Argand plane representation

Every complex number $z=a+ib$ has an algebraic life as a pair of real numbers and a geometric life as a single point. The Argand plane is the bridge between the two: plot $a$ along a horizontal real axis and $b$ along a vertical imaginary axis, and $z$ becomes the point $P(a,b)$ — or, just as often, the arrow from the origin $O$ to $P$ . Once this picture is in place, abstract facts about complex numbers turn into facts you can literally see: the modulus $|z|$ becomes a distance, the conjugate $\bar z$ becomes a mirror-image point, and addition becomes the familiar tip-to-tail rule for vectors. It also lets an algebraic fact about equality or inequality be read off at a glance — for instance, $|z|\ge0$ with equality only at $z=0$ simply says every point except the origin sits at some strictly positive distance from $O$ , and $|-z|=|z|$ says the points for $z$ and $-z$ are equally far from the origin, on exactly opposite sides of it. This section stays inside the CBSE Class 11 scope of plotting points and reading off distances between them — it does not venture into loci or complex-geometry problems, which come later.

The core insight

Identifying $z=a+ib$ with the point $(a,b)$ on the Argand plane turns modulus into distance from the origin, conjugation into reflection in the real axis, and addition into vector addition.

The core idea

The correspondence. Every complex number $z=a+ib$ is matched with exactly one ordered pair $(a,b)$ , and plotted as one point on a coordinate plane called the Argand plane (or complex plane). The horizontal axis is the real axis, carrying $\operatorname{Re}(z)=a$ ; the vertical axis is the imaginary axis, carrying $\operatorname{Im}(z)=b$ . This matching is one-to-one both ways: every complex number gives one point, and every point gives back one complex number.

Modulus as distance. Because $|z|=\sqrt{a^2+b^2}$ is exactly the distance formula from $(0,0)$ to $(a,b)$ , the modulus of $z$ is the distance $OP$ of its point from the origin:

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

where: $a=\operatorname{Re}(z)$ and $b=\operatorname{Im}(z)$ are the real and imaginary parts of $z$ , and $O$ is the origin $(0,0)$ .

Conjugate as reflection. Since $\bar z=a-ib$ corresponds to $(a,-b)$ , the point for $\bar z$ is the mirror image of the point for $z$ in the real axis — only the sign of the second coordinate flips.

Addition as vector addition. The point for $z_1+z_2$ has coordinates found by adding the coordinates of $z_1$ and $z_2$ separately — precisely the rule for adding two vectors tip-to-tail, so the Argand-plane picture of a sum is the vector sum of the two arrows.

Purely real and purely imaginary numbers. A real number $a+i0$ has zero imaginary part, so its point $(a,0)$ always lies on the real axis; a purely imaginary number $0+ib$ has zero real part, so its point $(0,b)$ always lies on the imaginary axis. Neither is "unplottable" — every complex number, however special, has some definite location on the Argand plane, even when that location happens to sit exactly on one of the two axes rather than off in a quadrant.

Figure

Three points plotted on the Argand plane in different quadrants and on an axis with arrows from the origin

Derivation ADistance between two points on the Argand plane

The goal here is to extend the single-point idea, distance of one point from the origin equals its modulus, to the distance between any two points on the Argand plane, which turns out to be the modulus of the difference of the two complex numbers they represent, and this small extension is what makes the geometric picture useful for more than plotting isolated points or measuring how far a single point sits from the origin.

Step 1. Let $z_1=a+ib$ and $z_2=c+id$ correspond to points $P_1(a,b)$ and $P_2(c,d)$ on the Argand plane. The ordinary coordinate-geometry distance formula gives the distance between these two points as $P_1P_2=\sqrt{(a-c)^2+(b-d)^2}$ .

Step 2. Now compute the complex number $z_1-z_2=(a-c)+i(b-d)$ by simple subtraction of real and imaginary parts, and take its modulus using the same rule used for a single point: $|z_1-z_2|=\sqrt{(a-c)^2+(b-d)^2}$ .

Step 3. Comparing the two expressions, they are identical, so

P_1P_2=|z_1-z_2|.

where: $z_1,z_2$ are the two complex numbers, $P_1,P_2$ are their Argand-plane points, and $|z_1-z_2|$ is the modulus of their difference.

Step 4. Read the result geometrically: the distance between any two points on the Argand plane is simply the modulus of the complex number obtained by subtracting one from the other, in exactly the same way $|z|$ alone gives the distance of $z$ from the origin $0$ (since taking $z_2=0$ recovers $P_1O=|z_1|$ ). It also matches ordinary intuition about vectors: $z_1-z_2$ is the arrow that carries $P_2$ to $P_1$ , and its length, its modulus, is naturally the separation between the two points, regardless of where the two points happen to sit relative to the axes.

Worked example

Problem: Using $z_1=5-2i$ and $z_2=1+2i$ , find the point representing $\bar z_1$ , the modulus $|z_1|$ , and the distance between the points representing $z_1$ and $z_2$ on the Argand plane.

Step 1. $z_1=5-2i$ corresponds to the point $(5,-2)$ . Its conjugate is $\bar z_1=5+2i$ , corresponding to $(5,2)$ — the reflection of $(5,-2)$ across the real axis, since only the imaginary part's sign changes.

Step 2. The modulus of $z_1$ : $|z_1|=\sqrt{5^2+(-2)^2}=\sqrt{25+4}=\sqrt{29}\approx5.39$ , so the point $(5,-2)$ sits $\sqrt{29}$ units from the origin.

Step 3. For the distance between $z_1$ and $z_2$ , first subtract: $z_1-z_2=(5-1)+i(-2-2)=4-4i$ . Then take the modulus: $|z_1-z_2|=\sqrt{4^2+(-4)^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt2$ .

Final answer: $\bar z_1$ is the point $(5,2)$ , $|z_1|=\sqrt{29}\approx5.39$ , and the distance between the points for $z_1=5-2i$ and $z_2=1+2i$ is $\boldsymbol{4\sqrt2\approx5.66}$ units.

Competitive stretch

Problem: For which point on the real axis is the distance from $z_1=5-2i$ equal to $\sqrt{29}$ , the same as $|z_1|$ itself?

Step 1. A general point on the real axis is $w=x+0i$ , so $x$ is real and its imaginary part is $0$ .

Step 2. Require $|z_1-w|=\sqrt{29}$ : $z_1-w=(5-x)-2i$ , so $|z_1-w|=\sqrt{(5-x)^2+(-2)^2}=\sqrt{(5-x)^2+4}$ .

Step 3. Set this equal to $\sqrt{29}$ and square both sides: $(5-x)^2+4=29$ , so $(5-x)^2=25$ , giving $5-x=\pm5$ , hence $x=0$ or $x=10$ .

Final answer: the points $\boldsymbol{w=0}$ and $\boldsymbol{w=10}$ on the real axis are each at distance $\sqrt{29}$ from $z_1=5-2i$ — one of them, $w=0$ , is the origin itself, matching $|z_1|=\sqrt{29}$ directly.

Common traps

Writing $a+ib$ as the point $(b,a)$ instead of $(a,b)$ — the real part is always the horizontal coordinate, the imaginary part is always the vertical coordinate.
Forgetting that a purely real number such as $z=5$ still plots as a definite point, $(5,0)$ on the real axis, not as "nothing to plot."
Computing the distance between two Argand-plane points as $|z_1|-|z_2|$ or $|z_1|+|z_2|$ instead of the correct $|z_1-z_2|$ .
Mixing up reflection in the real axis (which gives $\bar z$ , point $(a,-b)$ ) with reflection through the origin (which gives $-z$ , point $(-a,-b)$ ) — these are two different transformations.
Thinking the imaginary axis is graduated in $i,2i,3i,\dots$ as though $i$ itself were a length — it is graduated in ordinary real numbers $1,2,3,\dots$ , with the understanding that a mark at $b$ represents the imaginary part $ib$ .

Ask the tutor: want a walkthrough of plotting a fresh complex number and finding its distance from another point on the Argand plane? Ask the AI tutor for a step-by-step example.

Why does the distance formula from coordinate geometry also work for complex numbers?

Because a complex number and a point in the plane are just two names for the same ordered pair. Once $z=a+ib$ is identified with $(a,b)$ , any formula that works for points, like the distance formula, automatically works for complex numbers, since $z_1-z_2$ collects exactly the coordinate differences $(a-c)$ and $(b-d)$ that the distance formula squares and adds. There is no new geometry here, only a relabelling of familiar coordinate geometry in complex-number language, which is exactly why this section leans so heavily on skills you already have from plotting ordinary $(x,y)$ points.

Is the Argand plane the same as the ordinary xy-plane we use in coordinate geometry?

As a picture, yes — same two perpendicular axes, same grid, same distance formula. The only difference is what the axes mean: instead of two independent real-number coordinates, the horizontal axis is the real part and the vertical axis is the imaginary part of a single complex number. So every point still has an $x$ and $y$ value, but now that pair is read together as one number $a+ib$ , not as two separate quantities.

Do I need to worry about which quadrant a complex number lies in for this chapter?

You should be able to identify the quadrant, or axis, from the signs of $a$ and $b$ , since it is quick and reinforces the point-plotting idea, and worked examples often ask for it. Both parts positive puts the point in the first quadrant, real part negative with imaginary part positive puts it in the second, and so on around the plane, exactly as with ordinary $(x,y)$ points. But this chapter does not go further into quadrant-based reasoning, such as tracking how quadrants affect angles, which belongs to a later idea (argument of a complex number) outside this section's scope.

Argand plane representation

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads