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Complex Numbers Addition /Subtraction Visualizer


Complex Addition & Subtraction

Add component-wise, see the parallelogram
ReIm-5−5i-4−4i-3−3i-2−2i-1−1i11i22i33i44i55iz₁+z₂z₁−z₂−z₂z₁z₂
Show:
z₁
3 + i
z₂
1 + 3i
z₁ + z₂ — Sum
4 + 4i
|z₁+z₂| = 5.7
z₁ − z₂ — Difference
2 − 2i
|z₁−z₂| = 2.8 (distance between z₁ and z₂)
Step-by-Step
Addition:
(3 + i) + (1 + 3i)
= (3 + 1) + (1 + 3)i
= 4 + 4i
Subtraction:
(3 + i) − (1 + 3i)
= (3 − 1) + (1 − 3)i
= 2 − 2i
Triangle Inequality
|z₁ + z₂| ≤ |z₁| + |z₂|
5.7 ≤ 3.2 + 3.2 = 6.3 ✓
Equality holds when z₁ and z₂ point in the same direction (same argument).
Key Ideas
Addition is component-wise: add the real parts, add the imaginary parts. (a+bi) + (c+di) = (a+c) + (b+d)i. No cross-terms, no i² — it is just vector addition.
The parallelogram rule: z₁+z₂ is the diagonal of the parallelogram formed by z₁ and z₂. The dashed vectors show z₂ translated to the tip of z₁ (and vice versa) — “tip to tail.”
Subtraction gives the vector from z₂ to z₁. The purple dashed line between the two points has the same length and direction as z₁−z₂. The modulus |z₁−z₂| is the distance between the two points.
z₁ − z₂ = z₁ + (−z₂). Subtraction is the same as adding the negation. The ghost vector −z₂ is z₂ rotated by 180°. Adding it to z₁ via the parallelogram rule gives the same result.
Triangle inequality: |z₁+z₂| ≤ |z₁| + |z₂|. The sum can never be longer than both vectors placed end-to-end. Drag z₁ and z₂ to the same direction to see equality.
The two results are the two diagonals of one parallelogram — green through the interior, purple across the input points. Learn more about Both mode · Getting started

Getting Started — Drag Two Points
The Parallelogram Rule for Addition
Subtraction and the Difference Vector
Both Mode — Comparing Addition and Subtraction
The Mirror Pair Preset
The Axis Pair Preset
The Mixed-Signs Preset
Degenerate and Axis-Aligned Configurations
Reading the Step-by-Step Panel
The Triangle Inequality
How Complex Addition Works
How Complex Subtraction Works
Conjugate Pairs in Addition and Subtraction
Related Concepts and Tools




Getting Started — Drag Two Points

Two draggable points sit on the complex plane: z1z_1 (navy) and z2z_2 (orange). Grab either point and move it anywhere within the ±5\pm 5 grid to see the addition and subtraction results update instantly.

Alternatively, type precise values into the input fields on the right panel — each point has separate real and imaginary inputs accepting values from −5-5 to 55 in steps of 0.10.1. Five preset configurations are available below the plane: (3+i)(3+i) & (1+3i)(1+3i), (2+2i)(2+2i) & (−2+2i)(-2+2i), 44 & 3i3i, a conjugate pair, and (−1+3i)(-1+3i) & (2−i)(2-i). Click Random to generate two arbitrary points. Each preset has a dedicated section below with the tool frozen on it: the mirror pair, the axis pair, the conjugate pair, and the mixed-signs pair.

The Show toggle lets you display both operations simultaneously, or isolate addition or subtraction for a cleaner view. Start with Both to compare the two operations side by side, then switch to Addition or Subtraction to focus on the geometry of each one individually.

The Parallelogram Rule for Addition

Switch to Addition mode to see the parallelogram clearly. The green vector from the origin to the sum point z1+z2z_1 + z_2 forms the diagonal of the parallelogram whose sides are z1z_1 and z2z_2.

Two dashed "ghost" vectors complete the parallelogram. A dashed copy of z2z_2 (orange) is translated to the tip of z1z_1, and a dashed copy of z1z_1 (navy) is translated to the tip of z2z_2. Both ghost vectors arrive at the same destination — the sum point. This is the tip-to-tail construction: place one vector's tail at the other's tip, and the sum runs from the origin to where you end up.

