The Argand Plane

Complex Numbers as Points

Every ordered pair (x, y) of reals marks a unique point of the coordinate plane — and every complex number $z = x + iy$ IS essentially an ordered pair. Identify them:

Key Point: The complex number $x + iy$ is represented by the point P(x, y). The plane with a complex number attached to each of its points is called the complex plane or the Argand plane.

A gallery of examples: $2 + 4i \to (2, 4)$, $-2 + 3i \to (-2, 3)$, $i \to (0, 1)$, $2 \to (2, 0)$, $-5 - 2i \to (-5, -2)$, $1 - 2i \to (1, -2)$.

• Points on the x-axis are the real numbers $a + i0$ — so it is called the real axis.
• Points on the y-axis are the purely imaginary numbers $0 + ib$ — the imaginary axis.

[Board Important] Plotting complex numbers is a direct 1-mark skill; naming the axes correctly ("real axis", "imaginary axis") earns the second mark.

Modulus as Distance, Conjugate as Mirror

Distance

In the Argand plane, the modulus gets a geometric identity:

$$|z| = \sqrt{x^2 + y^2} = \text{distance of the point P(x, y) from the origin}$$

More generally, $|z_1 - z_2|$ is the distance BETWEEN the points $z_1$ and $z_2$ — which turns modulus equations into geometry: $|z| = 3$ describes the circle of radius 3 centred at the origin, and $|z - (1 + i)| = 2$ the circle of radius 2 centred at (1, 1).

Reflection

The conjugate also turns geometric:

Key Point: $z = x + iy$ sits at P(x, y) and $\bar{z} = x - iy$ at Q(x, $-y$) — the mirror image of P in the real axis.

Consequences read straight off the picture: $|z| = |\bar{z}|$ (reflection preserves distance), real numbers are their own mirror images, and the four numbers $z, -z, \bar{z}, -\bar{z}$ sit at the four "corners" equidistant from O.

[JEE Tip] Translate fluently in both directions: $|z - z_0| = r$ ↔ circle; $|z - a| = |z - b|$ ↔ perpendicular bisector of the segment ab; Im z > 0 ↔ upper half-plane. JEE locus questions are exactly this dictionary.

Solved Examples

Example 1: Plotting practice

Locate the points representing $2 + 4i$, $-2 + 3i$, $0 + 1i$, $2 + 0i$, $-5 - 2i$ and $1 - 2i$, stating the quadrant (or axis) of each.

Solution:

Step 1 — Convert each number to its coordinate pair. $z = x + iy$ sits at (x, y): $2 + 4i \to (2, 4)$, $-2 + 3i \to (-2, 3)$, $i \to (0, 1)$, $2 \to (2, 0)$, $-5 - 2i \to (-5, -2)$, $1 - 2i \to (1, -2)$.

Step 2 — Read the quadrant from the signs. $(2, 4)$: both positive — quadrant I. $(-2, 3)$: quadrant II. $(-5, -2)$: quadrant III. $(1, -2)$: quadrant IV.

Step 3 — Handle the axis cases. $(0, 1)$ lies on the imaginary axis; $(2, 0)$ on the real axis — neither belongs to a quadrant.

Takeaway: Sign of the real part ↔ left/right; sign of the imaginary part ↔ up/down — identical to ordinary coordinates.

Example 2: Distances from the origin

Find the distance of each from the origin: (i) $3 + 4i$ (ii) $-5 + 12i$ (iii) $-7$ (iv) $2i$.

Solution:

Step 1 — (i) Modulus = distance. $|3 + 4i| = \sqrt{9 + 16} = \sqrt{25} = 5$.

Step 2 — (ii) Same formula. $\sqrt{25 + 144} = \sqrt{169} = 13$.

Step 3 — (iii) A real number. $|-7| = 7$ — distance along the real axis.

Step 4 — (iv) A purely imaginary number. $|2i| = 2$ — distance along the imaginary axis.

Takeaway: Modulus = distance; the 3-4-5 and 5-12-13 triangles live in the Argand plane too.

Example 3: Distance between two complex numbers

Find the distance between $z_1 = 3 + 4i$ and $z_2 = -1 + i$.

Solution:

Step 1 — Subtract. $z_1 - z_2 = (3 - (-1)) + (4 - 1)i = 4 + 3i$.

Step 2 — Take the modulus. $|4 + 3i| = \sqrt{16 + 9} = 5$.

Takeaway: $|z_1 - z_2|$ IS the distance formula — no new machinery needed.

Example 4: Mirror images

Plot z = $3 + 2i$, $\bar{z}$, $-z$ and $-\bar{z}$, and describe the symmetry.

