

Same point or not

Decide whether the two names are the same point, and say why.

  1. \displaystyle \left(4, \frac{\pi}{5}\right) and \displaystyle \left(4, \frac{11\pi}{5}\right)
  2. \displaystyle \left(-3, \frac{\pi}{8}\right) and \displaystyle \left(3, \frac{9\pi}{8}\right)
  3. \displaystyle \left(2, \frac{\pi}{3}\right) and \displaystyle \left(-2, -\frac{\pi}{3}\right)
  4. \displaystyle \left(0, \frac{\pi}{4}\right) and \displaystyle \left(0, \frac{3\pi}{2}\right)

Screens

  1. On Yuina’s sonar screen (rings 1 km apart, angles counterclockwise from the polar axis), contact C is at \displaystyle \left(3, \frac{7\pi}{4}\right) and contact D is logged as \displaystyle \left(-1, \frac{3\pi}{4}\right). Where is D, and how far apart are C and D?
  2. On Lan’s air traffic control display an aircraft is at range 25 miles, bearing 300^\circ (clockwise from north). With the polar axis pointing east, give its polar coordinates with 0^\circ \le \theta < 360^\circ, and a second name with negative r.
  3. An aircraft at (35, 100^\circ) on the same display flies in a straight line directly over the radar station and keeps going until it is 15 miles past it. Give its new position with positive r, and with negative r and \theta = 100^\circ.