The light green shaded region fills the interior of the parallelogram, making the geometric relationship immediately visible. Try the preset (3+i)(3+i) & (1+3i)(1+3i) — the parallelogram tilts toward the imaginary axis because z2z_2 has a larger imaginary component. Then try 44 & 3i3i — one side lies along the real axis and the other along the imaginary axis, producing a rectangle.
ReIm-5−5i-4−4i-3−3i-2−2i-1−1i11i22i33i44i55iz₁+z₂z₁z₂
Addition mode — (3+i) + (1+3i), frozen

The teal diagonal reaches the sum 4 + 4i; the two dashed ghosts — z₂ shifted to the tip of z₁, and z₁ shifted to the tip of z₂ — both arrive at the same corner of the shaded parallelogram.

The frozen frame computes (3+i)+(1+3i)=4+4i(3+i) + (1+3i) = 4+4i: because the two inputs are mirror images across the 45° line, their sum lands exactly on it. The component-wise rule is visible in the coordinates — reals 3+13+1, imaginaries 1+31+3, no interaction between them.

Notice also what the picture does not show: the sum vector 4+4i4+4i has modulus 32≈5.66\sqrt{32} \approx 5.66, less than ∣z1∣+∣z2∣=210≈6.32|z_1|+|z_2| = 2\sqrt{10} \approx 6.32 — the triangle inequality at work, strict because the two vectors point in different directions.

Subtraction and the Difference Vector

Switch to Subtraction mode to isolate the subtraction geometry. The purple vector from the origin represents z1−z2z_1 - z_2, while a purple dashed line connects the tips of z2z_2 and z1z_1 directly.

This dashed line has the same length and direction as the difference vector — it literally shows "the vector from z2z_2 to z1z_1." The modulus ∣z1−z2∣|z_1 - z_2| equals the distance between the two points on the plane.

A faint dashed vector labeled −z2-z_2 appears opposite to z2z_2 (rotated 180°). This illustrates the key identity: z1−z2=z1+(−z2)z_1 - z_2 = z_1 + (-z_2). Subtraction is addition of the negation.

Try the conjugate pair preset: z1=3+2iz_1 = 3 + 2i and z2=3−2iz_2 = 3 - 2i. The difference is 4i4i (pure imaginary) and the dashed line between the points is vertical. The sum is 66 (pure real) and lies on the real axis — conjugate pairs always produce this split.
ReIm-5−5i-4−4i-3−3i-2−2i-1−1i11i22i33i44i55iz₁−z₂−z₂z₁z₂
Subtraction mode — (3+i) − (1+3i), frozen

The purple vector lands on 2 − 2i; the dashed line from z₂ to z₁ carries the same length and direction, and the faint −z₂ ghost points opposite the orange vector.

In the frozen frame, (3+i)−(1+3i)=2−2i(3+i) - (1+3i) = 2-2i: the purple vector drops into the fourth quadrant, and the dashed line from z2z_2 to z1z_1 is its exact translate. The distance between the two points is ∣2−2i∣=22≈2.83|2-2i| = 2\sqrt{2} \approx 2.83.

The faint −z2-z_2 ghost at −1−3i-1-3i completes the identity z1−z2=z1+(−z2)z_1 - z_2 = z_1 + (-z_2): run the parallelogram rule on z1z_1 and the ghost and you land on the same purple point. Full algebra in how complex subtraction works.

Both Mode — Comparing Addition and Subtraction

In Both mode, the addition parallelogram (green) and subtraction vector (purple) appear together. This side-by-side view reveals how the two operations relate geometrically.

The green sum vector and the purple difference vector are actually the two diagonals of the same parallelogram. The sum is the diagonal that starts at the origin and passes through the interior. The difference connects the two input points — it is the other diagonal, translated to start at the origin.

Try dragging z2z_2 to the exact opposite of z1z_1. For example, set z1=3+2iz_1 = 3 + 2i and z2=−3−2iz_2 = -3 - 2i. The sum collapses to zero (both vectors cancel) while the difference doubles to 6+4i6 + 4i. The parallelogram degenerates to a line segment because the two vectors are antiparallel.

Now set both points in the same direction — for example z1=2+iz_1 = 2 + i and z2=4+2iz_2 = 4 + 2i. The sum is 6+3i6 + 3i (aligned), the parallelogram again degenerates to a line, and the triangle inequality reaches equality.
ReIm-5−5i-4−4i-3−3i-2−2i-1−1i11i22i33i44i55iz₁+z₂z₁−z₂−z₂z₁z₂
Both mode — the opening pair, frozen

Green and purple share one parallelogram: the sum diagonal runs from the origin through the interior, while the difference matches the other diagonal, joining the two input points.

