Trig without Tears Part 10:
Trig without Tears Part 10:
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?
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
It turns out that for multiplying, dividing, and finding powers and roots, complex numbers are easier to work with in polar form. This means that instead of thinking about the real and imaginary components (the “−3” and “5” in −3+5i), you think of the length and direction of the line.
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 −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°.) You can write it this way:
−3+5i ≈ 5.83 e2.11i or 5.83[cos 2.11 + i sin 2.11] or 5.83 cis 2.11 or 5.83∠120.96°
There’s a nice trick that finds the angle in the correct quadrant automatically:
θ = 2 Arctan(y/(x + r))
This builds in the adjustment 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.
This 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, where the functions are expressed in terms of x, y, and r. A July 2003 article by Rob Johnson, archived here, inspired me to add this section.
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.)
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.
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
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
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 = eiπ
It’s interesting to take the natural log of both sides:
ln(−1) = ln eiπ
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.
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.
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