Precalculus · N.CN.B.4

Modulus, argument and polar form

A nonzero complex number is fixed by how far it sits from 0 and in which direction. Can the number be written with those two facts instead of its two parts, and how do you pass between the two ways of writing it?

You need

Converting between polar and rectangularPrecalculus Complex numbersGrades 10 to 12

Unlocks

Multiplying, powers and De Moivre's theoremPrecalculus

Where you are 3 in Polar form of a complex number

Trigonometry7

Polar coordinates and vectors10

Polar form of a complex number3

Modulus, argument and polar form

Multiplying, powers and De Moivre's theorem Roots of a complex number and roots of unity

Why this exists

The form a + bi says where a complex number sits by going across and then up. That suits adding, which works part by part, but it hides the two facts that the complex plane makes visible: how far the number is from 0, its modulus, and which way it points, its argument. For points in the plane this choice is old news. A point can be named by (x, y) or by a distance and an angle, and converting between polar and rectangular passes from one to the other with cosine and sine.

Complex numbers get the same second name. Writing z by its length and angle costs nothing when adding, but it turns multiplying, powers and roots from long algebra into short arithmetic. This lesson sets up that form and the conversions in both directions.

Definition

Let z = a + bi be a nonzero complex number, plotted at the point (a, b). Its modulus is \displaystyle r = |z| = \sqrt{a^2 + b^2}, the distance from 0. An argument of z is any angle \theta in standard position whose terminal side runs from 0 through z. Both were introduced in plotting complex numbers.

The point (a, b) lies at distance r along the terminal side of \theta, so by the definition of cosine and sine, a = r\cos\theta and b = r\sin\theta. Putting these into a + bi:

z = r(\cos\theta + i\sin\theta)

This is the polar form, also called the trigonometric form, of z. The bracket is often shortened to \operatorname{cis}\theta (for “cosine, i sine”), so z = r\operatorname{cis}\theta.

Many arguments, one principal argument. Turning a further whole turn either way reaches the same ray, so if \theta is an argument of z, so is \theta + 2\pi k for every integer k, and these are all of them. To name just one, this lesson follows the usual convention: the principal argument \operatorname{Arg} z is the one argument in the interval (-\pi, \pi], that is, -\pi < \operatorname{Arg} z \le \pi. Numbers above the real axis get a positive principal argument and numbers below it a negative one. Plotting complex numbers named arguments in [0, 2\pi) instead; both ranges name the same direction, and a stated range is what makes the answer unique.

Zero. 0 = 0(\cos\theta + i\sin\theta) for every \theta, so 0 has modulus 0 and no argument at all.

Intuition

The bracket \cos\theta + i\sin\theta is the point at angle \theta on the circle of radius 1 centered at 0: it carries the direction and has length 1. Multiplying it by the real number r > 0 stretches it out along the same ray to length r. So polar form reads as “go r units in direction \theta” (Figure 1).

The two forms of a number are two descriptions of one point. Rectangular form gives the legs of a right triangle, polar form gives its long side and its angle, and trigonometry passes between them. Since cosine and sine carry signs, the legs r\cos\theta and r\sin\theta come out with the signs of the quadrant automatically when going from polar to rectangular. Going the other way, the signs of a and b must be used to fix the quadrant, because a tangent alone cannot.

Results

From polar to rectangular. If z = r(\cos\theta + i\sin\theta), then

a = r\cos\theta, \qquad b = r\sin\theta

From rectangular to polar. If z = a + bi \ne 0, then \displaystyle r = \sqrt{a^2 + b^2}, and \theta is any angle with

\cos\theta = \frac{a}{r}, \qquad \sin\theta = \frac{b}{r}

The principal argument by position. With \displaystyle \alpha = \arctan\frac{b}{a} when a \ne 0:

Where z = a + bi sits	\operatorname{Arg} z
a > 0: quadrant I or IV, or the positive real axis	\alpha
a < 0,\ b \ge 0: quadrant II or the negative real axis	\alpha + \pi
a < 0,\ b < 0: quadrant III	\alpha - \pi
a = 0,\ b > 0	\displaystyle \frac{\pi}{2}
a = 0,\ b < 0	\displaystyle -\frac{\pi}{2}

When two polar forms are equal. For r_1, r_2 > 0, r_1\operatorname{cis}\theta_1 = r_2\operatorname{cis}\theta_2 exactly when r_1 = r_2 and \theta_1 - \theta_2 = 2\pi k for some integer k.