The frozen frame is the tool's opening state, (3+i)(3+i) & (1+3i)(1+3i), with everything switched on. The two diagonals tell the whole story: the green one, 4+4i4+4i, runs from the origin through the parallelogram's interior; the purple one, 2−2i2-2i, joins the two input points and reappears as a vector from the origin.

This double-diagonal reading is worth internalizing before visiting the presets — the mirror pair makes the two diagonals land on the two axes, and the conjugate pair splits them between real and imaginary in the opposite way.

The Mirror Pair Preset

The preset (2+2i)(2+2i) & (−2+2i)(-2+2i) loads two numbers that are reflections of each other across the imaginary axis: same height, opposite real parts.
ReIm-5−5i-4−4i-3−3i-2−2i-1−1i11i22i33i44i55iz₁+z₂z₁−z₂−z₂z₁z₂
Mirror pair — (2+2i) & (−2+2i), frozen

Sum 4i straight up, difference 4 straight right: real parts cancel in the sum, imaginary parts cancel in the difference. Equal moduli make the parallelogram a rhombus with perpendicular diagonals.

The cancellation is the reverse of the conjugate pair's: here the real parts cancel in the sum, (2+2i)+(−2+2i)=4i(2+2i) + (-2+2i) = 4i, and the imaginary parts cancel in the difference, (2+2i)−(−2+2i)=4(2+2i) - (-2+2i) = 4. One diagonal lands on each axis — the same clean split as conjugate pairs, rotated a quarter turn.

There is extra geometry hiding in this preset: both inputs have the same modulus 8\sqrt{8}, so the parallelogram is a rhombus — and the diagonals of a rhombus are always perpendicular. The frozen frame shows it directly: the vertical sum and horizontal difference cross at a right angle.

Reflection across the imaginary axis is the map z↦−zˉz \mapsto -\bar{z}: it negates the real part and keeps the imaginary part, which is exactly what pairs each input with the other.

The Axis Pair Preset

The preset 44 & 3i3i puts one number on each axis: z1z_1 purely real, z2z_2 purely imaginary. The parallelogram becomes a perfect rectangle.
ReIm-5−5i-4−4i-3−3i-2−2i-1−1i11i22i33i44i55iz₁+z₂z₁−z₂−z₂z₁z₂
Axis pair — 4 & 3i, frozen

A perfect rectangle: one side on each axis. The sum diagonal reaches 4 + 3i, whose modulus is exactly 5 — the 3-4-5 right triangle hiding inside the parallelogram rule.

With perpendicular sides, the sum 4+3i4 + 3i is the rectangle's diagonal, and its modulus is the classic 33–44–55 triangle: ∣4+3i∣=16+9=5|4+3i| = \sqrt{16+9} = 5. The difference 4−3i4 - 3i is the other diagonal, and a rectangle's diagonals are equal — so ∣z1−z2∣=5|z_1 - z_2| = 5 as well.

This preset is the cleanest possible view of the triangle inequality being strict: 5≤4+3=75 \leq 4 + 3 = 7, with a wide gap because the two vectors are perpendicular — as far from pointing the same way as they can be while staying in one quadrant.

It is also the natural bridge to the degenerate configurations: nudge z2z_2 down onto the real axis and the rectangle collapses to a line.

The Mixed-Signs Preset

The preset (−1+3i)(-1+3i) & (2−i)(2-i) places the two inputs in opposite quadrants — second and fourth — so the vectors pull against each other.
ReIm-5−5i-4−4i-3−3i-2−2i-1−1i11i22i33i44i55iz₁+z₂z₁−z₂−z₂z₁z₂
Mixed signs — (−1+3i) & (2−i), frozen

Vectors pulling in nearly opposite directions: the sum shrinks to 1 + 2i while the difference −3 + 4i grows longer than either input — cancellation working in reverse.

Opposing directions make addition cancel and subtraction amplify. The sum shrinks to 1+2i1 + 2i, with modulus 5≈2.24\sqrt{5} \approx 2.24 — shorter than z1z_1 itself. The difference stretches to −3+4i-3 + 4i, with modulus exactly 55 (another 33–44–55 triangle, mirrored into the second quadrant): longer than either input.

This is the triangle inequality near its other extreme — 5\sqrt{5} against 10+5≈5.4\sqrt{10} + \sqrt{5} \approx 5.4. Push it all the way by making the vectors exactly antiparallel, as in the Both mode example, and the sum collapses to zero.

