MATH 117 The Polar Form of Complex Numbers

At the end of the 18th century, Caspar Wessel (1745 – 1818), Jean Robert Argand (1768 – 1822), and Carl Friedrich Gauss (1777 – 1855) independently came up with a geometric interpretation of a complex number z = a + bi . The complex number could simply be considered as an ordered pair (a , b ) in the x y -plane. Then the point (a , b ) could be further represented in polar form by an angle θ and a radius r . The angle θ is always measured counterclockwise from the positive x -axis with 0 ≤ θ < 2π. The radius is given by

r = a2 + b2 , and we always take r ≥ 0.

rθ

The Polar Form

First recall Euler’s Formula which states that ei x = cos x + i sin x . Now because cosθ = adjhyp

= ar

, then a = r cosθ . Likewise since sinθ = opphyp

= br

, then b = rsin θ . Thus, z =

a + bi = r cosθ + (rsin θ) i = r (cosθ + i sinθ ) , and so by Euler’s Formula:

z = r e iθ

Conversion: Polar to Rectangular Given the radius and angle (r , θ ), we can easily convert back to rectangular form z = a + bi since a = r cosθ and b = rsin θ . Example 1. Convert to rectangular form (a) z = 4 e 5 π i / 3 (b) z = 10e 7 π i /6 . Solution. (a) Here r = 4 and θ = 5π/3, so a = 4 cos(5π / 3) = 2 and b = 4 sin(5π / 3) = – 2 3 . Thus, z = 2 − 2 3 i .

(b) z = 10cos(7π / 6) + 10sin(7π / 6) i = 10 × − 3

2

+ 10 × −1

2

i = −5 3 − 5 i .

To obtain the above results in decimal form with your calculator, simply enter 4�^(5π�/3). (Make sure that you are in radian mode.)

Conversion: Rectangular to Polar

Given z = a + bi , we also can find the radius and angle (r , θ ). We know that

r = a2 + b2 . Also tan θ = b / a . To find θ , compute tan−1(b /a) and adjust the quadrant

appropriately. Recall that

−π

2< tan−1 x <

π

2. By definition, tan −1 x is the angle θ between –π/2 and

π/2 such that tan θ = x . For instance tan −11 = π/4 becasue tan(π / 4) = 1.

f (x) = tan x

Domain: –π/2 < x < π/2 Range: –∞ < y < ∞

f (x) = tan −1 x

Domain: –∞ < x < ∞ Range: –π/2 < y < π/2

x tan x

x tan −1 x

– π

2

–∞

–∞ – π

2

– π

3

– 3

– 3 – π

3

– π

4

–1

–1 – π

4

– π

6 –

13

–

13

– π

6

0

0

0

0

π

6

13

13

π

6

π

4

1

1 π

4

π

3

3

3 π

3

π

2

∞

∞ π

2

Note that angles in the 4th quadrant are written as negative angles. However, for the angle θ in the polar form of z , we want 0 ≤ θ < 2π; therefore we must adjust tan −1 x by adding π or 2π appropriately.

Example 2. Find the polar form of the complex numbers.

(a) z = −3 3 + 3i (b) z = −6 − 6i (c) z = 4 − 4 3 i Solution. (a) The point ( −3 3 , 3 ) is in the 2nd quadrant. The radius is r = (−3 3 )2 + 32 = 36 = 6. Next b / a = −1 / 3 and

tan−1(−1 / 3) = –π/6. So we add π to obtain the angle in the

2nd quadrant. Thus θ = 5π/6 and z = 6e 5π i / 6 . (b) The point (–6, –6) is in the 3rd quadrant. The radius is

r = (−6)2 + (−6)2 = 6 2 . Next b / a = 1 and tan−1(1) = π/4.

But we add π to obtain the angle in the 3rd quadrant. Thus θ = 5π/4 and z = 6 2 e 5π i / 4 .

(c) The point (4, − 4 3 ) is in the 4th quadrant. The radius is

r = 42 + (−4 3)2 = 64 = 8. Next b / a = – 3 and tan−1(− 3)

= –π/3. But we now add 2π to obtain an angle θ in the 4th quadrant with 0 ≤ θ < 2π. Thus θ = 5π/3 and z = 8e 5π i / 3. Most often though, we do not have complex numbers that give one of the special unit circle angles. In these cases, we usually write the angle in degree form. For example, if z = −4 + 2i , then r = 20 and tan

−1(b / a) = tan−1(−1 / 2) ≈ −26.565

(using Degree mode on calculator). So in the 2nd quadrant, θ ≈ 153.435 .

The Axes Angles

Consider the complex numbers 8, 6i , –10, – 20i . In the x y plane, these values are points on the axes: (8, 0), (0, 6), (–10, 0), and (0, –20). The radii are the positive values 8, 6, 10, and 20, and the angles are 0, π/2, π, and 3π/2. Thus, by Euler’s Formula, we have 8 = 8e0 i , 6i = 6 e π i / 2 , –10 = 10 e π i , and – 20i = 20e3 π i / 2 .

Geometric Interpretations Powers: Let z = r e iθ be the polar form of a complex number. Then we can compute various integral powers of z and see the effect on the radius and the angle. Since

zn = re i θ( )n = rn e i (nθ ),

we see that zn raises the radius to the n th power and rotates the angle by a factor of n .

For example, if z = 2 ei (20 ) , then z2 = 4 e

i (40 ) , z3 = 8ei (60 ) , and so on.

z , z2 , and z3

Reciprocal: If z = r e iθ , then z

−1 = r−1 e i (− θ) . Thus, we obtain the reciprocal of the radius and we have the same the angle but rotated in the negative direction.

For example, if z = 3ei (45 ) , then

z−1 =

13e i (−45 ) .

z and z−1

Products and Quotients: Let z1 = r1 e

iθ1 and z2 = r2 ei θ2 . Then

z1 × z2 = r1 eiθ1 × r2 e

i θ2 = ( r1 r2 ) ei (θ1+θ2 ) .

For a product, we multiply the radii and

add the angles. If z1 = 2 ei (20 ) and z2 =

3ei (45 ) , then z1 × z2 = 6e

i (65 ) .

z1 and z2 z1 × z2

Also

z1z2

=

r1 ei θ1

r2 ei θ2

=

r1r2

ei (θ1−θ2 ) .

Thus for a quotient, we divide the radii and subtract the angles.

Exercises

1. Find the rectangular form of the complex numbers.

(a) z = 8e π i (b) z = 4 eπi / 3 (c) z = 12e 11π i /6

2. Convert the complex numbers to the polar form r e iθ .

(a) z = −5 2 + 5 2 i (b) z = −3 − 3 3 i (c) 7 − 16 i 3. Let z1 = 4 e

πi / 3 and z2 = −3 − 3 3 i . (a) Multiply in rectangular form. (b) Multiply in polar form. (c) Convert the polar product to rectangular form to verify that the result is the same as in (a).