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4. Polar Form of a Complex Number

by M. Bourne

We can think of complex numbers as vectors, as in our earlier example. [See more on Vectors in 2-Dimensions].

We have met a similar concept to "polar form" before, in Polar Coordinates, part of the analytical geometry section.

Our aim in this section is to write complex numbers in terms of a distance from the origin and a direction (or angle) from the positive horizontal axis.

From Pythagoras, we have: r² = x² + y² and basic trigonometry gives us:

x = r cos θy = r sin θ

Multiplying the last expression throughout by j gives us:

yj = jr sin θ

So we can write the polar form of a complex number as:

x + yj = r(cos θ + j sin θ)

r is the absolute value (or modulus) of the complex number

θ is the argument of the complex number.

There are two other ways of writing the polar form of a complex number:

r cis θ [means r (cos θ + j sin θ)]

r ∠ θ [means once again, r (cos θ + j sin θ)]

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NOTE: When writing a complex number in polar form, the angle θ can be in DEGREES or RADIANS.

Example 1

Find the polar form and represent graphically the complex number 7 - 5j.

Answer:

We need to find r and θ.

To find θ, we first find the acute angle α (see Trigonometric Functions of Any Angle if you are rusty on this):

Now, 7 - 5j is in the fourth quadrant, so

θ = 360° - 35.54° = 324.46°

So, expressing 7 - 5j in polar form, we have:

7 - 5j = 8.6 (cos 324.5° +j sin 324.5°)

We could also write this answer as 7 - 5j = 8.6 cis 324.5°.

Also we could write: 7 - 5j = 8.6 ∠ 324.5°

The graph is as follows:

Now let's see how it looks in Livemath.

LIVEMath

 

Example 2:

Express 3(cos 232°+ j sin 232°) in rectangular form.

We simply multiply out the expression:

3(cos 232° +j sin 232°)

= 3 cos 232° + j (3sin 232°)

= -1.85 - 2.36j

 

Exercises

1. Represent 1 + j √3 graphically and write it in polar form.

Answer:

We recognise this triangle as our 30-60 triangle from before.

θ = 60°, r = 2

So

1 + j√3 = 2 ∠ 60° = 2(cos60° + jsin60°)

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2. Represent √2 - j√2 graphically and write it in polar form.

Answer:

To find θ, we first find the acute angle α:

The complex number is in the 4th quadrant, so

θ = 360° - 45° = 315°

√2 - j√2 = 2∠ 315° = 2(cos315° + jsin315°)

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3. Represent graphically and give the rectangular form of
6(cos180°+ jsin180°).

Answer:

6(cos180° + jsin180°) = -6

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4. Represent graphically and give the rectangular form of 7.32 ∠ -270°

Answer

7.32 ∠ -270°= 7.32j

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Need a break? Play a math game. Well, they all involve math... No, really!

Chapter Contents

• Complex Numbers
• 1. Basic Definitions
• 2. Basic Operations
• 3. Graphical Representation
• 4. Polar Form
• 5. Exponential Form
• 6. Products and Quotients
• 7. Powers and Roots
• 8. AC Circuit Definitions
• 9. Impedance and Phase Angle
• 10. Reactance and Angular Velocity
• 11. AC Circuit Exercises
• 12. Parallel AC Circuits
• Vocabulary
• Fractals

• Comments, Questions?

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