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Complex Numbers
Rectangular & Polar Form

Learn to represent complex numbers as \(a + bi\) and \(r(\cos\theta + i\sin\theta)\), plot them on the complex plane, and convert fluently between both forms.

Algebra 2

What Is a Complex Number?

A complex number combines a real part and an imaginary part. The imaginary unit \(i\) is defined so that \(i^2 = -1\), which lets us take square roots of negative numbers.

Real Part
The value \(a\) plotted on the horizontal axis.
Imaginary Part
The value \(b\) plotted on the vertical axis (multiplied by \(i\)).

The Complex Plane

Every complex number \(z = a + bi\) corresponds to a unique point \((a, b)\) on the complex plane (also called the Argand diagram). The horizontal axis represents real values; the vertical axis represents imaginary values.

Re
Im
1
2
-1
-2
\(i\)
\(-i\)
\(z = 3 + 2i\)
\(a = 3\)
\(b = 2\)
\(r\)
\(\theta\)

Why Two Forms?

Rectangular form \(a + bi\) makes addition and subtraction easy: just combine real and imaginary parts. Polar form \(r(\cos\theta + i\sin\theta)\) makes multiplication, division, and powers easy: multiply moduli and add angles.

Rectangular
Best for \(z_1 + z_2\) and \(z_1 - z_2\). Components add directly.
Polar
Best for \(z_1 \cdot z_2\), \(z_1 / z_2\), and \(z^n\). Angles add, moduli multiply.

The Conversion Formulas

These relationships link rectangular coordinates \((a,b)\) to polar coordinates \((r,\theta)\) and back.

Rectangular → Polar
\(r = \sqrt{a^2 + b^2}\)
\(\theta = \arctan\!\left(\dfrac{b}{a}\right)\)
(adjust \(\theta\) for quadrant)
Polar → Rectangular
\(a = r\cos\theta\)
\(b = r\sin\theta\)
so \(z = r\cos\theta + ir\sin\theta\)

Polar Form: Formal Definition

Re
Im
\(z = a + bi\)
\(a = r\cos\theta\)
\(b = r\sin\theta\)
\(r\)
\(\theta\)

Rectangular vs. Polar

Rectangular \(a + bi\) Polar \(r(\cos\theta + i\sin\theta)\)
Coordinates \((a, b)\) \((r, \theta)\)
Axes Used Real & Imaginary Distance & Angle
Addition Easy: add parts Hard: must convert
Multiplication Requires FOIL Easy: multiply \(r\), add \(\theta\)
Powers Very tedious Easy: De Moivre's theorem
Visualization Grid position Direction & distance

How to Convert: Step by Step

1
Identify your starting form. Is the number given as \(a + bi\) (rectangular) or \(r(\cos\theta + i\sin\theta)\) (polar)?
2
Rectangular → Polar: Compute \(r = \sqrt{a^2 + b^2}\). Then find \(\theta = \arctan(b/a)\), adjusting for the correct quadrant.
3
Polar → Rectangular: Compute \(a = r\cos\theta\) and \(b = r\sin\theta\). Write the answer as \(a + bi\).
4
Verify. Check that \(a^2 + b^2 = r^2\) and the point lies in the expected quadrant.

Quadrant Adjustment for \(\theta\)

The basic \(\arctan(b/a)\) only returns values in \((-90^\circ, 90^\circ)\). You must adjust based on which quadrant the point \((a, b)\) falls in.

Q I: \(a > 0, b > 0\)
\(\theta = \arctan(b/a)\). No change needed.
Q II: \(a < 0, b > 0\)
\(\theta = \arctan(b/a) + 180^\circ\)
Q III: \(a < 0, b < 0\)
\(\theta = \arctan(b/a) + 180^\circ\)
Q IV: \(a > 0, b < 0\)
\(\theta = \arctan(b/a) + 360^\circ\)

Example 1: Rectangular → Polar

Convert \(z = 3 + 4i\) to polar form.

1
Find \(r\): \(r = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)
2
Find \(\theta\): \(\theta = \arctan(4/3) \approx 53.13^\circ\). Since \(a > 0\) and \(b > 0\) (Q I), no adjustment needed.
3
Write polar form: \(z = 5(\cos 53.13^\circ + i\sin 53.13^\circ)\)
Re
Im
\(3 + 4i\)
\(3\)
\(4i\)
\(r = 5\)
\(53.1^\circ\)
1
2
3

Example 2: Polar → Rectangular

Convert \(z = 6(\cos 120^\circ + i\sin 120^\circ)\) to rectangular form.

1
Find \(a\): \(a = 6\cos 120^\circ = 6 \times (-0.5) = -3\)
2
Find \(b\): \(b = 6\sin 120^\circ = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3} \approx 5.20\)
3
Write rectangular form: \(z = -3 + 3\sqrt{3}\,i\)

Example 3: A Q III Conversion

Convert \(z = -2 - 2i\) to polar form.

1
Find \(r\): \(r = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}\)
2
Reference angle: \(\arctan\!\left(\frac{-2}{-2}\right) = \arctan(1) = 45^\circ\)
3
Quadrant adjustment: Both \(a\) and \(b\) are negative (Q III), so \(\theta = 45^\circ + 180^\circ = 225^\circ\)
4
Result: \(z = 2\sqrt{2}\,(\cos 225^\circ + i\sin 225^\circ)\)

Visualizing \(-2 - 2i\) on the Plane

Re
Im
\(-2 - 2i\)
\(-2\)
\(-2i\)
\(r = 2\sqrt{2}\)
\(225^\circ\)
-1
-2
1
2
\(-i\)
\(-2i\)
\(i\)

Special Cases on the Axes

Some complex numbers lie directly on an axis. Their angles are exact multiples of \(90^\circ\).

Pure Real (\(b = 0\))
Positive: \(\theta = 0^\circ\)
Negative: \(\theta = 180^\circ\)
Example: \(5 = 5(\cos 0^\circ + i\sin 0^\circ)\)
Pure Imaginary (\(a = 0\))
Positive: \(\theta = 90^\circ\)
Negative: \(\theta = 270^\circ\)
Example: \(3i = 3(\cos 90^\circ + i\sin 90^\circ)\)

Common Mistakes to Avoid

✕ Forgetting Quadrant Adjustment
Using \(\arctan(b/a)\) alone gives the wrong angle in Q II, III, or IV. Always check which quadrant \((a, b)\) falls in.
✕ Mixing Up \(a\) and \(b\)
In \(z = a + bi\), the coefficient of \(i\) is \(b\), not \(a\). Writing \(\arctan(a/b)\) instead of \(\arctan(b/a)\) flips the angle.
✕ Negative Under the Radical
When computing \(r = \sqrt{a^2 + b^2}\), remember both terms are squared, so \(r\) is always non-negative. Do not write \(\sqrt{(-3)^2}\) as \(\sqrt{-9}\).

Summary Reference

Task Formula Watch Out
Modulus \(r\) \(\sqrt{a^2 + b^2}\) Always \(\geq 0\)
Argument \(\theta\) \(\arctan(b/a)\) Adjust for quadrant
Real part \(a = r\cos\theta\) Can be negative
Imaginary part \(b = r\sin\theta\) Can be negative
Verify \(a^2 + b^2 = r^2\) Must hold exactly

Question 1 of 5

What is the modulus of \(z = 5 + 12i\)?

Question 2 of 5

Convert \(z = 4(\cos 60^\circ + i\sin 60^\circ)\) to rectangular form.

Question 3 of 5

The complex number \(z = -1 + i\) lies in which quadrant, and what is its argument \(\theta\)?

Question 4 of 5

Which statement best explains why rectangular and polar forms represent the same complex number?

Question 5 of 5

Convert \(z = -3 - 3\sqrt{3}\,i\) to polar form. Which is correct?