Why it is true

The conversions. On the terminal side of \theta, the point at distance r from the origin has coordinates (r\cos\theta, r\sin\theta); that is how cosine and sine are defined for any angle. The point of z is at distance r = |z| on that side, so its coordinates are a = r\cos\theta and b = r\sin\theta, which gives both directions at once. Dividing by r gives \displaystyle \cos\theta = \frac{a}{r} and \displaystyle \sin\theta = \frac{b}{r}, and the two together fix the ray: cosine gives the left or right side, sine the upper or lower.

The table. The arctangent returns an angle in \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right), a ray pointing right (the inverse tangent is cut down this way so that it has an inverse). Its line through 0 has slope \displaystyle \frac{b}{a}, the same as the line through z, so z is on that line. When a > 0, z is on the right-pointing ray itself, and \alpha already lies in (-\pi, \pi]. When a < 0, z is on the opposite ray, half a turn away, so the argument is \alpha \pm \pi. If b \ge 0 then \displaystyle -\frac{\pi}{2} < \alpha \le 0 and \alpha + \pi lies in \displaystyle \left(\frac{\pi}{2}, \pi\right]; if b < 0 then \displaystyle 0 < \alpha < \frac{\pi}{2} and \alpha - \pi lies in \displaystyle \left(-\pi, -\frac{\pi}{2}\right). Each choice lands inside (-\pi, \pi]. The axis cases are read off the plane.

Equality. The modulus is the distance from 0, which both forms must share, so r_1 = r_2. Then \cos\theta_1 = \cos\theta_2 and \sin\theta_1 = \sin\theta_2, so both angles have the same terminal side, and angles share a terminal side exactly when they differ by whole turns.

Worked examples

Example 1

From polar form to a + bi

Write each number in the form a + bi: (a) \displaystyle 6\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right) (b) \displaystyle 10\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right).

Solution

(a) The angle \displaystyle \frac{5\pi}{6} is in quadrant II with reference angle \displaystyle \frac{\pi}{6}, so \displaystyle \cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2} and \displaystyle \sin\frac{5\pi}{6} = \frac{1}{2}. Multiplying by r = 6:

6\left(-\tfrac{\sqrt{3}}{2}\right) + 6\left(\tfrac{1}{2}\right)i = -3\sqrt{3} + 3i

Figure 1 shows the point: 6 units out, \displaystyle \frac{5\pi}{6} around, with legs \displaystyle 3\sqrt{3} to the left and 3 up.

Figure 1Example 1. The modulus 6 and the angle \tfrac{5\pi}{6} fix the point; 6\cos\tfrac{5\pi}{6} and 6\sin\tfrac{5\pi}{6} are its two signed legs.

(b) The angle \displaystyle -\frac{\pi}{3} turns clockwise into quadrant IV, where cosine is positive and sine negative: \displaystyle \cos\left(-\frac{\pi}{3}\right) = \frac{1}{2} and \displaystyle \sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}. So the number is \displaystyle 5 - 5\sqrt{3}\,i.

Check (a): \displaystyle \left|-3\sqrt{3} + 3i\right| = \sqrt{27 + 9} = \sqrt{36} = 6, the modulus we started with.

Example 2

From a + bi to polar form, special angles

Write 6 - 6i and \displaystyle -\sqrt{3} - i in polar form with the principal argument.

Solution

For z = 6 - 6i: \displaystyle r = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}. Then \displaystyle \cos\theta = \frac{6}{6\sqrt{2}} = \frac{\sqrt{2}}{2} and \displaystyle \sin\theta = -\frac{\sqrt{2}}{2}: cosine positive, sine negative, quadrant IV, reference angle \displaystyle \frac{\pi}{4}. The table agrees, since a > 0 gives \displaystyle \operatorname{Arg} z = \arctan(-1) = -\frac{\pi}{4}. So

6 - 6i = 6\sqrt{2}\operatorname{cis}\left(-\tfrac{\pi}{4}\right)

For \displaystyle w = -\sqrt{3} - i: \displaystyle r = \sqrt{3 + 1} = 2, and \displaystyle \cos\theta = -\frac{\sqrt{3}}{2}, \displaystyle \sin\theta = -\frac{1}{2}, quadrant III with reference angle \displaystyle \frac{\pi}{6}. The arctangent gives \displaystyle \arctan\frac{-1}{-\sqrt{3}} = \frac{\pi}{6}, a ray into quadrant I; a < 0 and b < 0, so subtract \pi:

\begin{aligned} \operatorname{Arg} w &= \tfrac{\pi}{6} - \pi = -\tfrac{5\pi}{6} \\ w &= 2\operatorname{cis}\left(-\tfrac{5\pi}{6}\right) \end{aligned}

Figure 2Example 2. The principal argument of \displaystyle -\sqrt{3} - i turns clockwise, so it is negative. Turning counterclockwise instead gives \tfrac{7\pi}{6}, another argument of the same number.

Figure 2 shows the clockwise turn. Turning counterclockwise instead gives \displaystyle \frac{7\pi}{6}, and \displaystyle 2\operatorname{cis}\frac{7\pi}{6} is the same number, but \displaystyle \frac{7\pi}{6} is outside (-\pi, \pi].

Example 3

Numbers on the axes, and zero

Write -8 and -2i in polar form with the principal argument, and say what can be said of 0.

Solution

-8 is 8 units from 0 on the negative real axis, the direction \pi. Since \pi is in (-\pi, \pi] (the right end is included), \operatorname{Arg}(-8) = \pi and -8 = 8(\cos\pi + i\sin\pi). Check: 8(-1 + 0i) = -8. The angle -\pi points the same way but is excluded, which is why the interval is closed on one side only: the negative real axis needs exactly one name.

-2i is 2 units straight down, so \displaystyle \operatorname{Arg}(-2i) = -\frac{\pi}{2} and \displaystyle -2i = 2\left(\cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right)\right); check: 2(0 - i) = -2i. No arctangent is used here, since \displaystyle \frac{b}{a} would divide by 0.

0 has modulus 0, and 0(\cos\theta + i\sin\theta) = 0 for every \theta, so no angle is singled out: 0 has no argument.

Example 4

Arguments that are not special angles

Write -24 + 7i and -12 - 5i in polar form, the principal argument in radians to the nearest hundredth.

Solution

For z = -24 + 7i: \displaystyle r = \sqrt{576 + 49} = \sqrt{625} = 25. The arctangent gives \displaystyle \arctan\frac{7}{-24} \approx -0.2838, a ray pointing right and down, but a < 0 and b > 0 put z in quadrant II, so add \pi:

\begin{aligned} \operatorname{Arg} z &\approx -0.2838 + \pi \approx 2.86 \\ z &\approx 25\operatorname{cis} 2.86 \end{aligned}

For w = -12 - 5i: \displaystyle r = \sqrt{144 + 25} = 13. Now \displaystyle \arctan\frac{-5}{-12} \approx 0.3948, and a < 0, b < 0 put w in quadrant III, so subtract \pi:

\begin{aligned} \operatorname{Arg} w &\approx 0.3948 - \pi \approx -2.75 \\ w &\approx 13\operatorname{cis}(-2.75) \end{aligned}

Both numbers sit just off the negative real axis, one above and one below. Their principal arguments are close to \pi and to -\pi, on opposite ends of the interval.

Check for z: 25\cos 2.86 \approx -24.0 and 25\sin 2.86 \approx 6.9, close to -24 and 7 (the gap is the rounding of the angle).

Example 5

Many names for one number

Find the principal argument of \displaystyle z = 2\left(\cos\frac{11\pi}{4} + i\sin\frac{11\pi}{4}\right), list all its arguments, and write z as a + bi.

Solution

The angle \displaystyle \frac{11\pi}{4} is more than a full turn. Subtracting one turn, \displaystyle \frac{11\pi}{4} - 2\pi = \frac{3\pi}{4}, which lies in (-\pi, \pi], so \displaystyle \operatorname{Arg} z = \frac{3\pi}{4}. Every argument differs from this one by whole turns: \displaystyle \frac{3\pi}{4} + 2\pi k for any integer k (the given \displaystyle \frac{11\pi}{4} is k = 1). Then

z = 2\left(-\tfrac{\sqrt{2}}{2}\right) + 2\left(\tfrac{\sqrt{2}}{2}\right)i = -\sqrt{2} + \sqrt{2}\,i

Example 6

Not yet in polar form

Zikora meets \displaystyle -3\left(\cos\frac{\pi}{5} + i\sin\frac{\pi}{5}\right) and \displaystyle 5\left(\cos\frac{\pi}{7} - i\sin\frac{\pi}{7}\right). Neither is in polar form. Rewrite each with a positive modulus and the principal argument.

