Precalculus · N.CN.B.4 · N.CN.B.5

The complex plane, modulus and argument

A real number is a point on a line. A complex number has two parts. Where does it live, how big is it, and which way does it point?

You need

Inverse trig functionsPrecalculus Adding, subtracting and multiplying complex numbersAlgebra 2

Unlocks

Polar form of a complex numberGrades 11 to 12

Where you are 5 in Complex numbers

Number9

Complex numbers5

Adding, subtracting and multiplying complex numbers Conjugates and dividing complex numbers Imaginary and complex numbers Complex roots come in pairs

The complex plane, modulus and argument

When you reach for it

A real number is a point on a number line. A complex number a + bi in standard form carries two real numbers, its real part a and its imaginary part b, and one line cannot hold two independent numbers. A plane can. Once each complex number is a point, questions that sounded strange get plain answers. How big is -6 + 2.5i? It is a distance from 0. In which direction does it lie? It is an angle. How far apart are two complex numbers? It is the length of a segment.

You reach for the complex plane whenever size, direction or distance matter: to compare the sizes of complex numbers, to see what adding and subtracting does, to describe a set of complex numbers by a condition such as |z| = 3, and as the first step toward writing a complex number by its length and angle.

The procedure

Plotting. Draw a horizontal axis, the real axis, and a vertical axis, the imaginary axis. The number a + bi is plotted at the point (a, b): real part across, imaginary part up. Two complex numbers are equal exactly when both parts match, which is exactly when they are the same point, so every complex number gets one point and every point names one complex number. A real number a = a + 0i lands on the real axis, and the real axis is the old number line sitting inside the plane. A pure imaginary number bi lands on the imaginary axis.

Modulus. The modulus of z = a + bi is its distance from 0. The distance formula from (0, 0) to (a, b) gives

|z| = \sqrt{a^2 + b^2}

This is the magnitude of the vector \langle a, b \rangle, and Figure 2 shows it as the long side of a right triangle with legs |a| and |b|. For a real number, \displaystyle |a + 0i| = \sqrt{a^2}, which is the usual absolute value of a. That is why the same bars are used: the modulus extends absolute value from the line to the plane. It also matches the conjugate’s product, since z\overline{z} = a^2 + b^2 = |z|^2.

Argument. An argument of z \ne 0 is an angle in standard position whose terminal side runs from 0 through z: start on the positive real axis and turn counterclockwise until you reach the point. Turning a further whole number of turns either way reaches the same point, so the arguments of z are \theta + k \cdot 360^\circ for any integer k. This lesson names the one in [0^\circ, 360^\circ), or in [0, 2\pi) in radians. Some books choose (-180^\circ, 180^\circ] instead; it is the same direction under a different name.

Since (a, b) lies on the terminal side, \displaystyle \tan\theta = \frac{b}{a} when a \ne 0. The arctangent key cannot simply be trusted with this. Tangent repeats every 180^\circ, so to be inverted it was cut down to the slice (-90^\circ, 90^\circ), and the arctangent only ever names a ray pointing right. It always finds the correct line through 0; the sign of a says which ray of that line z is on. Writing \displaystyle \alpha = \arctan\frac{b}{a}:

Where z = a + bi sits	\arg z in [0^\circ, 360^\circ)
a > 0,\ b \ge 0: quadrant I or the positive real axis	\alpha
a < 0: quadrant II or III, or the negative real axis	\alpha + 180^\circ
a > 0,\ b < 0: quadrant IV	\alpha + 360^\circ
a = 0,\ b > 0: the positive imaginary axis	90^\circ
a = 0,\ b < 0: the negative imaginary axis	270^\circ

This is the same rule that gives a vector its direction angle in magnitude and direction, and the signs of a and b are the two signs that fix a quadrant.

The conjugate and the negative. The conjugate \overline{z} = a - bi keeps a and flips b, so it is the reflection of z in the real axis, pictured in dividing complex numbers. The negative -z = -a - bi flips both signs, a half turn about 0. Neither move changes the distance from 0.

Sums and differences. Adding works part by part: (a + bi) + (c + di) = (a + c) + (b + d)i. That is exactly how the vectors \langle a, b \rangle and \langle c, d \rangle add, so the sum is drawn by vector addition: the arrow to z + w is the diagonal of the parallelogram with sides the arrows to z and to w. For the difference, w + (z - w) = z, so z - w is the trip that starts at w and ends at z. Its length is the distance between the two points:

|z - w| = \sqrt{(a - c)^2 + (b - d)^2}

which is the distance formula written in complex numbers.

Worked examples

Example 1

Plotting six numbers

Plot 4 + 2i, -3 + 5i, -5 - 2i, 1 - 4i, -6 and 3i, and say where each lies.

Solution

Each number a + bi goes to the point (a, b) (Figure 1).

Figure 1Example 1. Real part across, imaginary part up. Real numbers sit on the real axis and pure imaginary numbers on the imaginary axis.

4 + 2i goes to (4, 2), quadrant I. -3 + 5i goes to (-3, 5), quadrant II. -5 - 2i goes to (-5, -2), quadrant III. 1 - 4i goes to (1, -4), quadrant IV. The real number -6 is -6 + 0i, the point (-6, 0) on the real axis. The pure imaginary number 3i is 0 + 3i, the point (0, 3) on the imaginary axis.

Example 2

Exact modulus and argument

Find the modulus and the argument, in degrees and in radians, of (a) -3 + 3i, (b) \displaystyle -2\sqrt{3} - 2i and (c) -7i.

Solution

(a) \displaystyle |-3 + 3i| = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}. The key gives \displaystyle \arctan\frac{3}{-3} = \arctan(-1) = -45^\circ, a ray pointing down and to the right. The number has a < 0, so it sits on the opposite ray, in quadrant II:

\arg z = -45^\circ + 180^\circ = 135^\circ = \tfrac{3\pi}{4}

(b) \displaystyle |-2\sqrt{3} - 2i| = \sqrt{12 + 4} = \sqrt{16} = 4. Here \displaystyle \frac{b}{a} = \frac{-2}{-2\sqrt{3}} = \frac{1}{\sqrt{3}} and the key gives 30^\circ, a ray into quadrant I. Both parts are negative, so the point is in quadrant III and

\arg z = 30^\circ + 180^\circ = 210^\circ = \tfrac{7\pi}{6}

(c) -7i = 0 - 7i sits on the imaginary axis, 7 below 0. Its modulus is \displaystyle \sqrt{0 + 49} = 7, and \displaystyle \frac{b}{a} does not exist, so read the direction straight off the plane: straight down, \displaystyle \arg z = 270^\circ = \frac{3\pi}{2}.

Check for (b): \displaystyle 4\cos 210^\circ = 4\left(-\frac{\sqrt{3}}{2}\right) = -2\sqrt{3} and \displaystyle 4\sin 210^\circ = 4\left(-\frac{1}{2}\right) = -2, the two parts.

Example 3

Arguments that are not special angles

Find the modulus and argument of -6 + 2.5i and of 5 - 11i, the argument to the nearest tenth of a degree.

Solution

For z = -6 + 2.5i (Figure 2):

|z| = \sqrt{36 + 6.25} = \sqrt{42.25} = 6.5

Figure 2Example 3. The modulus is the long side of a right triangle with legs 6 and 2.5. The argument is measured from the positive real axis, all the way around to the point.

The key gives \displaystyle \arctan\frac{2.5}{-6} \approx -22.6^\circ, but a < 0 puts z in quadrant II, so \arg z \approx -22.6^\circ + 180^\circ = 157.4^\circ. The figure agrees: the point is a little above the negative real axis, so the angle is a little less than 180^\circ.

For w = 5 - 11i, \displaystyle |w| = \sqrt{25 + 121} = \sqrt{146} \approx 12.08. Now a > 0 and b < 0, quadrant IV. The key’s ray \displaystyle \arctan\frac{-11}{5} \approx -65.6^\circ is the right ray already, but its angle is negative, and adding a full turn names it inside [0^\circ, 360^\circ): \arg w \approx -65.6^\circ + 360^\circ = 294.4^\circ.

Example 4

A number, its conjugate and their negatives

Let z = 2 + 5i. Plot z, \overline{z}, -z and -\overline{z}, and find the modulus and argument of each, the arguments to the nearest tenth of a degree.

Solution

\overline{z} = 2 - 5i, -z = -2 - 5i and -\overline{z} = -2 + 5i. Every one has parts \pm 2 and \pm 5, and squaring removes signs, so all four have modulus \displaystyle \sqrt{4 + 25} = \sqrt{29} \approx 5.39. They are the corners of a rectangle centered at 0 (Figure 3).

