Topic 3.13B

Trigonometry and Polar Coordinates

A complex number $a + bi$ has two parts, a real part a and an imaginary part $b$ , where $i^{2} = -1$ . Two numbers call for two coordinates, so every complex number can be drawn as a point in a plane. And once it is a point, Part A tells us how to describe it with a distance and an angle instead.

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01
$a + bi$ ↔ point $(a, b)$ ↔ $r(\cos \theta + i \sin \theta)$ : $a = r \cos \theta$ , $b = r \sin \theta$ .

Practice 41 problems Next: Chapter 3.14A Polar Function Graphs

The Complex Plane

CONCEPT

Complex numbers as points

The complex number $a + bi$ is plotted as the point $(a, b)$ .

The horizontal axis is the real axis (Re) and the vertical axis is the imaginary axis (Im).

Real numbers such as 4 lie on the real axis; pure imaginary numbers such as −3i lie on the imaginary axis.

The modulus $|a + bi| = \sqrt{a^{2} + b^{2}}$ is the distance from 0 to the point, just like $r$ in Part A. For example, $|5 + 12i| = \sqrt{169} = 13$ .

Because $a + bi$ sits at the point $(a, b)$ , the conversions from Part A apply directly: $a = r \cos \theta$ and $b = r \sin \theta$ . Substituting gives the polar form of a complex number.

z=a+bi=r(\cos\theta+i\,\sin\theta),\qquad r=|z|=\sqrt{a^2+b^2}

Here $r$ is the modulus, and $\theta$ is an argument of $z$ : the angle from the positive real axis to the point. Like polar points, a complex number has many arguments that differ by multiples of $2\pi$ .

Left: $-2 + 2\sqrt{3}\,i$ has $r = 4$ and $\theta = 2\pi /3$ . Right: $4 - 3i$ has $r = 5$ and $\theta \approx 5.640$ .

Rectangular Form to Polar Form

Worked example

Example 1. Write $z = -2 + 2\sqrt{3}\,i$ in polar form with $0 \le \theta < 2\pi$ .

01
Modulus: $r = \sqrt{(-2)^{2} + (2\sqrt{3})^{2}} = \sqrt{4 + 12} = \sqrt{16} = 4$ .
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Quadrant: real part negative, imaginary part positive, so the point is in Quadrant II.
03
$\cos \theta = a/r = -2/4 = -1/2$ and $\sin \theta = b/r = 2\sqrt{3}/4 = \sqrt{3}/2$ . The Quadrant II angle with these values is $\theta = 2\pi /3$ .
04
$z = 4(\cos (2\pi /3) + i \sin (2\pi /3))$ .

COMMON MISTAKE

Using $\theta = \arctan (b/a) = \arctan (2\sqrt{3}/(-2)) = \arctan (-\sqrt{3}) = -\pi /3$ .

$-\pi /3$ points into Quadrant IV, but $z$ is in Quadrant II. The same fix as in Part A applies: when the real part is negative, add $\pi$ . $-\pi /3 + \pi = 2\pi /3$ ✓. Finding $\cos \theta = a/r$ and $\sin \theta = b/r$ together avoids the problem entirely, because both signs are used.

Worked example

Example 2. Write $w = 4 - 3i$ in polar form with $0 \le \theta < 2\pi$ . Round $\theta$ to three decimal places.

01
$r = \sqrt{16 + 9} = 5$ .
02
The point $(4, -3)$ is in Quadrant IV. The reference angle is $\arctan (3/4) \approx 0.644$ .
03
In Quadrant IV, $\theta = 2\pi - 0.644 \approx 5.640$ . ( $\arctan (-3/4) \approx -0.644$ is also a correct argument, just not in $[0, 2\pi )$ .)
04
$w \approx 5(\cos 5.640 + i \sin 5.640)$ .

Quick check

Write $z = -1 + i$ in polar form with $0 \le \theta < 2\pi$ .

Polar Form to Rectangular Form

To go back, just evaluate: $a = r \cos \theta$ and $b = r \sin \theta$ .

Worked example

Example 3. Write $z = 6(\cos (7\pi /6) + i \sin (7\pi /6))$ and $w = 10(\cos 1.2 + i \sin 1.2)$ in the form $a + bi$ .

01
$\cos (7\pi /6) = -\sqrt{3}/2$ and $\sin (7\pi /6) = -1/2$ (Quadrant III). So $z = 6(-\sqrt{3}/2) + 6(-1/2)i = -3\sqrt{3} - 3i$ .
02
1.2 is not a special angle, so use a calculator in RADIAN mode: $10 \cos$ $1.2 \approx 3.624$ and $10 \sin$ $1.2 \approx 9.320$ . So $w \approx 3.624 + 9.320i$ .
03
Check the modulus: $\sqrt{3.624^{2} + 9.320^{2}} \approx 10$ ✓.

COMMON MISTAKE

Writing $r(\cos \theta + i \sin \theta)$ with the i on the cosine term, as r(i $\cos \theta + \sin \theta$ ).

The real part goes with cosine because cosine is the horizontal (x) coordinate, and the imaginary part goes with sine because sine is the vertical (y) coordinate — exactly as on the unit circle (3.2B).

KEY RULE

$a + bi$ ↔ point $(a, b)$ ↔ $r(\cos \theta + i \sin \theta)$ : $a = r \cos \theta$ , $b = r \sin \theta$ .

Quick check

Write $4\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$ in the form $a + bi$ .

Common slips

Using $\theta = \arctan (b/a) = \arctan (2\sqrt{3}/(-2)) = \arctan (-\sqrt{3}) = -\pi /3$ .

$-\pi /3$ points into Quadrant iv, but $z$ is in Quadrant ii. The same fix as in Part A applies: when the real part is negative, add $\pi$ . $-\pi /3 + \pi = 2\pi /3$ ✓. Finding $\cos \theta = a/r$ and $\sin \theta = b/r$ together avoids the problem entirely, because both signs are used.
Writing $r(\cos \theta + i \sin \theta)$ with the i on the cosine term, as r(i $\cos \theta + \sin \theta$ ).

The real part goes with cosine because cosine is the horizontal (x) coordinate, and the imaginary part goes with sine because sine is the vertical (y) coordinate — exactly as on the unit circle (3.2B).

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Recap card

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01
The complex number $a + bi$ is the point $(a, b)$ in the complex plane: real axis horizontal, imaginary axis vertical.
02
The modulus $|a + bi| = \sqrt{a^{2} + b^{2}}$ is the distance from 0.
03
Polar form: $a + bi = r(\cos \theta + i \sin \theta)$ , with $r = |z|$ and $\theta$ an argument of $z$ .
04
Choose $\theta$ by the quadrant of the point, not by $\arctan$ alone.
05
Coming up in 3.14: graphing polar functions $r = f(\theta )$ .

Practice 41 problems Next: Chapter 3.14A Polar Function Graphs

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Trigonometry and Polar Coordinates

The Complex Plane

Rectangular Form to Polar Form

Polar Form to Rectangular Form

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