Solution

(a) Polar form needs r > 0, and here the front number is -3. A minus sign is a half turn: \cos(\theta + \pi) = -\cos\theta and \sin(\theta + \pi) = -\sin\theta, so -(\cos\theta + i\sin\theta) = \cos(\theta + \pi) + i\sin(\theta + \pi). With \displaystyle \theta = \frac{\pi}{5}:

-3\operatorname{cis}\tfrac{\pi}{5} = 3\operatorname{cis}\tfrac{6\pi}{5} = 3\operatorname{cis}\left(-\tfrac{4\pi}{5}\right)

since \displaystyle \frac{6\pi}{5} - 2\pi = -\frac{4\pi}{5} is the argument in (-\pi, \pi]. The modulus is 3, not -3: a distance is never negative.

(b) Polar form needs a plus sign between the terms. Cosine is even and sine is odd, so \displaystyle \cos\left(-\frac{\pi}{7}\right) = \cos\frac{\pi}{7} and \displaystyle \sin\left(-\frac{\pi}{7}\right) = -\sin\frac{\pi}{7}. Therefore

5\left(\cos\tfrac{\pi}{7} - i\sin\tfrac{\pi}{7}\right) = 5\operatorname{cis}\left(-\tfrac{\pi}{7}\right)

and \displaystyle -\frac{\pi}{7} is already the principal argument.

Example 7

A number fixed by two conditions

Find every complex number z with |z| = 2 and imaginary part 1, in both forms.

Solution

Write z = 2(\cos\theta + i\sin\theta), which builds in |z| = 2. Its imaginary part is 2\sin\theta, so 2\sin\theta = 1 and \displaystyle \sin\theta = \frac{1}{2}. In (-\pi, \pi] this has two solutions, \displaystyle \theta = \frac{\pi}{6} and \displaystyle \theta = \frac{5\pi}{6}. So

\begin{aligned} z &= 2\operatorname{cis}\tfrac{\pi}{6} = \sqrt{3} + i \\ z &= 2\operatorname{cis}\tfrac{5\pi}{6} = -\sqrt{3} + i \end{aligned}

Figure 3 shows why there are two: the circle |z| = 2 meets the horizontal line of imaginary part 1 twice.

Figure 3Example 7. Modulus 2 puts z on the circle; imaginary part 1 puts it on the line. They meet twice.

Check in rectangular form: with z = a + i, |z| = 2 gives a^2 + 1 = 4, so \displaystyle a = \pm\sqrt{3}.

Boundaries

0 has no polar form of its own. Its modulus is 0 and every \theta works, so it has no argument, and statements about arguments are for z \ne 0 only.

The modulus must be positive. In r(\cos\theta + i\sin\theta), r > 0. An expression with a negative front number is a complex number, but not written in polar form; a half turn fixes it (Example 6). Polar coordinates of points allow a negative r; polar form of a complex number does not.

An argument is never unique. Only a stated range makes it so. \operatorname{Arg} z means the one in (-\pi, \pi]; some books use [0, 2\pi), and the two disagree for every number below the real axis. Two answers that differ by 2\pi k name the same number.

\displaystyle \arctan\frac{b}{a} alone is the principal argument only when a > 0. For a < 0 it is off by \pi, and for a = 0 it does not exist.

The pattern must be exactly cosine plus i sine of the same angle. Expressions such as \cos\theta - i\sin\theta, \sin\theta + i\cos\theta or \cos 2\theta + i\sin\theta have to be rewritten first, or converted through a + bi.

Where this goes next

Polar form earns its keep in multiplying, powers and De Moivre’s theorem: multiplying two numbers in polar form multiplies their moduli and adds their arguments, which makes a high power a one-line computation. The fact that an argument is only fixed up to 2\pi k, a nuisance here, becomes the key in roots of a complex number, where those extra turns produce all n of the nth roots. Both sit under polar form of a complex number.

Exercises

From polar form to a + bi

Write \displaystyle 4\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) as a + bi.
Write \displaystyle 8\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right) as a + bi.
Write \displaystyle 3\left(\cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right)\right) as a + bi.
Write \displaystyle 12\operatorname{cis}\frac{4\pi}{3} as a + bi.
Write 5(\cos 1.2 + i\sin 1.2) as a + bi, each part to the nearest hundredth.

From a + bi to polar form, exact

Give the modulus exactly and the principal argument exactly.

2 + 2i
\displaystyle -3 + 3\sqrt{3}\,i
\displaystyle \sqrt{6} - \sqrt{2}\,i
-10
-11i
-4 - 4i

From a + bi to polar form, approximate

Give the modulus and the principal argument in radians to the nearest hundredth.

20 - 21i
-9 + 40i
-3 - 2i

Arguments and rewriting

Find the principal argument of \displaystyle 1 - \sqrt{3}\,i and list all of its arguments.
Find the principal argument of \displaystyle 7\left(\cos\frac{13\pi}{5} + i\sin\frac{13\pi}{5}\right).
Find the principal argument of \displaystyle 3\operatorname{cis}\left(-\frac{9\pi}{4}\right).
Write \displaystyle -2\left(\cos\frac{\pi}{9} + i\sin\frac{\pi}{9}\right) in polar form with the principal argument.
Write \displaystyle 6\left(\cos\frac{2\pi}{5} - i\sin\frac{2\pi}{5}\right) in polar form with the principal argument.
Are \displaystyle 4\operatorname{cis}\left(-\frac{\pi}{6}\right) and \displaystyle 4\operatorname{cis}\frac{23\pi}{6} the same number? Explain.

Using the form

The number z has modulus 25, real part -15, and lies above the real axis. Find z as a + bi and its principal argument to the nearest hundredth.
Find every z with |z| = 4 and real part -2, in both forms.

Reasoning

Amaka gives \displaystyle \operatorname{Arg}(-1 - i) = \frac{5\pi}{4}. Is \displaystyle \frac{5\pi}{4} an argument of -1 - i? Is it the principal argument? Explain.
Explain why every nonzero complex number has exactly one principal argument, and why 0 has none.
Chibuike claims that if z = r\operatorname{cis}\theta, then its conjugate is \overline{z} = r\operatorname{cis}(-\theta). Prove it, and decide whether \operatorname{Arg}\overline{z} = -\operatorname{Arg} z for every nonzero z.