Reasoning

  1. Using turning and walking, explain why (-r, \theta) and (r, \theta + \pi) are the same point.
  2. Riko says that (-r, \theta) is the mirror image of (r, \theta) across the line of the polar axis. Is she right? Give an example, and say where (-r, \theta) really is.
  3. Show that if r \ne 0 and (r, \theta) and (r, \varphi) are the same point, then \theta - \varphi is a multiple of 2\pi. Why does this fail when r = 0?
  4. Explain why every point other than the pole has exactly one name with r > 0 and 0 \le \theta < 2\pi, and why the pole has none.
Answers
  1. Ray \displaystyle \theta = \frac{5\pi}{3}, 4 units out, quadrant IV. The angle is \displaystyle \frac{\pi}{3} short of a full turn.
  2. Ray \displaystyle \theta = \frac{3\pi}{2}, 1 unit out, straight below the pole. Three quarter turns counterclockwise face straight down.
  3. Ray \displaystyle \theta = \frac{7\pi}{6}, 3 units out, quadrant III. Turning clockwise \displaystyle \frac{5\pi}{6} is coterminal with \displaystyle -\frac{5\pi}{6} + 2\pi = \frac{7\pi}{6}.
  4. Ray \displaystyle \theta = \frac{3\pi}{4}, 2 units out, quadrant II. \displaystyle \frac{11\pi}{4} - 2\pi = \frac{3\pi}{4}.
  5. Ray \theta = \pi, 2.5 units out, straight left of the pole on the line of the polar axis.
  6. Ray \displaystyle \theta = \frac{7\pi}{6}, 1 unit out, quadrant III. Walking backward lands on \displaystyle \frac{\pi}{6} + \pi.
  7. Ray \displaystyle \theta = \frac{5\pi}{3}, 4 units out, quadrant IV. \displaystyle \frac{2\pi}{3} + \pi = \frac{5\pi}{3}.
  8. Ray \displaystyle \theta = \frac{\pi}{4}, 3 units out, quadrant I. \displaystyle -\frac{3\pi}{4} + \pi = \frac{\pi}{4}.
  9. Ray \theta = 0, 2 units out, on the polar axis. Face left, walk 2 backward, to the right.
  10. (a) \displaystyle \left(5, -\frac{7\pi}{4}\right), since \displaystyle \frac{\pi}{4} - 2\pi = -\frac{7\pi}{4}. (b) \displaystyle \left(-5, \frac{5\pi}{4}\right), since \displaystyle \frac{\pi}{4} + \pi = \frac{5\pi}{4}.
  11. \displaystyle \left(2, \frac{\pi}{3}\right). Change the sign of r and subtract a half turn: \displaystyle \frac{4\pi}{3} - \pi = \frac{\pi}{3}.
  12. \displaystyle \left(3, \frac{7\pi}{6}\right). Subtract a full turn: \displaystyle \frac{19\pi}{6} - 2\pi = \frac{7\pi}{6}.
  13. \displaystyle \left(6, \frac{\pi}{2}\right). \displaystyle -\frac{\pi}{2} + \pi = \frac{\pi}{2}: face down, walk backward up.
  14. \displaystyle \left(1, \frac{2\pi}{3}\right), \displaystyle \left(1, -\frac{4\pi}{3}\right), \displaystyle \left(-1, \frac{5\pi}{3}\right), \displaystyle \left(-1, -\frac{\pi}{3}\right). The angles are \displaystyle \frac{2\pi}{3} + 2\pi k for r = 1 and \displaystyle \frac{5\pi}{3} + 2\pi k for r = -1.
  15. Same point. Same r, and \displaystyle \frac{11\pi}{5} - \frac{\pi}{5} = 2\pi.
  16. Same point. Opposite r, and \displaystyle \frac{9\pi}{8} - \frac{\pi}{8} = \pi, an odd multiple of \pi.
  17. Not the same point. Opposite r, but \displaystyle -\frac{\pi}{3} - \frac{\pi}{3} = -\frac{2\pi}{3} is not an odd multiple of \pi. In fact \displaystyle \left(-2, -\frac{\pi}{3}\right) = \left(2, \frac{2\pi}{3}\right), in quadrant II, while \displaystyle \left(2, \frac{\pi}{3}\right) is in quadrant I.
  18. Same point: both are the pole, since r = 0 walks nowhere whatever the angle.
  19. \displaystyle D = \left(1, \frac{7\pi}{4}\right), on the same ray as C, since \displaystyle \frac{3\pi}{4} + \pi = \frac{7\pi}{4}. Both are on one ray, so they are 3 - 1 = 2 km apart.
  20. (25, 150^\circ) and (-25, 330^\circ). \theta = 90^\circ - 300^\circ = -210^\circ, coterminal with 150^\circ. For negative r, 150^\circ + 180^\circ = 330^\circ.
  21. (15, 280^\circ), or (-15, 100^\circ). Passing over the station puts the aircraft on the opposite ray, 100^\circ + 180^\circ = 280^\circ.
  22. Walking |r| units backward along the ray at \theta is walking |r| units forward along the opposite ray, which is a half turn round, at \theta + \pi. That is exactly the walk for (r, \theta + \pi). (If r < 0 both walks reverse and still agree; if r = 0 both are the pole.)
  23. No. The mirror image of \displaystyle \left(2, \frac{\pi}{6}\right) across the polar axis is \displaystyle \left(2, -\frac{\pi}{6}\right), on the ray \displaystyle \frac{11\pi}{6}, but \displaystyle \left(-2, \frac{\pi}{6}\right) is on the ray \displaystyle \frac{7\pi}{6}. The point (-r, \theta) is the same distance out on the opposite ray, a half turn about the pole; the mirror image is (r, -\theta). They agree only on the vertical line through the pole.
  24. Both walks go |r| > 0 units the same way (forward or backward), so they end at the same point only if they face the same ray, that is, if \theta and \varphi are coterminal: \theta - \varphi = 2\pi k. When r = 0 there is no walk, so every angle gives the pole.
  25. A point other than the pole lies d > 0 out on exactly one ray. A name with r > 0 must have r = d and an angle coterminal with that ray’s angle, and such angles are spaced 2\pi apart, so exactly one lies in 0 \le \theta < 2\pi. The pole is 0 out, so no name of it has r > 0.
kg1128 · updated 2026-10-20 · CC-BY-SA-4.0