The complex plane, modulus and conjugation

A complex number corresponds to one point with two real coordinates

The preceding lesson constructed z = a + bi from the real pair (a, b). Represent that pair by a point on two perpendicular axes: the horizontal real axis records a = Re z and the vertical imaginary axis records b = Im z. Positive real coordinates point right and positive imaginary coefficients point up. Thus −2 + 3i is the point (−2, 3), while 3 − 2i is a different point. The imaginary coordinate is the real coefficient b, not the term bi.

Use the same unit length on both axes. With equal scales, a horizontal change of one and a vertical change of one have the same drawn length, which makes ordinary Euclidean distance agree with the modulus formula. Unequal display scales can still label coordinates but distort circles, distances and perpendicularity. Complex number one lies at (1, 0), i lies at (0, 1) and zero lies at the origin (0, 0). Real numbers lie on the real axis; purely imaginary numbers lie on the imaginary axis.

This representation is called the complex plane or an Argand diagram. Each complex number gives exactly one point and each point gives exactly one complex number. Here a and b are independent coordinates, rather than one being a function value determined by the other. A modulus condition can describe a whole circle or region, not necessarily a graph b = f(a). If coordinates represent physical displacements, both axes must use compatible length units and a stated common scale; unrelated quantities such as time and price do not acquire a Euclidean length merely by being placed on two axes.

The modulus is a nonnegative real distance

The modulus of z = a + bi is |z| = √(a² + b²), using the principal nonnegative real square root. The right triangle from the origin to (a, b) has horizontal and vertical leg lengths |a| and |b|, so Pythagoras gives this distance. The formula also covers points on either axis and the origin, where a triangle may be degenerate. For −3 + 4i the modulus is 5; for 1 + 2i it is √5, an exact value even without a decimal approximation.

The squared modulus is |z|² = a² + b² = z z̄. It is a different quantity from |z| and from z². For 1 + i, the modulus is √2, its square is 2, and the complex square is 2i. The modulus is zero exactly when both a and b are zero, because a² and b² are nonnegative real numbers. Otherwise it is positive. For a real number a, this definition gives √(a²) = |a| and agrees with real absolute value.

The modulus map accepts every complex input and has complete range [0, ∞): all its outputs are nonnegative, and any proposed real ρ ≥ 0 is attained at z = ρ + 0i. It is not one-to-one; for example, 3, −3, 3i and −3i all have modulus 3. The complete preimage of a positive modulus is a circle, established later in this lesson. When coordinates represent lengths, modulus has the same length unit and squared modulus has its square; do not report a squared distance as a distance.

The difference points from the second endpoint to the first

Let z = a + bi and w = c + di. Their difference z − w = (a − c) + (b − d)i records the directed displacement from point w to point z. Its modulus is their undirected distance: |z − w| = √((a − c)² + (b − d)²). This is the Cartesian distance formula from Pythagoras. Reversing the endpoints negates the displacement, but leaves its distance unchanged because both coordinate differences are squared.

The distance is nonnegative and is zero exactly when z = w. Adding any fixed complex translation v to both endpoints preserves it, since (z + v) − (w + v) = z − w. Thus a displacement arrow can be drawn from another tail while retaining its coordinate change and length. Distinguish this free displacement from a point's position arrow, which starts at the origin. The same number can describe either when the starting point is stated.

A straight endpoint distance also differs from the length of a chosen route. Moving horizontally then vertically from w to z has route length |a − c| + |b − d|, which need not equal √((a − c)² + (b − d)²). The complex distance formula records the direct Euclidean separation, not travel constrained to a grid or a forecast of an actual path. Its exact radical can be retained without measuring pixels in a drawing.

Coordinate arithmetic becomes head-to-tail displacement addition

Adding z = a + bi and w = c + di gives the point (a + c, b + d). Begin with the position arrow from zero to z and append a translated arrow with displacement w; its endpoint is z + w. Appending the displacements in the other order gives the same endpoint. The four points 0, z, w and z + w form the corresponding parallelogram, with a degenerate version when the displacements lie along one line or one is zero.

Subtraction is addition of the opposite displacement: z − w = z + (−w). Its position arrow runs from zero to the point z − w, whereas the arrow from w to z is a translated copy with the same coordinate change. For z = 2 − i and w = −1 + 3i, z − w = 3 − 4i; the endpoint distance is 5. Reversing the subtraction gives −3 + 4i, reversing the directed arrow rather than changing its length.

The parallelogram is justified by the coordinate formulas, rather than by the appearance of a drawing alone. At each vertex, the parallel translated edges have exactly the same two coordinate differences. A graphical arrow does not change complex equality or permit treating nonreal complex values as positive or negative. When vectors represent a context, carry the same axis directions and compatible units through both additions.

Midpoints average coordinates and real scalars scale distances

The midpoint of z and w is m = (z + w)/2, whose coordinates are ((a + c)/2, (b + d)/2). The two differences are z − m = (z − w)/2 and w − m = −(z − w)/2. Both therefore have modulus |z − w|/2. The averaging point lies on the joining segment and is equally far from the endpoints, including the coincident-endpoint case where all three points are the same.

For a real scalar λ, the point λz has coordinates (λa, λb), and |λz|² = λ²(a² + b²). Taking nonnegative roots gives |λz| = |λ||z|. For nonzero z, positive scaling multiplies the origin displacement's length and keeps its direction, while negative scaling also reverses its direction. The scalar λ = 0 sends every point to the origin, and z = 0 stays at the origin for every real scalar; the zero displacement has no direction. This real-scalar statement does not introduce polar angles or the geometric effect of multiplication by an arbitrary complex factor.

In particular, −z = −a − bi is the point (−a, −b), the reflection of z through the origin. Its modulus equals |z|, and the distance between z and −z is |2z| = 2|z|. Do not confuse this opposite point with the conjugate (a, −b), which retains the real coordinate. They agree exactly when a = 0, including zero; this will follow again from comparing the two reflection formulas.

Conjugation reflects in the real axis and preserves paired distances

Conjugation changes z = a + bi to z̄ = a − bi, so it maps (a, b) to (a, −b). This is reflection in the real axis: horizontal position is unchanged and the vertical displacement reverses. Conjugating twice returns z. Its fixed points are precisely the real axis, where b = 0. Its modulus is unchanged because a² + (−b)² = a² + b².

Reflecting both endpoints preserves their distance: |z̄ − w̄|² = (a − c)² + (−b + d)² = |z − w|², and nonnegative square roots give |z̄ − w̄| = |z − w|. The claim concerns reflection of both endpoints; reflecting only one while leaving a nonreal second endpoint fixed can change the distance. The distance from z to its own conjugate is |z − z̄| = |2bi| = 2|b|.

Compare z̄ = a − bi with −z = −a − bi. Their imaginary coefficients agree, and their real coefficients agree exactly when a = 0. Conjugation therefore agrees with the additive opposite only on the imaginary axis. Reflection sends a complete circle or disk with centre c to the corresponding circle or disk with centre c̄ and the same radius, because |z̄ − c̄| = |z − c|. An off-axis centre itself must be reflected; it is not generally left fixed.

Lengths satisfy real inequalities without ordering complex values

The triangle inequality is |z + w| ≤ |z| + |w|. It expresses that the direct displacement length is no greater than the sum of two successive displacement lengths. For an algebraic proof, put A = a² + b², B = c² + d² and S = ac + bd. The identity AB − S² = (ad − bc)² ≥ 0 gives |S| ≤ √(AB) = |z||w|, hence S ≤ |z||w|. All quantities in these comparisons are real.

Now |z + w|² = A + B + 2S ≤ A + B + 2|z||w| = (|z| + |w|)². Both candidate lengths are nonnegative, so taking principal real square roots preserves the comparison and proves the inequality for all complex inputs, including zero. Equality occurs when one input is zero or the two nonzero displacements have the same direction, equivalently one is a positive real multiple of the other. It is not a general rule that the modulus of a sum equals the sum of moduli.

Apply the same inequality to z = (z − w) + w to get |z| − |w| ≤ |z − w|, then exchange z and w. Combining the two gives the reverse distance bound ||z| − |w|| ≤ |z − w|. These inequalities compare real-valued lengths, not arbitrary complex numbers under an ordered-field rule. They also justify the geometric statement that a two-segment route cannot be shorter than the direct endpoint distance.

A modulus equation gives every point of its circle

A locus is the complete set of points satisfying a stated condition. For real R > 0, the condition |z| = R is equivalent to a² + b² = R² because both sides of the original equality are nonnegative. This is exactly the set of points at distance R from the origin, a circle. It contains every point satisfying that equation, not only real-axis, imaginary-axis or selected control points.

For a fixed complex centre c = h + ki, |z − c| = R is equivalent to (a − h)² + (b − k)² = R² when R ≥ 0. If R > 0 this is the circle centred at (h, k) with radius R. An explicit complete coordinate description is a = h + u and b = k ± √(R² − u²), for every real −R ≤ u ≤ R. Every such pair satisfies the circle equation, and any point on the equation has that u and one of those b values. At u = ±R the two signs give the same point, counted once.

At R = 0 the condition instead gives the single centre c, because a sum of nonnegative real squares is zero only when both differences are zero. If R < 0 the original modulus equality has no points at all. Squaring a negative right side without preserving that guard would incorrectly produce a nonempty circle equation. Read the centre from the complete subtraction z − c: for example, |z + 1 − 2i| = 3 has centre −1 + 2i, not 1 − 2i.

An included boundary makes a closed disk

For real R > 0, the condition |z − c| ≤ R is the closed disk centred at c: its interior and its boundary circle are both included. The strict condition |z − c| < R is the open disk, excluding the boundary. Because modulus is nonnegative, the equivalent squared conditions are (a − h)² + (b − k)² ≤ R² and < R² on the retained guard R ≥ 0. These equivalences classify every possible coordinate pair.

There is also a complete coordinate description. For a closed disk, choose any real u with −R ≤ u ≤ R, set a = h + u and allow every b between k − √(R² − u²) and k + √(R² − u²), endpoints included. For an open disk with R > 0, require −R < u < R and make both b bounds strict. Each constructed pair meets its distance bound, and every permitted disk point falls within exactly those coordinate restrictions. Samples cannot fill in this completeness argument.

At R = 0, the closed disk is the single centre and the open disk is empty; no nonnegative distance is less than zero. At R < 0 both original conditions are empty. When drawing an open disk, a dashed circle can mark the excluded boundary, while a solid boundary can mark inclusion in a closed disk. The drawn marks explain membership; stroke thickness and pixel rounding do not alter the exact inequalities.

Equal distances give a perpendicular bisector only for distinct centres

Let p = h + ki and q = u + vi be fixed. The condition |z − p| = |z − q| can always be squared reversibly because both distances are nonnegative. Its exact coordinate form is (a − h)² + (b − k)² = (a − u)² + (b − v)². Expanding and cancelling yields 2(u − h)a + 2(v − k)b = u² + v² − h² − k². Every point satisfying that linear equation satisfies the original equal-distance condition, and conversely.

If p ≠ q, at least one coefficient u − h or v − k is nonzero, so the equation is a full straight line. It passes through the midpoint ((h + u)/2, (k + v)/2). Its direction vector (−(v − k), u − h) has component-product sum zero with the joining direction (u − h, v − k), the Euclidean perpendicularity condition. Hence it is exactly the perpendicular bisector of the segment joining p and q. This description also covers horizontal and vertical joining segments without dividing by a possibly zero slope denominator.

If p = q, both distances are identical at every complex z. The expanded equation is 0 = 0, and the complete locus is the whole plane rather than a particular perpendicular-bisector line. Comparing the centres and deriving the reversible coordinate equation establishes the complete locus here. If an additional condition is imposed, such as a circle equation, take the intersection of the two full loci and verify every resulting branch.

The explorer draws selected geometry while exact statements determine it

The explorer has four views: paired distance with midpoint, conjugate versus opposite, addition and subtraction. Choose z and w independently from its exact coordinate presets. The named real and imaginary axes use equal unit scales on a fixed square window from −7 to 7 in each coordinate. Read the exact point coordinates, moduli and squared moduli alongside the drawing; the finite preset list is not the domain of the complex plane.

The paired distance remains an endpoint quantity |z − w| even when another view is selected. In the distance view, inspect the coordinate-average midpoint (z + w)/2 and verify that its two endpoint distances are each half |z − w|. Equal distances alone describe the full equidistance locus, so they are not a proof that an arbitrary point is the midpoint. In the conjugate view, distinguish reflection in the real axis from opposition through the origin. In the addition view, inspect the parallelogram endpoint z + w; in the subtraction view, distinguish the position arrow to z − w from the displacement arrow running from w to z. Coincident points or zero arrows have exact meanings even when their plotted marks overlap.

Drawing coordinates may use numerical positioning, while equality and length statements use exact real fractions, squared distances and radicals. A graph window is a visible portion of the plane, not a contextual restriction on complex inputs or on a complete locus. The circle, disk and equidistance proofs in this article establish their full point sets independently of these four selected views. Polar angles, trigonometric form and geometric complex multiplication come later; none is required for the current coordinate and distance checks.

Worked examples

Try it yourself

1. Represent z = −4 + 3i and w = 1/2 + (1/2)i as plane points. Find |z|, |z|², |w| and |w|² exactly. Explain why the vertical coordinates are 3 and 1/2 rather than 3i and i/2, and why |z|² is not z².

   Show the answer and reasoning

   The points are (−4, 3) and (1/2, 1/2). Their moduli are 5 and √2/2, with squared moduli 25 and 1/2. In contrast, z² = 7 − 24i.

   Coordinates are the real coefficients Re and Im. Pythagoras gives |z|² = 16 + 9 = 25 and |w|² = 1/4 + 1/4 = 1/2; taking principal nonnegative real roots gives the stated lengths. Squaring the complex expression z gives (−4)² + 2(−4)(3i) + (3i)² = 7 − 24i, a different operation from multiplying z by its conjugate.

2. For z = 2 − i and w = −1 + 3i, find the directed displacement from w to z, the endpoint distance, z + w and the midpoint. Give the reverse displacement and both midpoint distances.

   Show the answer and reasoning

   The displacement from w to z is 3 − 4i, with distance 5. The sum is 1 + 2i and the midpoint is 1/2 + i. The reverse displacement is −3 + 4i, and each midpoint distance is 5/2.

   Subtract coordinates in the stated direction: (2 − (−1), −1 − 3) = (3, −4), whose squared length is 25. Addition gives (1, 2), and averaging gives (1/2, 1). The endpoint-minus-midpoint differences are 3/2 − 2i and its opposite, each with squared modulus 9/4 + 4 = 25/4. Reversing direction negates the displacement without changing its length.

3. For z = −2 + 3i, give z̄ and −z as complex numbers and plane points, find their moduli and calculate the distances |z − z̄| and |z − (−z)|. For which complex inputs do conjugation and opposition give the same point?

   Show the answer and reasoning

   z̄ = −2 − 3i is (−2, −3), and −z = 2 − 3i is (2, −3). Both have modulus √13. The two requested distances are 6 and 2√13. Conjugation and opposition coincide exactly when Re z = 0.

   Conjugation preserves the real coefficient, while opposition negates both coefficients. Each squared modulus is 4 + 9 = 13. The difference z − z̄ = 6i has modulus 6; z − (−z) = 2z has modulus 2|z| = 2√13. Comparing a − bi and −a − bi requires a = −a, hence a = 0, including zero. Real-axis reflection and origin reflection are therefore different except on the imaginary axis.

4. Compare |z + w| with |z| + |w| for (z, w) = (3, 4i), (3, 4) and (3, −4). Explain which pair gives equality and why none violates the triangle inequality.

   Show the answer and reasoning

   The comparisons are 5 < 7, 7 = 7 and 1 < 7, respectively. The same-direction pair (3, 4) gives equality.

   The first sum is 3 + 4i, whose modulus is 5 while the two lengths sum to 7. The second sum is the positive real 7 and both nonzero displacements point in the same direction. The third sum is −1 with modulus 1 because the real displacements partly cancel, while their separate lengths still sum to 7. Every comparison concerns nonnegative real lengths; no ordering of nonreal complex values is being asserted.

5. Describe the complete locus |z| = 3 in coordinates, find all its real- and imaginary-axis intersections and check z = √5 + 2i. Does the selected point list prove the full locus? State the membership of those points in |z| ≤ 3 and |z| < 3.

   Show the answer and reasoning

   The full circle is a² + b² = 9. Its axis intersections are ±3 and ±3i, and √5 + 2i is also on it. All these points belong to the closed disk |z| ≤ 3 and none to the open disk |z| < 3. A finite list does not prove the full locus.

   The nonnegative modulus equality is equivalent to a² + b² = 9. Setting b = 0 gives a = ±3, and setting a = 0 gives b = ±3. The additional point has squared modulus 5 + 4 = 9. Completeness follows from the coordinate equivalence, or from all real −3 ≤ a ≤ 3 with b = ±√(9 − a²), rather than from those examples. Each checked point has distance exactly 3, so it lies on the included closed boundary and the excluded open boundary.

6. Describe |z + 1 − 2i| ≤ 2 and test z = −1 + 2i, z = 1 + 2i and z = 2 + 2i. How does replacing ≤ by < change their membership? Then describe the same closed and open conditions when the right-side bound is zero or −2.

   Show the answer and reasoning

   The closed disk has centre −1 + 2i and coordinate condition (a + 1)² + (b − 2)² ≤ 4. The three points are respectively inside, on the boundary and outside. The strict disk keeps the centre, excludes the boundary point and still excludes the outside point. At bound zero the closed set is the centre and the open set is empty; at bound −2 both are empty.

   Rewrite the argument as z − (−1 + 2i) before identifying the centre. The three exact centre distances are 0, 2 and 3. A strict inequality excludes distance 2. Zero can bound a nonnegative distance only through equality at the centre, while no distance is less than zero or at most a negative bound. Squaring −2 and using a positive squared-radius disk would change the original condition, so retain the radius guard first.

7. Find the complete locus |z| = |z − 4i| and its intersection with |z| = √5. State the midpoint of the two centres, 0 and 4i, and explain what the first locus becomes if 4i is replaced by zero.

   Show the answer and reasoning

   The full equidistance locus is Im z = 2, or z = a + 2i for every real a. The midpoint of the two centres is 2i, and the line's circle intersection is exactly {1 + 2i, −1 + 2i}. Replacing 4i by zero makes the equidistance locus the whole plane.

   For z = a + bi, squaring both nonnegative distances gives a² + b² = a² + (b − 4)², equivalent to 8b = 16 and b = 2. Every real a then satisfies the original equality, giving the full horizontal perpendicular bisector of the vertical segment from 0 to 4i. Intersecting with a² + b² = 5 yields a² + 4 = 5, so a = ±1 and no other branches exist. Coincident endpoints instead have identical distances at every point; their midpoint remains zero, but no particular bisector line is selected.

What to carry forward

• Represent a + bi by the unique point (a, b) on named perpendicular real and imaginary axes with equal scales. The imaginary coordinate is the real coefficient b, and geometric distance requires compatible axis units.
• The modulus |z| = √(a² + b²) is a nonnegative real distance and is zero exactly at z = 0. Its square is z z̄, which differs from both the modulus itself and the complex square z².
• The difference z − w is the directed displacement from w to z; |z − w| is their symmetric endpoint distance. Translations preserve paired distances, and (z + w)/2 is their midpoint.
• Addition and subtraction correspond to head-to-tail displacements. Real scaling satisfies |λz| = |λ||z|, while opposition maps (a, b) to (−a, −b) and conjugation maps it to (a, −b).
• Conjugation preserves modulus and distances when both endpoints are reflected. It equals opposition exactly on the imaginary axis. Triangle and reverse-distance inequalities compare real lengths rather than imposing an ordered-field order on complex numbers.
• For nonnegative radius bounds, exact coordinate equivalences describe full circles and open or closed disks. At radius zero the circle and closed disk are one point and the open disk is empty; negative bounds give empty original loci.
• Equidistance from two distinct centres gives their full perpendicular-bisector line. Coincident centres give the whole plane, and extra conditions require intersecting complete loci rather than trusting plotted samples or a cropped view.