Answers

\displaystyle 2 + 2\sqrt{3}\,i. \displaystyle 4 \cdot \frac{1}{2} = 2 and \displaystyle 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}.
\displaystyle -4\sqrt{2} + 4\sqrt{2}\,i. \displaystyle \cos\frac{3\pi}{4} = -\frac{\sqrt{2}}{2}, \displaystyle \sin\frac{3\pi}{4} = \frac{\sqrt{2}}{2}.
-3i. \displaystyle \cos\left(-\frac{\pi}{2}\right) = 0 and \displaystyle \sin\left(-\frac{\pi}{2}\right) = -1.
\displaystyle -6 - 6\sqrt{3}\,i. \displaystyle \frac{4\pi}{3} is in quadrant III: \displaystyle \cos = -\frac{1}{2}, \displaystyle \sin = -\frac{\sqrt{3}}{2}.
1.81 + 4.66i. 5\cos 1.2 \approx 1.812 and 5\sin 1.2 \approx 4.660 (calculator in radians).
\displaystyle 2\sqrt{2}\operatorname{cis}\frac{\pi}{4}. \displaystyle r = \sqrt{8}; a > 0, so \displaystyle \arctan 1 = \frac{\pi}{4}.
\displaystyle 6\operatorname{cis}\frac{2\pi}{3}. \displaystyle r = \sqrt{9 + 27} = 6; \displaystyle \arctan(-\sqrt{3}) = -\frac{\pi}{3} and a < 0, b > 0, so add \pi.
\displaystyle 2\sqrt{2}\operatorname{cis}\left(-\frac{\pi}{6}\right). \displaystyle r = \sqrt{6 + 2} = 2\sqrt{2}; \displaystyle \frac{b}{a} = -\frac{\sqrt{2}}{\sqrt{6}} = -\frac{1}{\sqrt{3}} and a > 0.
10\operatorname{cis}\pi. On the negative real axis; \pi is in (-\pi, \pi].
\displaystyle 11\operatorname{cis}\left(-\frac{\pi}{2}\right). Straight down from 0.
\displaystyle 4\sqrt{2}\operatorname{cis}\left(-\frac{3\pi}{4}\right). \displaystyle r = \sqrt{32}; \displaystyle \arctan 1 = \frac{\pi}{4}, quadrant III, so \displaystyle \frac{\pi}{4} - \pi.
29\operatorname{cis}(-0.81). \displaystyle r = \sqrt{400 + 441} = 29; a > 0, so \displaystyle \arctan\left(-\frac{21}{20}\right) \approx -0.8098.
41\operatorname{cis} 1.79. \displaystyle r = \sqrt{81 + 1600} = 41; \displaystyle \arctan\left(-\frac{40}{9}\right) + \pi \approx -1.3495 + 3.1416.
\displaystyle \sqrt{13} \approx 3.61 and \operatorname{Arg} \approx -2.55. Quadrant III: \displaystyle \arctan\frac{2}{3} - \pi \approx 0.5880 - 3.1416.
\displaystyle \operatorname{Arg} = -\frac{\pi}{3}; all arguments are \displaystyle -\frac{\pi}{3} + 2\pi k, k any integer. r = 2, \displaystyle \cos\theta = \frac{1}{2}, \displaystyle \sin\theta = -\frac{\sqrt{3}}{2}.
\displaystyle \frac{3\pi}{5}. \displaystyle \frac{13\pi}{5} - 2\pi = \frac{3\pi}{5}, inside (-\pi, \pi].
\displaystyle -\frac{\pi}{4}. \displaystyle -\frac{9\pi}{4} + 2\pi = -\frac{\pi}{4}.
\displaystyle 2\operatorname{cis}\left(-\frac{8\pi}{9}\right). The minus sign is a half turn: \displaystyle \frac{\pi}{9} + \pi = \frac{10\pi}{9}, and \displaystyle \frac{10\pi}{9} - 2\pi = -\frac{8\pi}{9}.
\displaystyle 6\operatorname{cis}\left(-\frac{2\pi}{5}\right). Cosine is even and sine is odd.
Yes. The moduli match and \displaystyle \frac{23\pi}{6} - \left(-\frac{\pi}{6}\right) = 4\pi, two whole turns.
z = -15 + 20i, \operatorname{Arg} z \approx 2.21. b^2 = 625 - 225 = 400 and b > 0, so b = 20; \displaystyle \arctan\left(-\frac{20}{15}\right) + \pi \approx -0.9273 + 3.1416.
\displaystyle -2 + 2\sqrt{3}\,i = 4\operatorname{cis}\frac{2\pi}{3} and \displaystyle -2 - 2\sqrt{3}\,i = 4\operatorname{cis}\left(-\frac{2\pi}{3}\right). 4\cos\theta = -2 gives \displaystyle \cos\theta = -\frac{1}{2}, so \displaystyle \theta = \pm\frac{2\pi}{3} in (-\pi, \pi]; check b^2 = 16 - 4 = 12.
It is an argument: \displaystyle \sqrt{2}\operatorname{cis}\frac{5\pi}{4} = -1 - i, since \displaystyle \frac{5\pi}{4} points into quadrant III at reference angle \displaystyle \frac{\pi}{4}. It is not the principal argument, because \displaystyle \frac{5\pi}{4} > \pi. Subtracting a turn gives \displaystyle \operatorname{Arg}(-1 - i) = -\frac{3\pi}{4}.
A nonzero z fixes one ray from 0, and the angles with that terminal side are \theta + 2\pi k, spaced exactly 2\pi apart. The interval (-\pi, \pi] has length 2\pi and includes exactly one of its two ends, so it contains exactly one of them. The point 0 lies on every ray, so no ray, and no angle, belongs to it.
Because cosine is even and sine is odd, r\operatorname{cis}(-\theta) = r\cos(-\theta) + ir\sin(-\theta), which equals r\cos\theta - ir\sin\theta, and that is a - bi = \overline{z}. Usually \operatorname{Arg}\overline{z} = -\operatorname{Arg} z, but not always: if z is a negative real number, \operatorname{Arg} z = \pi, while \overline{z} = z has \operatorname{Arg} = \pi too, and -\pi is outside (-\pi, \pi].