Figure 3Example 4. The conjugate reflects z in the real axis, -z is a half turn about 0, and -\overline{z} reflects z in the imaginary axis. All four are the same distance from 0.

z is in quadrant I, so \displaystyle \arg z = \arctan\frac{5}{2} \approx 68.2^\circ. The other three follow from the picture without another arctangent. The reflection \overline{z} makes the same angle below the positive real axis, so \arg\overline{z} \approx 360^\circ - 68.2^\circ = 291.8^\circ. The half turn -z adds 180^\circ: \arg(-z) \approx 248.2^\circ. And -\overline{z} makes 68.2^\circ with the negative real axis, on the upper side: \arg(-\overline{z}) \approx 180^\circ - 68.2^\circ = 111.8^\circ.

Check the last one with the table: \displaystyle \arctan\frac{5}{-2} \approx -68.2^\circ and a < 0, so -68.2^\circ + 180^\circ = 111.8^\circ.

Example 5

Adding and subtracting as arrows

For z = 1 + 3i and w = 5 + 2i, find and draw z + w and z - w, and find the distance between z and w.

Solution

z + w = (1 + 5) + (3 + 2)i = 6 + 5i. In Figure 4 the arrows from 0 to z and to w are two sides of a parallelogram, and 6 + 5i is its fourth corner: from z, move the way w moves, 5 right and 2 up.

Figure 4Example 5. The sum is the far corner of the parallelogram on z and w. The difference z - w is the arrow from w to z, so its length is the distance between them.

z - w = (1 - 5) + (3 - 2)i = -4 + i. Since w + (-4 + i) = z, the difference is the trip from w to z: 4 left and 1 up. Drawn from 0 it is the dashed arrow to -4 + i, parallel to the arrow from w to z and just as long. So the distance from w to z is

|z - w| = |-4 + i| = \sqrt{16 + 1} = \sqrt{17} \approx 4.12

Check with the distance formula on (1, 3) and (5, 2): \displaystyle \sqrt{(5 - 1)^2 + (2 - 3)^2} = \sqrt{17}. The other order, w - z = 4 - i, is the trip back, with the same length.

Example 6

Distance and midpoint

Find the distance between 9 - 2i and 1 + 4i, and the complex number halfway between them.

Solution

The distance is the modulus of the difference:

|(9 - 2i) - (1 + 4i)| = |8 - 6i| = \sqrt{64 + 36} = 10

The midpoint of (9, -2) and (1, 4) averages the coordinates, which here means averaging the two numbers:

\frac{(9 - 2i) + (1 + 4i)}{2} = \frac{10 + 2i}{2} = 5 + i

Check. |(9 - 2i) - (5 + i)| = |4 - 3i| = 5 and |(5 + i) - (1 + 4i)| = |4 - 3i| = 5: half the distance on each side.

Example 7

A set described by a modulus

Describe every complex number z with |z - (2 - i)| = 5. Then decide whether each of 6 + 2i, 5 - 5i, 3 + 2i and -2 + 3i lies on the set, inside it or outside it.

Solution

|z - (2 - i)| is the distance from z to the point 2 - i, so the condition says: z is exactly 5 from 2 - i. Those points make the circle with center (2, -1) and radius 5 (Figure 5). In coordinates, with z = x + yi the difference is (x - 2) + (y + 1)i, and squaring its modulus gives (x - 2)^2 + (y + 1)^2 = 25, the equation of that circle.

Figure 5Example 7. Every z with |z - (2 - i)| = 5 is 5 units from 2 - i, so together they make a circle.

Measure each point’s distance from the center:

z	z - (2 - i)	Distance	Where
6 + 2i	4 + 3i	5	on the circle
5 - 5i	3 - 4i	5	on the circle
3 + 2i	1 + 3i	\displaystyle \sqrt{10} \approx 3.16	inside
-2 + 3i	-4 + 4i	\displaystyle 4\sqrt{2} \approx 5.66	outside

Boundaries

0 has no argument. Its modulus is 0 and its point is the origin, which lies on every ray and points along none. Every other complex number has exactly one argument in [0^\circ, 360^\circ), and infinitely many outside it.

An argument needs a stated range. \displaystyle \arctan\frac{b}{a} is the argument only in quadrant I and on the positive real axis. For a < 0 it is off by 180^\circ, and in quadrant IV it is negative.

Complex numbers have no order. The real line runs from left to right, but the plane has no single direction for “bigger”, so 3i < 2 and 3i > 2 are both meaningless. Moduli are real and can be compared: |3i| = 3 > |2| says 3i is farther from 0, not that it is larger.

Moduli do not add. In general |z + w| \ne |z| + |w|. For z = 1 + 3i and w = 5 + 2i, \displaystyle |z + w| = \sqrt{61} \approx 7.81, while \displaystyle |z| + |w| = \sqrt{10} + \sqrt{29} \approx 8.55. The triangle inequality says |z + w| \le |z| + |w| always.

Multiplying is not the parallelogram. The vector picture covers adding, subtracting and multiplying by a real number. Multiplying two complex numbers has a picture of its own, a turn and a stretch, which needs the modulus and argument together.

Where this goes next

A nonzero complex number is fixed by its modulus and argument, just as a point is fixed by its polar coordinates. Modulus, argument and polar form writes z by its length and angle and converts between that form and a + bi; there multiplying turns out to multiply moduli and add arguments, and polar form makes powers and roots of complex numbers practical. The mirror picture of the conjugate is why complex roots of real polynomials sit symmetrically about the real axis.

Exercises

Plotting and reading

Give the point for each number and say which quadrant or axis it is on: (a) -7 + 2i (b) 3 - 8i (c) -5i (d) 9.
Name the complex number plotted at each point: (a) (-4, -1) (b) (0, 6) (c) (2.5, -3).
Describe the set of all complex numbers with real part 2, and the set of all with imaginary part -3.

Modulus

Find |-9 - 12i|.
Find |6 - 3i| exactly.
Find |-2.4 + 0.7i|.
Find \displaystyle |\sqrt{7} - 3i|.

Argument

Give the modulus exactly and the argument in [0^\circ, 360^\circ), exactly (in degrees and radians) where possible and otherwise to the nearest tenth of a degree.

5 + 5i
\displaystyle -1 + \sqrt{3}\,i
\displaystyle -4 - 4\sqrt{3}\,i
8 - 3i
-1 - 6i
-9

Conjugates, sums and differences

Let z = 7 - i. Find the modulus and argument of z and of \overline{z}, the arguments to the nearest tenth of a degree.
Let z = -1 + 2i and w = 3 + 2i. Find z + w and name the four corners of the parallelogram that shows the sum.
Let z = 4 + 6i and w = 1 + 2i. Find z - w and the distance between z and w.
Find the distance between -6 + 2i and 1 - 3i, exactly and to the nearest hundredth.
Find the distance between 4i and -5, exactly and to the nearest hundredth.

Sets and unknowns

Describe the set of all z with |z| = 6.
Describe the set of all z with |z + 1 - 3i| = 4. Is 3 + 3i on it? Is 2 + 6i?
Ulan looks for every z = x + yi with |z - 2| = |z - 4i|. Describe the set and give its equation in x and y.
The number z = a - 15i has modulus 17 and lies in quadrant III. Find a and \arg z to the nearest tenth of a degree.

Reasoning

Talgat gives the argument of -2 - 2i as \displaystyle \arctan\frac{-2}{-2} = 45^\circ. Explain what went wrong and give the correct argument.
Explain why |\overline{z}| = |z| and |-z| = |z| for every z, and why \arg\overline{z} = 360^\circ - \arg z whenever 0^\circ < \arg z < 360^\circ.
Samal says that the parallelogram picture proves |z + w| \le |z| + |w|. Explain her argument, and say when the two sides are equal.

