Polar representation

Every complex number $z=a+ib$ can be plotted as a point $P(a,b)$ on the Argand plane, and Cartesian coordinates $(a,b)$ are not the only address for that point. Polar representation describes $P$ instead by its distance $r$ from the origin and the angle $\theta$ it makes with the positive real axis, giving the polar form $z=r(\cos\theta+i\sin\theta)$ . This is a genuinely useful second language for complex numbers — it is the same information as $a+ib$ , repackaged so that geometry (distance and direction) becomes visible at a glance. CBSE Class 11 treats this topic as formative only: you must be comfortable converting between $a+ib$ and $r(\cos\theta+i\sin\theta)$ and reading off the right quadrant for $\theta$ , but nothing heavier — such as De Moivre's theorem or $n$ -th roots — is examined at this stage.

The core insight

Writing $z=a+ib$ as $z=r(\cos\theta+i\sin\theta)$ , with $r=|z|=\sqrt{a^2+b^2}$ and $\theta$ the angle $OP$ makes with the positive real axis, swaps Cartesian coordinates for a distance-and-direction description of the same point on the Argand plane.

The core idea

Plot $z=a+ib$ as the point $P(a,b)$ on the Argand plane, with the horizontal axis carrying the real part and the vertical axis carrying the imaginary part. Two numbers now fix the same point $P$ just as well as $(a,b)$ does:

$r$ , the distance of $P$ from the origin $O$ — this is exactly the modulus of $z$ , so $r=|z|$ ;
$\theta$ , the angle the segment $OP$ makes with the positive real axis, called the argument of $z$ .

Drop a perpendicular from $P$ to the real axis and you get a right triangle with hypotenuse $r$ and angle $\theta$ at the origin. Ordinary trigonometry of that triangle gives:

a=r\cos\theta, \qquad b=r\sin\theta,

where: $a$ and $b$ are the real and imaginary parts of $z$ , $r=\sqrt{a^2+b^2}$ is the modulus, and $\theta$ is the argument, measured anticlockwise from the positive real axis when positive. Substituting these two equations back into $z=a+ib$ produces the polar form itself:

z=r(\cos\theta+i\sin\theta).

By convention $\theta$ is chosen in the range $-\pi<\theta\le\pi$ (equivalently, $-180^\circ$ to $180^\circ$ ), which fixes it uniquely; this restricted choice is the principal argument. For a nonzero $z$ , the size of $\theta$ comes from $\tan\theta=\dfrac{b}{a}$ , but the sign and the actual quadrant come only from checking the signs of $a$ and $b$ separately — the geometric picture below makes this concrete.

Figure

The Argand plane showing point P at distance r from the origin at angle theta from the positive real axis with the right triangle legs a and b marked

Two facts round out the picture. For $z=0$ , the modulus is $0$ and the argument is undefined — there is no meaningful direction from the origin to itself. And the two forms are fully interchangeable: given $(a,b)$ you get $(r,\theta)$ via $r=\sqrt{a^2+b^2}$ and a quadrant check, and given $(r,\theta)$ you get $(a,b)$ back via $a=r\cos\theta,\ b=r\sin\theta$ . This chapter also notes, without proof, that the Fundamental Theorem of Algebra guarantees every polynomial equation has a root among the complex numbers; polar form becomes far more powerful once you meet De Moivre's theorem later, but at the CBSE Class 11 level the goal is simply fluent conversion and a clear geometric picture of what $r$ and $\theta$ mean. It is worth pausing on why this second description is worth learning at all. Cartesian coordinates $(a,b)$ tell you how far to move right or left, and how far to move up or down, to reach the point $P$ from the origin — two independent, perpendicular instructions. Polar coordinates $(r,\theta)$ instead give a single distance to travel outward and a single direction to travel in, which is often the more natural way to think about rotation, symmetry or distance-based problems. Both descriptions pin down exactly the same point, so neither is "more correct" than the other; polar form is simply the version that makes $r$ and $\theta$ — the two quantities that matter for geometric reasoning on the Argand plane — visible directly in the formula instead of hidden inside a sum of squares and an inverse tangent.

Derivation AConverting a+ib into polar form r(cos theta + i sin theta)

The goal is to derive, from nothing but the right-triangle definition of sine and cosine and the Pythagorean theorem, exactly how the two real numbers $a$ and $b$ that specify $z=a+ib$ translate into the two polar numbers $r$ and $\theta$ that specify the same point, and to state precisely how the quadrant of $(a,b)$ fixes which $\theta$ is correct rather than just its reference angle.

Step 1. Plot $z=a+ib$ as the point $P(a,b)$ on the Argand plane and drop a perpendicular from $P$ to the real axis, meeting it at $N(a,0)$ . Triangle $OPN$ is right-angled at $N$ , with $ON=a$ , $NP=b$ and hypotenuse $OP=r$ .

Step 2. Apply the Pythagorean theorem to triangle $OPN$ : $r^2=ON^2+NP^2=a^2+b^2$ , so $r=\sqrt{a^2+b^2}$ . This is just the distance formula from the origin to $(a,b)$ , and it matches the modulus $|z|$ you already know from Cartesian form.

Step 3. Read off the trigonometric ratios of angle $\theta=\angle NOP$ in the right triangle: $\cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}}=\dfrac{a}{r}$ and $\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}=\dfrac{b}{r}$ , which rearrange to $a=r\cos\theta$ and $b=r\sin\theta$ .

