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Trig without Tears Part 10:

Fun with Complex Numbers

Copyright © 1997–2024 by Stan Brown,

Summary: Trig can answer some of the questions your algebra teacher wouldn’t answer, like what’s √i? and what’s the log of a negative number?


Polar Form of a Complex Number

complex addition: 3+4i plus 2-7i equals 5-3i As you may remember, complex numbers like 3+4i and 2−7i are often plotted on a complex plane. This makes it easier to visualize adding and subtracting. The illustration at right shows that

(3+4i) + (2−7i) = 5−3i

complex number: negative 3 plus 5 i It turns out that for multiplying, dividing, and finding powers and roots, complex numbers are easier to work with in polar form. The graph at left shows −3+5i, but in polar form you don’t think about the real and imaginary components (the “−3” and “5” in −3+5i). Instead, you think about the length and direction of the line that connects the origin to the plotted point.

The length is easy, just the good old Pythagorean theorem in fact:

r = √3² + 5² = √34 ≈ 5.83

(The length r is called the absolute value or modulus of the complex number.)

But what about the direction? You can see from the picture that θ is about 120° or about 2 radians (measured counterclockwise from the positive real axis, which points east), but what is it exactly? Well, you know that tan θ = −5/3. If you take Arctan(−5/3) on your calculator, you get about −1.03, and adding π to get into the proper quadrant gives θ ≈ 2.11 radians. (In degrees, Arctan(−5/3) ≈ −59.04°, and adding 180 gives θ ≈ 120.96°.)

Here’s a nice trick to find the angle in the correct quadrant automatically:

θ = 2 Arctan(y/(x + r))

with x, y, and r as defined here.

This formula automatically adjusts to the correct quadrant, so you never have to worry about whether to add or subtract 180° or π. If you test 2 Arctan(5/(−3 + √34)) on your calculator, you get 2.11 radians or 120.96° as before.

The formula is particularly handy when you’re writing a program or spreadsheet formula to find θ. It flows directly from the formulas for tangent of half an angle.

A July 2003 article by Rob Johnson, archived here, inspired me to add this section.

A lot of ways to write a given number in polar form exist, and all of them can use degrees or radian measure.

Remember, these are just notation. They all mean the same thing and follow the same rules.

Multiplying and Dividing Complex Numbers in Polar Form

It’s not exactly hard to multiply complex numbers in rectangular form, but it’s not exactly easy either:

(a+bi) × (c+di) = (acbd) + (bc+ad)i

Division is harder because of eliminating imaginaries from the denominator:

(a+bi) / (c+di) = [(a+bi)(cdi)] / [(c+di)(cdi)] = [(ac+bd) + (bcad)i] / (c² + d²)

But in polar form, it’s much easier because of the laws of exponents:

(81) r e × s e = rs ei(θ+φ)      and      r e / s e = (r/s)ei(θ−φ)

Let’s make an example, with 11+24i and 22−17i. (Be careful! d is negative.)

(11+24i) × (22−17i) = (11×22+24×17) + (24×22−11×17)i = 650+341i

And dividing:

(11+24i) / (22−17i) = [(11×22−24×17) + (24×22+11×17)i] / (22²+17²)

(11+24i) / (22−17i) = (−166+715i)/773 ≈ −0.215+0.925i

Now put 11+24i into polar form:

r = √11²+24² ≈ 26.401

θ ≈ 2 Arctan(24/(11+26.401)) ≈ 1.141 (radians)

And 22−17i:

r = √22²+17² ≈ 27.803

θ ≈ 2 Arctan(17/(−17+27.803)) ≈ −0.658 (radians)

Multiplying and dividing is a piece of cake:

26.401 e1.141i × 27.803 e−0.658i = 26.401×27.803 e(1.141−0.658)i ≈ 734.017 e0.483i

26.401 e1.141i / 27.803 e−0.658i = (26.401/27.803) e(1.141+0.658)i ≈ 0.950 e1.799i

Once the numbers are in polar form, multiplying and dividing them is very easy. But if you have numbers in rectangular form, getting them into polar form is not exactly trivial, so in that case you may opt to do your arithmetic in rectangular form after all.

Powers and Roots of a Complex Number

One of the applications of Euler’s formula (equation 47) is finding any power or root of any complex number. (Sawyer doesn’t do this, or at least not in the same book.) It’s not hard to develop the formula, using just Euler’s formula and the laws of exponents.

Start by writing down Euler’s formula, multiplied left and right by a scale factor r:

r (cos x + i sin x) = r eix

Next, raise both sides to the nth power:

[r (cos x + i sin x)]n = [r eix]n

Let’s leave the left-hand side alone for a while, and simplify the right-hand side. First, by the laws of exponents

[r eix]n = rn einx

Apply Euler’s formula to the right-hand side and you have

[r eix]n = rn (cos nx + i sin nx)

Connect that right-hand side to the original left-hand side and you have DeMoivre’s theorem:

(82) [r(cos x + i sin x)]n = rn (cos nx + i sin nx)

This tells you that if you put a number into polar form, you can find any power or root of it. (Remember that the nth root of a number is the same as the 1/n power.)

Square Root of i

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For instance, you know the square roots of −1 are i and −i, but what’s the square root of i? You can use DeMoivre’s theorem to find it.

First, put i into eix form (polar form), using the above technique. i = 0+1i. cos x = 0 and sin x = 1 when x = 90° or π/2. Therefore

i = cos(π/2) + i sin(π/2)

i = i½ = [cos(π/2) + i sin(π/2)]½

And by equation 82

i = cos(½×π/2) + i sin(½×π/2)

i = cos(π/4) + i sin(π/4)

i = 1/√2 + i×1/√2

i = (1+i)/√2

The other square root is minus that, as usual.

Principal Root of Any Number

You can find any root of any complex number in a similar way, but usually with one preliminary step.

For instance, suppose you want the cube roots of 3+4i. The first step is to put that number into polar form. The absolute value is √3²+4² = 5, and you use the Arctan technique given above to find the angle θ = 2 Arctan(4/(3+5)) ≈ 2 Arctan(½) ≈ 53.13° or 0.9273 radians. The polar form is

3+4i ≈ 5(cos 53.13° + i sin 53.13°)

To take a cube root of that, use equation 82 with n = 1/3:

cube root of (3+4i) ≈ 51/3 [cos(53.13°/3) + i sin(53.13°/3)]

The cube root of 5 is about 1.71. Dividing the angle θ by 3, 53.13°/3 ≈ 17.71°; the cosine and sine of that are about 0.95 and 0.30. Therefore

cube root of (3+4i) ≈ 1.71 × (cos 17.71° + i sin 17.71°)

cube root of (3+4i) ≈ 1.63+0.52i

Multiple Roots

You may have noticed that I talked about “the cube roots [plural] of 3+4i” and what we found was “a cube root”. Even with the square root of i, I waved my hand and said that the “other” square root was minus the first one, “as usual”.

You already know that every positive real has two square roots. In fact, every complex number has n nth roots.

How can you find them? It depends on the periodic nature of the cosine and sine functions. To start, look back at Euler’s formula,

eix = cos x + i sin x

What happens if you add 2π or 360° to x? You have

ei(x+2π) = cos(x + 2π)  + i sin(x + 2π)

But taking sine or cosine of 2π plus an angle is exactly the same as taking sine or cosine of the original angle. So the right-hand side is equal to cos x + i sin x, which is equal to eix. Therefore

ei(x+2π) = eix

In fact, you can keep adding 2π or 360° to x as long as you like, and never change the value of the result. Symbolically,

ei(x+2πk) = eix   for all integer k

When you take an nth root, you simply use that identity.

Getting back to the cube roots of 3+4i, recall that that number is the same as 5eiθ, where θ is about 53.13° or 0.9273 radians. The three cube roots of eiθ are

ei(θ+2πk)/3 for k = 0,1,2

Expanding that, you have

eiθ/3, ei(θ+2π)/3, and ei(θ+4π)/3  


eiθ/3, ei(θ/3+2π/3), and ei(θ/3+4π/3)  

Compute those as cos x + i sin x in the usual way, and then multiply by the (principal) cube root of 5. I get these three roots:

cube roots of 3+4i ≈ 1.63+0.52i, −1.26+1.15i, −0.36−1.67i

More generally, we can extend DeMoivre’s theorem (equation 82) to show the n nth roots of any complex number:

(83) [r (cos x + i sin x)](1/n) = r(1/n) [cos(x/n+2πk/n) + i sin(x/n+2πk/n)]  for k = 0,1,2,…,n−1

Logarithm of a Negative Number

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See also: If you need a refresher on logarithms, please see It’s the Law Too — the Laws of Logarithms.

Another application flows from a famous special case of Euler’s formula. Substitute x = π or 180° in equation 47. Since sin 180° = 0, the imaginary term drops out. And cos 180° = −1, so you have the famous formula

(84) −1 = e

It’s interesting to take the natural log of both sides:

ln(−1) = ln e

ln(−1) = iπ

It’s easy enough to find the logarithm of any other negative number. Since

ln ab = ln a + ln b

then for all real a you have

ln(−a) = ln(a × −1) = ln a + ln(−1) = ln a + iπ

(85) ln(−a) = ln a + iπ

I don’t honestly know whether all of this has any practical application. But if you’ve ever wondered about the logarithm of a negative number, now you know.

Logarithm of a Complex Number

But why stop there? You can play the same game in taking the logarithm of a complex number.

But wait! A number like 3+4i isn’t ready to have its logarithm taken. First you have to convert it to polar form. That’s nothing new: look back at the previous section. I started with −1 = e, and what is that but converting −1 to polar form?

(86) ln(r eiθ) = ln r + iθ

There’s really no need to memorize that formula; it’s just ln(xy) = ln x + ln y and ln(ez) = z.

Example: Earlier, in Principal Root of a Complex Number, we found that 3+4i ≈ 5 e0.9273i. Therefore,

ln(3+4i) ≈ ln(5 e0.9273i) = ln 5 + 0.9273i ≈ 1.609 + 0.9273i

Practice Problems

To get the most benefit from these problems, work them without first looking at the solutions. Refer back to the chapter text if you need to refresh your memory.

Recommendation: Work them on paper — it’s harder to fool yourself about whether you really understand a problem completely.

You’ll find full solutions for all problems. Don’t just check your answers, but check your method too.

1 Express in a+bi form: (a) 62∠240° (b) 100e1.17i
2 Express in polar form, in both degrees and radian measure: (a) −42+17i (b) 100i (c) −14.7
3 Find the three cube roots of −i, in a+bi form. You’ll be able to give an exact answer, not rounded decimals.
4 Find (1.04−0.10i)16.
5 In Multiplying and Dividing Complex Numbers in Polar Form we found a product of 650+341i and a quotient of about −0.215+0.925i from two complex numbers in rectangular form, then translated the original numbers to polar form and found a product of about 734.017 e0.483i and a quotient of about 0.950 e1.799i. Convert the rectangular product and quotient to polar form to verify that they match the polar ones.
6 Find the logarithms, to four decimal places: (a) -67.12; (b) 1−i; (c) 734.017 e0.483i.

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