The lesson of the preset: subtraction measures separation, not size. Two modest numbers far apart in direction produce a small sum and a large difference — ∣z1−z2∣|z_1 - z_2| is the distance between the points, whatever their moduli.

Degenerate and Axis-Aligned Configurations

Several input combinations produce visually distinct special cases worth capturing.

Pure real inputs: click preset 44 & 3i3i, then change z2z_2 to a real number like 22. Both vectors lie on the real axis, the parallelogram collapses to a horizontal line, and both the sum and difference are real numbers.

Pure imaginary inputs: set z1=0+3iz_1 = 0 + 3i and z2=0+2iz_2 = 0 + 2i. Both vectors are vertical, the parallelogram collapses to a vertical line, and results are purely imaginary.

One vector zero: set z2=0z_2 = 0. The sum equals z1z_1 itself, the difference also equals z1z_1, and the parallelogram disappears because one side has zero length.

Collinear vectors: when z1z_1 and z2z_2 point in the same direction (same argument), the parallelogram degenerates to a line and the triangle inequality becomes an equality. Drag both points to, say, the positive real axis to see this.

Off-screen result: when the sum exceeds the ±5\pm 5 range, the green vector is replaced by a dashed ray pointing toward the edge with an arrow and the label "z1+z2→z_1 + z_2 \to", indicating the result is beyond the visible area.

Reading the Step-by-Step Panel

The Step-by-Step section on the right breaks down each operation into three lines.

For addition, it shows: the original expression (z1)+(z2)(z_1) + (z_2), the component-wise grouping (a1+a2)+(b1+b2)i(a_1 + a_2) + (b_1 + b_2)i, and the final simplified result highlighted in green.

For subtraction, the same three-line pattern appears with minus signs: the original expression, the component-wise grouping (a1−a2)+(b1−b2)i(a_1 - a_2) + (b_1 - b_2)i, and the result in purple.

All values update dynamically as you drag or type. When "Both" mode is active, both operations appear stacked. Switching to "Addition" or "Subtraction" mode hides the irrelevant calculation, giving a cleaner view.

This panel makes the component-wise nature of complex addition explicit — the real and imaginary parts are handled independently, with no cross-terms or powers of ii involved.

The Triangle Inequality

When addition is visible, a dedicated panel displays the triangle inequality for the current configuration:

∣z1+z2∣≤∣z1∣+∣z2∣|z_1 + z_2| \leq |z_1| + |z_2|


The panel substitutes the actual modulus values and verifies the inequality with a green checkmark. This states that the length of the sum vector can never exceed the sum of the individual lengths — geometrically, the diagonal of a triangle is always shorter than the sum of the other two sides.

Equality holds when z1z_1 and z2z_2 have the same argument (point in the same direction). Drag both points onto the positive real axis — for example z1=3z_1 = 3 and z2=2z_2 = 2 — and watch the left side equal the right side exactly: ∣5∣=∣3∣+∣2∣=5|5| = |3| + |2| = 5.

Now rotate z2z_2 perpendicular to z1z_1 and the gap widens. The farther apart the two vectors' directions, the shorter the sum vector relative to the individual lengths.

How Complex Addition Works

Adding two complex numbers is straightforward — combine real parts and imaginary parts separately:

(a+bi)+(c+di)=(a+c)+(b+d)i(a + bi) + (c + di) = (a + c) + (b + d)i


There are no cross-terms because the real and imaginary components are independent. This is identical to adding two-dimensional vectors component-wise: if z1=(a,b)z_1 = (a, b) and z2=(c,d)z_2 = (c, d), then z1+z2=(a+c,b+d)z_1 + z_2 = (a + c, b + d).

The geometric interpretation is the parallelogram rule. Place z1z_1 and z2z_2 as arrows from the origin. The sum is the diagonal of the parallelogram they form. Equivalently, translate z2z_2 so its tail sits at the tip of z1z_1 — the arrow from the origin to the new tip of z2z_2 is z1+z2z_1 + z_2.

Complex addition is commutative (z1+z2=z2+z1z_1 + z_2 = z_2 + z_1) and associative. It also preserves the structure of the complex plane — the sum of two complex numbers is always another complex number.

How Complex Subtraction Works

Subtracting complex numbers also works component-wise:

(a+bi)−(c+di)=(a−c)+(b−d)i(a + bi) - (c + di) = (a - c) + (b - d)i


The key geometric insight is that z1−z2z_1 - z_2 equals the vector from the point z2z_2 to the point z1z_1. Its modulus ∣z1−z2∣|z_1 - z_2| is the Euclidean distance between the two points on the complex plane.