Step 4. Substitute these two expressions into $z=a+ib$ : $z=r\cos\theta+i(r\sin\theta)=r(\cos\theta+i\sin\theta)$ , which is the polar form. To actually find $\theta$ for a given $z$ , first compute the reference angle from $\tan(\text{ref})=\left|\dfrac{b}{a}\right|$ , then place it in the correct quadrant using the signs of $a$ and $b$ : both positive gives $\theta$ in $\left(0,\dfrac{\pi}{2}\right)$ ; $a$ negative, $b$ positive gives $\theta=\pi-\text{ref}$ ; both negative gives $\theta=-(\pi-\text{ref})$ ; $a$ positive, $b$ negative gives $\theta=-\text{ref}$ — always keeping $\theta$ within the principal range $-\pi<\theta\le\pi$ .

Worked example

Problem: Write $z=1+i$ in polar form, then use the same method on $z=-\sqrt3+i$ .

Step 1. For $z=1+i$ : here $a=1$ , $b=1$ , so $r=\sqrt{1^2+1^2}=\sqrt2$ .

Step 2. Both $a>0$ and $b>0$ , so $(1,1)$ sits in the first quadrant and $\theta$ lies between $0$ and $\dfrac{\pi}{2}$ . Since $\cos\theta=\dfrac{1}{\sqrt2}$ and $\sin\theta=\dfrac{1}{\sqrt2}$ , $\theta=\dfrac{\pi}{4}$ . So $z=1+i=\sqrt2\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)$ .

Step 3. For $z=-\sqrt3+i$ : here $a=-\sqrt3$ , $b=1$ , so $r=\sqrt{(-\sqrt3)^2+1^2}=\sqrt{3+1}=2$ . Since $a<0$ and $b>0$ , the point lies in the second quadrant.

Step 4. The reference angle satisfies $\cos(\text{ref})=\dfrac{\sqrt3}{2}$ , giving $\text{ref}=\dfrac{\pi}{6}$ ; placed in the second quadrant, $\theta=\pi-\dfrac{\pi}{6}=\dfrac{5\pi}{6}$ . So $z=-\sqrt3+i=2\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right)$ .

Competitive stretch

Problem: Convert the polar form $z=4\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right)$ back into standard $a+ib$ form, to check the round trip works both ways.

Step 1. Here $r=4$ and $\theta=\dfrac{\pi}{3}$ , so use $a=r\cos\theta$ and $b=r\sin\theta$ directly. $\cos\dfrac{\pi}{3}=\dfrac{1}{2}$ and $\sin\dfrac{\pi}{3}=\dfrac{\sqrt3}{2}$ .

Step 2. Substitute: $a=4\times\dfrac{1}{2}=2$ and $b=4\times\dfrac{\sqrt3}{2}=2\sqrt3$ . So $z=2+2\sqrt3\,i$ , and you can check $r=\sqrt{2^2+(2\sqrt3)^2}=\sqrt{4+12}=\sqrt{16}=4$ , matching the given modulus exactly.

Common traps

Computing $r$ correctly but guessing $\theta$ from the reference angle alone, without checking the quadrant of $(a,b)$ — the same reference angle can correspond to four different values of $\theta$ .
Writing " $r,\theta$ " as the polar form without the $\cos\theta+i\sin\theta$ structure — polar form is still a complex number, $z=r(\cos\theta+i\sin\theta)$ , not just a coordinate pair.
Allowing $r$ to be negative — the modulus $r=|z|$ must always be non-negative; a negative $r$ is never valid in this form.
Reaching for De Moivre's theorem or $n$ -th roots on a polar-form problem — these lie outside the CBSE Class 11 syllabus, where polar form is used only for basic conversion and geometric understanding.

Ask the tutor: Want to practise converting a few more complex numbers to and from polar form, or double-check which quadrant your angle falls in?

Why do we even need polar form if a+ib already describes the complex number completely?

Cartesian form $a+ib$ is complete, but it hides geometric information that polar form makes obvious at a glance: how far the point is from the origin ( $r$ ) and in which direction ( $\theta$ ). For CBSE Class 11 this connects complex numbers cleanly to the Argand-plane picture you already draw, and it is the natural stepping stone toward more powerful tools like De Moivre's theorem that you meet later, even though those are not tested yet.

How do I know which quadrant to place theta in without a diagram in the exam?

Look only at the signs of $a$ and $b$ in $z=a+ib$ : both positive means quadrant one ( $0$ to $\pi/2$ ), $a$ negative with $b$ positive means quadrant two ( $\pi/2$ to $\pi$ ), both negative means quadrant three (use negative angles, $-\pi$ to $-\pi/2$ ), and $a$ positive with $b$ negative means quadrant four ( $-\pi/2$ to $0$ ). Compute the reference angle from $\tan(\text{ref})=|b/a|$ first, then place it using this sign check.

Is polar form going to be examined with heavy questions, like finding roots or powers of complex numbers, and does the argument change if I measure clockwise instead of anticlockwise?

No, not at the CBSE Class 11 level. The syllabus marks polar representation as formative only, meaning it builds conceptual understanding rather than being drilled with rigorous problems. You are expected to convert simple complex numbers like $1+i$ or $-\sqrt3+i$ between $a+ib$ and $r(\cos\theta+i\sin\theta)$ , and to know what $r$ and $\theta$ represent geometrically — heavier machinery such as De Moivre's theorem or $n$ -th roots is outside this syllabus. On the sign question: the principal argument is measured anticlockwise from the positive real axis for a positive angle and clockwise for a negative angle, always staying within $-\pi<\theta\le\pi$ ; a point in the third or fourth quadrant is usually easiest to describe with a negative $\theta$ rather than adding $2\pi$ to push a positive angle beyond $\pi$ , since the principal range never allows $\theta$ to exceed $\pi$ or drop below $-\pi$ .

Polar representation

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads