Polar Form of Complex Numbers

Every complex number \(z = a + bi\) can be plotted as a single point on the complex plane, where \(a\) is the horizontal (real) coordinate and \(b\) is the vertical (imaginary) coordinate. Instead of describing that point with a horizontal and vertical distance, polar form describes it with two different pieces of information: how far the point is from the origin, and what angle it makes with the positive real axis.

These two quantities are called the modulus \(r\) and the argument \(\theta\). Once you know them, the complex number can be written as:

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

This expression is often abbreviated as \(z = r\,\)cis\(\,\theta\). If you have already studied the angle and absolute value of complex numbers, you already know how to find \(r\) and \(\theta\) individually. Polar form simply combines them into one compact expression for the number itself.

Picture the complex number \(a + bi\) as a point on the plane. Draw a segment from the origin to that point: its length is \(r\), and the angle it sweeps from the positive real axis is \(\theta\). This turns the point into the vertex of a right triangle, with legs \(a\) and \(b\) and hypotenuse \(r\).

Given \(z = a + bi\), the right triangle relationships give:

\( r = \sqrt{a^2 + b^2} \)

\( \theta = \tan^{-1}\left(\dfrac{b}{a}\right) \)

This distance formula for \(r\) works the same way as the distance formula you use to find the distance between two points, since the origin and \((a, b)\) are just two points on a coordinate grid. Finding \(\theta\) uses the tangent ratio, and you must check which quadrant the point lies in, since a calculator's inverse tangent only returns angles between negative 90 and 90 degrees.

Example 1: Convert \(3 + 4i\) to polar form

Here \(a = 3\) and \(b = 4\), both positive, so the point lies in the first quadrant.

\( r = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \)

\( \theta = \tan^{-1}\left(\dfrac{4}{3}\right) \approx 53.13^\circ \)

So \(3 + 4i = 5(\cos 53.13^\circ + i\sin 53.13^\circ)\).

Example 2: Convert \(-1 + i\) to polar form

Here \(a = -1\) and \(b = 1\), so the point lies in the second quadrant. The reference angle comes from a 45-45-90 triangle, since the two legs are equal in size (ignoring sign). This is the same triangle you use when solving expressions using 45-45-90 special right triangles.

\( r = \sqrt{(-1)^2 + 1^2} = \sqrt{2} \)

Since the reference angle is \(45^\circ\) and the point is in the second quadrant, \(\theta = 180^\circ - 45^\circ = 135^\circ\).

So \(-1 + i = \sqrt{2}(\cos 135^\circ + i\sin 135^\circ)\).

To go the other direction, evaluate the cosine and sine and distribute \(r\):

\( a = r\cos\theta \)

\( b = r\sin\theta \)

Example 3: Convert \(2(\cos 60^\circ + i\sin 60^\circ)\) to rectangular form

This angle appears in a 30-60-90 triangle, the same kind used when solving expressions using 30-60-90 special right triangles.

\( a = 2\cos 60^\circ = 2 \times \dfrac{1}{2} = 1 \)

\( b = 2\sin 60^\circ = 2 \times \dfrac{\sqrt{3}}{2} = \sqrt{3} \)

So \(2(\cos 60^\circ + i\sin 60^\circ) = 1 + \sqrt{3}i\).

Polar form is especially useful when multiplying, dividing, or raising complex numbers to powers, since those operations become simple rules involving the moduli and arguments instead of full algebraic expansion. Being comfortable moving between rectangular form \(a + bi\) and polar form \(r(\cos\theta + i\sin\theta)\) is the key skill that makes all of these later techniques possible.

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