Subtraction can always be rewritten as adding the negation: z1−z2=z1+(−z2)z_1 - z_2 = z_1 + (-z_2), where −z2=−c−di-z_2 = -c - di is z2z_2 reflected through the origin (rotated by 180°180°). The visualizer shows this as the faint dashed −z2-z_2 ghost vector.

Two important special cases: when z2z_2 is the conjugate of z1z_1 (same real part, opposite imaginary part), the difference is purely imaginary (2bi2bi). When z1=z2z_1 = z_2, the difference is zero.

Conjugate Pairs in Addition and Subtraction

Click the Conjugate pair preset to load z1=3+2iz_1 = 3 + 2i and z2=3−2iz_2 = 3 - 2i. These are complex conjugates — they share the same real part but have opposite imaginary parts.

Adding conjugates always cancels the imaginary components: (3+2i)+(3−2i)=6(3 + 2i) + (3 - 2i) = 6. The sum is a pure real number lying on the real axis. In the visualizer, the green sum point sits exactly on the horizontal axis.

Subtracting conjugates always cancels the real components: (3+2i)−(3−2i)=4i(3 + 2i) - (3 - 2i) = 4i. The difference is a pure imaginary number on the vertical axis. The purple vector points straight up.

The dashed line connecting z1z_1 and z2z_2 is vertical — conjugates are always symmetric about the real axis. This pair produces one of the cleanest illustrations: real-axis sum, imaginary-axis difference, and a perfectly symmetric parallelogram.
ReIm-5−5i-4−4i-3−3i-2−2i-1−1i11i22i33i44i55iz₁+z₂ →z₁−z₂−z₂z₁z₂
Conjugate pair — (3+2i) & (3−2i), frozen

The sum 6 has left the ±5 window, so a dashed teal ray points along the real axis toward it; the purple difference 4i climbs the imaginary axis. One result per axis — the conjugate split.

The frozen frame also demonstrates the tool's off-screen behavior: the sum 66 exceeds the ±5\pm 5 window, so the green vector is replaced by a dashed ray aimed along the positive real axis — the direction is exact even when the endpoint is out of view.

The general rule behind the split: z+zˉ=2 Re(z)z + \bar{z} = 2\,\text{Re}(z) and z−zˉ=2 Im(z) iz - \bar{z} = 2\,\text{Im}(z)\,i. Sum and difference of a conjugate pair are the real and imaginary parts of zz, each doubled — which is why one diagonal always lies on each axis. The mirror pair preset produces the same axis-split with the roles reversed.

Related Concepts and Tools

Explore the full suite of complex number visual tools — each one targets a different operation or concept on the complex plane.

Complex Numbers — foundational theory covering the imaginary unit ii, rectangular form a+bia + bi, modulus, argument, and the structure of the complex plane. Start here if you are new to the topic.

Complex Number Explorer — a general-purpose interactive tool for plotting complex numbers, adjusting their real and imaginary parts, and observing how they behave under basic operations.

Polar & Rectangular Converter — switch between rectangular form a+bia + bi and polar form reihetar e^{i heta}. Polar form simplifies multiplication and division; rectangular form, used in this tool, simplifies addition and subtraction.

Multiplication Visualizer — see how multiplying two complex numbers stretches and rotates vectors on the plane. The moduli multiply and the arguments add.

Division Visualizer — visualize complex division as the inverse of multiplication: the moduli divide and the arguments subtract.

Complex Conjugate Explorer — explore the conjugate zˉ=a−bi\bar{z} = a - bi and its geometric meaning as a reflection across the real axis. Conjugates appear throughout complex arithmetic and are essential for division.

Distance & Midpoint Tool — compute and visualize the distance ∣z1−z2∣|z_1 - z_2| and midpoint between two complex numbers, connecting complex arithmetic to Euclidean geometry.

De Moivre's Theorem Visualizer — animate the identity (reiheta)n=rneinheta(r e^{i heta})^n = r^n e^{in heta} and watch how raising a complex number to a power scales its modulus and multiplies its angle.

Powers of i Calculator — compute ini^n for any integer using the mod 4 cycle i,−1,−i,1i, -1, -i, 1 and see where each power lands on the unit circle.

Euler's Formula Explorer — visualize eiheta=cosheta+isinhetae^{i heta} = cos heta + isin heta and see how the exponential function traces the unit circle, connecting trigonometry, complex numbers, and the number ee.