Answers

(a) (-7, 2), quadrant II. (b) (3, -8), quadrant IV. (c) (0, -5), the imaginary axis below 0. (d) (9, 0), the real axis right of 0.
(a) -4 - i. (b) 6i. (c) 2.5 - 3i. Real part is the first coordinate, imaginary part the second.
Real part 2 means points (2, b): the vertical line through 2 on the real axis. Imaginary part -3 means points (a, -3): the horizontal line through -3i.
15. \displaystyle \sqrt{81 + 144} = \sqrt{225}.
\displaystyle 3\sqrt{5}. \displaystyle \sqrt{36 + 9} = \sqrt{45} = \sqrt{9 \cdot 5}.
2.5. \displaystyle \sqrt{5.76 + 0.49} = \sqrt{6.25}.
4. \displaystyle \sqrt{7 + 9} = \sqrt{16}.
\displaystyle 5\sqrt{2} and \displaystyle 45^\circ = \frac{\pi}{4}. \displaystyle \sqrt{50}; quadrant I, \arctan 1.
2 and \displaystyle 120^\circ = \frac{2\pi}{3}. \displaystyle \sqrt{1 + 3} = 2; a < 0, so \displaystyle \arctan(-\sqrt{3}) + 180^\circ = -60^\circ + 180^\circ.
8 and \displaystyle 240^\circ = \frac{4\pi}{3}. \displaystyle \sqrt{16 + 48} = 8; quadrant III, so \displaystyle \arctan\sqrt{3} + 180^\circ = 60^\circ + 180^\circ.
\displaystyle \sqrt{73} \approx 8.54 and 339.4^\circ. Quadrant IV: \displaystyle \arctan\left(-\frac{3}{8}\right) + 360^\circ \approx -20.6^\circ + 360^\circ.
\displaystyle \sqrt{37} \approx 6.08 and 260.5^\circ. Quadrant III: \arctan 6 + 180^\circ \approx 80.5^\circ + 180^\circ.
9 and 180^\circ = \pi. -9 is on the negative real axis: \arctan 0 + 180^\circ.
Both have modulus \displaystyle \sqrt{50} = 5\sqrt{2} \approx 7.07. \arg z \approx 351.9^\circ, from \displaystyle \arctan\left(-\frac{1}{7}\right) + 360^\circ \approx -8.1^\circ + 360^\circ, and \arg\overline{z} \approx 8.1^\circ, the mirror angle; the two add to 360^\circ.
z + w = 2 + 4i. The corners are 0, -1 + 2i, 3 + 2i and 2 + 4i.
z - w = 3 + 4i, and the distance is \displaystyle |3 + 4i| = \sqrt{9 + 16} = 5.
\displaystyle \sqrt{74} \approx 8.60. The difference is (-6 - 1) + (2 - (-3))i = -7 + 5i, and \displaystyle \sqrt{49 + 25} = \sqrt{74}.
\displaystyle \sqrt{41} \approx 6.40. \displaystyle |4i - (-5)| = |5 + 4i| = \sqrt{25 + 16}.
The circle with center 0 and radius 6: every point 6 units from 0.
|z + 1 - 3i| = |z - (-1 + 3i)|, so the set is the circle with center -1 + 3i and radius 4. 3 + 3i is on it: |4 + 0i| = 4. 2 + 6i is not, it is outside: \displaystyle |3 + 3i| = 3\sqrt{2} \approx 4.24 > 4.
The points equally far from 2 and from 4i: the perpendicular bisector of the segment from (2, 0) to (0, 4), the line \displaystyle y = \frac{1}{2}x + \frac{3}{2}. Squaring both moduli, (x - 2)^2 + y^2 = x^2 + (y - 4)^2, so -4x + 4 = -8y + 16.
a = -8 and \arg z \approx 241.9^\circ. a^2 + 225 = 289 gives a = \pm 8, and quadrant III needs a < 0; then \displaystyle \arctan\frac{15}{8} + 180^\circ \approx 61.9^\circ + 180^\circ.
The key returns a ray pointing right, into quadrant I, which is where 2 + 2i lies. The number -2 - 2i is on the opposite ray, in quadrant III, so add 180^\circ: 45^\circ + 180^\circ = 225^\circ.
\overline{z} = a - bi and -z = -a - bi have moduli \displaystyle \sqrt{a^2 + (-b)^2} and \displaystyle \sqrt{(-a)^2 + (-b)^2}, both equal to \displaystyle \sqrt{a^2 + b^2} because squaring removes signs. In the plane, reflecting in the real axis and a half turn about 0 both keep distances from 0. The reflection sends a ray at angle \theta above the positive real axis to the ray at \theta below it, which counterclockwise is 360^\circ - \theta; for \theta = 0^\circ that would be 360^\circ, outside the range, which is why the positive real axis is left out.
The arrow to z + w is one side of the triangle with corners 0, z and z + w, whose other two sides have lengths |z| and |w| (the side from z to z + w is a copy of the arrow to w). A side of a triangle is never longer than the other two together, so |z + w| \le |z| + |w|. Equality holds only when the triangle flattens with z between 0 and z + w: when z and w point the same way (one is a positive real multiple of the other), or when one of them is 0.