Solution:

Step 1 — Compute the four points. z at (3, 2); $\bar{z} = 3 - 2i$ at (3, $-2$); $-z = -3 - 2i$ at ($-3$, $-2$); $-\bar{z} = -3 + 2i$ at ($-3$, 2).

Step 2 — Name each reflection. $\bar{z}$ is z reflected in the REAL axis; $-z$ is z reflected through the ORIGIN; $-\bar{z}$ is z reflected in the IMAGINARY axis.

Step 3 — Observe the common modulus. All four have modulus $\sqrt{9 + 4} = \sqrt{13}$ — corners of a rectangle centred at O.

Takeaway: The four sign-conjugate variants of z make a rectangle; each reflection has its own algebraic name.

Example 5: Which numbers lie on an axis?

For what real x does $z = (x - 2) + i(x^2 - 4)$ lie (i) on the real axis (ii) on the imaginary axis?

Solution:

Step 1 — (i) Real axis means Im z = 0. $x^2 - 4 = 0$ gives $x = \pm 2$.

Step 2 — (ii) Imaginary axis means Re z = 0. $x - 2 = 0$ gives $x = 2$.

Step 3 — Notice the overlap. At x = 2 both parts vanish: z is the ORIGIN, which lies on both axes simultaneously.

Takeaway: Axis membership is a one-part condition; the origin satisfies both.

Example 6: The circle |z| = 2

Describe the set of points z with $|z| = 2$, and check whether $1 + i$ and $\sqrt{2} + \sqrt{2}i$ belong to it.

Solution:

Step 1 — Translate the condition. $|z| = 2$ collects all points at distance 2 from the origin — the circle of radius 2 centred at O.

Step 2 — Test the first candidate. $|1 + i| = \sqrt{1 + 1} = \sqrt{2} \neq 2$ — not on the circle (it lies inside).

Step 3 — Test the second. $|\sqrt{2} + \sqrt{2}i| = \sqrt{2 + 2} = 2$ ✓ — on the circle.

Takeaway: Modulus conditions are membership tests for circles — compute and compare.

Example 7: A locus as a perpendicular bisector

Describe the set of z with $|z - 1| = |z + 1|$.

Solution:

Step 1 — Read the two moduli as distances. $|z - 1|$ is the distance from z to the point 1 (at (1, 0)); $|z + 1| = |z - (-1)|$ is the distance to $-1$ (at ($-1$, 0)).

Step 2 — Interpret the equality. Points equidistant from two fixed points form the PERPENDICULAR BISECTOR of the segment joining them — here, the imaginary axis (Re z = 0).

Step 3 — Confirm by algebra. $(x-1)^2 + y^2 = (x+1)^2 + y^2$ gives $-2x = 2x$, so $x = 0$ ✓.

Takeaway: Equal distances = perpendicular bisector; algebra and geometry give the same one-line answer.

Example 8: Conjugate geometry in a computation

z = $x + iy$ satisfies $z + \bar{z} = 6$ and $z - \bar{z} = 4i$. Find z and plot it.

Solution:

Step 1 — Use the sum. $z + \bar{z} = 2x = 6$, so $x = 3$.

Step 2 — Use the difference. $z - \bar{z} = 2iy = 4i$, so $y = 2$.

Step 3 — Assemble and place. $z = 3 + 2i$ — the point (3, 2) in quadrant I.

Takeaway: $z + \bar{z}$ and $z - \bar{z}$ isolate the two coordinates — the conjugate is a coordinate-extraction tool.

Example 9: Regions in the plane

Shade-describe the sets: (i) Im z > 0 (ii) $|z| \leq 1$ (iii) Re z = 2.

Solution:

Step 1 — (i) Translate. Im z is the y-coordinate; y > 0 is the OPEN upper half-plane (boundary excluded — the inequality is strict).

Step 2 — (ii) Translate. Distance from O at most 1: the CLOSED unit disc — the circle $|z| = 1$ together with its interior.

Step 3 — (iii) Translate. Re z = 2 fixes x = 2: the vertical line through (2, 0).

Takeaway: Equations give curves; inequalities give regions. The boundary belongs to the set exactly when the inequality is non-strict.

Example 10: Modulus from the plane, quadrant from signs

z = $-1 + \sqrt{3}i$. Find |z|, the quadrant of z, and the mirror image of z in the real axis.

Solution:

Step 1 — Modulus. $|z| = \sqrt{1 + 3} = 2$.

Step 2 — Quadrant. The point ($-1$, $\sqrt{3}$) has negative x, positive y: quadrant II.

Step 3 — Mirror image. Reflection in the real axis is the conjugate: $\bar{z} = -1 - \sqrt{3}i$, in quadrant III.

Takeaway: This z is a JEE regular ($2\omega$ in disguise!); its geometry — modulus 2, 120° from the positive real axis — previews the polar form in the JEE Corner.