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6.5. Complex Numbers in Polar Form

Now that we have introduced polar coordinates, we can apply them to an important topic from Chapter 2, namely complex numbers. As we have seen, the complex number z = x + yi may be identified with the point with rectangular coordinates (x, y) in the complex plane. But we may use Equations (6.3.2) to rewrite z in polar coordinates. Upon making these substitutions, we see that the complex number z takes on the form

**(6.5.1)**z = r(cos θ + i sin θ)

where r and θ are given in terms of x and y by Equations (6.3.3). Note that by virtue of Equation (2.1.3), namely

**(6.5.2)**|z| = √x^{2}+ y^{2}

we see that r = √x^{2} + y^{2} is nothing but the modulus, |z|. What about θ? This quantity is known as the * argument* of z, denoted as arg(z). By virtue of Equation (6.3.3b) we see that the argument θ of z satisfies the equation

**(6.5.3)**tan θ = y / x

But as in the case of computing θ for points in the complex plane, we need to be careful in derminining the argument θ of a complex number. Equatioin (6.5.3) only tells us the value θ up to the possible addition of π. To determine whether or not we need to add π, we must look at the sign of y (add π to arctan(y/x) if y is negative, otherwise no). We also must be careful about the case in which x=0. In this case, the argument is equal to π/2 if y is positive or 3π/2 if x is negative. Finally, the argument of the complex number 0 is undefined.

**Example 1: **Calculate the modulus and argument of the following complex numbers and rewrite the complex numbers in polar form.

- (a) 8
- (b) -4
- (c) 3i
- (d) -2i
- (e) 1 + i
- (f) -3 + 3i
- (g) 1 - √3
^{}i - (h) -√3 - 3
^{}i

**Solution: **

- (a) The modulus of 8 is √8
^{2}+ 0^{2}= 8 and its argument θ satisfies tan θ = 0/8 = 0, whence θ is equal to either 0 or π. Since 8 is positive, we have θ = 0. The polar form of 8 is thus 8(cos 0 + i sin 0). - (b) The modulus of -4 is √(-4)
^{2}+ 0^{2}= 4 and its argument θ satisfies tan θ = 0/(-4) = 0, whence θ is equal to either 0 or π. Since -4 is negative, we have θ = π. The polar form of 8 is thus 8(cos π + i sin π). - (c) The modulus of 3i is √0
^{2}+ 3^{2}= 4 and its argument is equal to π/2 since x=0 and y is positive. The polar form of 3i is thus 3(cos π/2 + i sin π/2). - (d) The modulus of -2i is √0
^{2}+ (-2)^{2}= 2 and its argument is equal to 3π/2 since x=0 and y is negaitive. The polar form of -2i is thus 2(cos 3π/2 + i sin 3π/2). - (e) The modulus of 1 + i is √1
^{2}+ 1^{2}= √2 and its argument θ satisfies tan θ = 1/1 = 1, whence θ is equal to either π/4 or 5π/4. Since y=1 is positive, we have θ = π/4. The polar form of 1 + i is thus √2 (cos π/4 + i sin π/4). - (f) The modulus of -3 + 3i is √(-3)
^{2}+ 3^{2}= 3√2 and its argument θ satisfies tan θ = 3/(-3) = -1, whence θ is equal to either 3π/4 or 7π/4. Since y=3 is positive, we have θ = 3π/4. The polar form of -3 + 3i is thus 3√2 (cos 3π/4 + i sin 3π/4). - (g) The modulus of 1 - √3
^{}i is √1^{2}+ (√3^{})^{2}= 2 and its argument θ satisfies tan θ = - √3/1 = - √3, whence θ is equal to either 2π/3 or 5π/3. Since y=-√3^{}is negative, we have θ = 5π/3. The polar form of -3 + 3i is thus 2 (cos 5π/3 + i sin 5π/3). - (h) The modulus of -√3 - 3i
^{}i is √(-√3)^{2}+ 3^{2}= 2√3 and its argument θ satisfies tan θ = -3/(- √3) = √3, whence θ is equal to either π/3 or 4π/3. Since y=-3^{}is negative, we have θ = 4π/3. The polar form of -√3 - 3i^{}is thus 2√3(cos 4π/3 + i sin 4π/3).

**Example 2: ** Rewrite the following complex numbers, given in polar form, in standard (rectangular) form.

- (a) 14(cos 0 + i sin 0)
- (b) e(cos π + i sin π)
- (c) √3(cos π/2 + i sin π/2)
- (d) 55(cos 3π/2 + i sin 3π/2)
- (e) 1(cos π/4 + i sin π/4)
- (f) 14.7(cos 1.8 + i sin 1.8)
- (g) 6(cos 5π/3 + i sin 5π/3)
- (h) 5.00(cos 4.069 + i sin 4.069)

**Solution:**

- (a) Since cos 0 = 1 and sin 0 = 0, we see that 14(cos 0 + i sin 0) = 14.
- (b) Since cos π = -1 and sin π = 0, we see that e(cos π + i sin π) = -e.
- (c) Since cos π/2 = 0 and sin π/2 = 1, we see that √3(cos π/2 + i sin π/2) = √3 i.
- (d) Since cos 3π/2 = 0 and sin 3π/2 = -1, we see that 55(cos 3π/2 + i sin 3π/2) = -55i.
- (e) Since cos π/4 = sin π/4 = √2/2, we see that 1(cos π/4 + i sin π/4) = √2/2 + √2/2 i.
- (f) We have x = 14.7 cos 1.8 = (14.7)(0.227) = -3.3 and 14.7 sin 1.8 = (14.7)(0.974) = 14.3, whence 14.7(cos 1.8 + i sin 1.8) = -3.3 + 14.3 i.
- (g) Since cos 5π/3 = 1/2 and sin 5π/3 = -√3/2, we see that 6(cos 5π/3 + i sin 5π/3) = 3 - √3 i.
- (h) We have x = 5.00 cos 4.069 = (5.00)(-0.600) = -3.00 and 5.00 sin 4.069 = (5.00)(-0.800) = -4.00, whence 5.00(cos 4.069 + i sin 4.069) = -3.00 - 4.00 i.

At this point you may be wondering why we are bothering with all this. Is the polar form of a complex number useful in any special way? Well as it turns out, it is, as the following beautiful theorem shows.

**Theorem 6.5.4: **Let z_{1} and z_{2} be arbitrary nonzero complex numbers. Then the following equations hold:

- (a) |z
_{1}z_{2}| = |z_{1}| |z_{2}| - (b) arg(z
_{1}z_{2}) = arg(z_{1}) + arg(z_{2})

In other words, the modulus of the product of two complex numbers is equal to the product of the moduli, and the argument of the product is equal to the sum of the arguments.

**Proof: **We let r_{k} and θ_{k} denote the modulus and argument of z_{k}, where k = 1 or 2. Then we have z_{k} = r_{k}(cos θ_{k} + i sin θ_{k}), whence

z_{1}z_{2} = r_{1}r_{2} (cos θ_{1} + i sin θ_{1})(cos θ_{2} + i sin θ_{2}) .

Using the complex multiplication formula (2.2.5), we see that this is equal to

z_{1}z_{2} = r_{1}r_{2} [(cos θ_{1} cos θ_{2} - sin θ_{1} sin θ_{2}) + i (sin θ_{1} cos θ_{2} + cos θ_{1} sin θ_{2})].

But from the sum formulas for sine and cosine, we recognize this as just

z_{1}z_{2} = r_{1}r_{2} [cos (θ_{1} + θ_{2}) + i sin(θ_{1} + θ_{2})],

Thus we see that the modulus |z_{1}z_{2}| of z_{1}z_{2} is r_{1}r_{2} = |z_{1}| |z_{2}| and the argument arg(z_{1}z_{2}) of z_{1}z_{2} is θ_{1} + θ_{2} = arg(z_{1}) + arg(z_{2}), proving the theorem.

**QED**

This is quite a powerful theorem! Besides providing a nice geometric interpretation of the product of two complex numbers, it allows one to easily multiply two complex numbers given in polar form. Also, as we will see in the following section, it will allow us to prove another very useful theorem known as De Moivre's Theorem.

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