De Moivre's Theorem
Chapter Nineteen
Syllabus topic Module 1, "1.1 Complex Numbers"
Pages 44 to 46 of 303
In one line
Raising a complex number to the power n raises the modulus to the power n and multiplies the argument by n.
The statement
For any integer n:
(cos(t) + i sin(t))^3 = cos(3t) + i sin(3t)
(cos(t) + i sin(t))^4 = cos(4t) + i sin(4t)
and in general, for a number of modulus r:
(r(cos(t) + i sin(t)))^3 = r^3 (cos(3t) + i sin(3t))
That is De Moivre's theorem, after Abraham de Moivre, who worked with the result from 1707 onwards.
Why it is true, for a positive whole number
By induction, which is the proof an examination wants.
The base case. For n = 1 the statement says the number equals itself, which is true.
The step. Suppose it holds for n = k. Multiply both sides by cos t + i sin t once more and use the multiplication rule of the previous chapter, which adds the arguments.
(cos(t) + i sin(t))^2 (cos(t) + i sin(t)) = (cos(2t) + i sin(2t))(cos(t) + i sin(t))
(cos(2t) + i sin(2t))(cos(t) + i sin(t)) = cos(3t) + i sin(3t)
So if it holds for k it holds for k plus one, and since it holds for one it holds for every positive whole number.
Negative powers
The theorem holds for negative integers too, and the proof is one line: a negative power is the reciprocal of a positive one, and taking a reciprocal negates the argument.
(cos(t) + i sin(t))^(-1) = cos(-t) + i sin(-t)
(cos(t) + i sin(t))^(-3) = cos(-3t) + i sin(-3t)
And for n = 0 both sides are 1, so the theorem holds for every integer.
A fractional index, which needs care
For a fraction the statement needs a warning, and the warning is the whole content of the next two chapters.
If n is a fraction p over q, then cos(nt) + i sin(nt) is one of the values of the left-hand side, but it is not the only one. Raising to the power one over q means taking a qth root, and a complex number has q different qth roots.
So for a fractional index the theorem reads: cos(nt) + i sin(nt) is one of the values of (cos t + i sin t) to the power n. Getting all of them is what the chapter on the n roots of a complex number is about.
What it is for: powers
The direct use. Compute (1 + i) to the tenth power.
Cartesian expansion means the binomial theorem with eleven terms. In polar form it is three lines. The modulus of 1 + i is root two and its argument is pi over four.
(1 + i)^10 = (sqrt(2))^10 (cos(10 pi/4) + i sin(10 pi/4))
(sqrt(2))^10 = 32
(1 + i)^10 = 32(cos(5 pi/2) + i sin(5 pi/2)) = 32i
De Moivre's Theorem
Five pi over two is 450 degrees, which is a full turn plus 90 degrees, so the cosine is zero and the sine is one, and the answer is 32i.
Another, with a negative index.
(1 - i)^(-6) = (sqrt(2))^(-6)(cos(6 pi/4) + i sin(6 pi/4))
(1 - i)^(-6) = (1/8)(cos(3 pi/2) + i sin(3 pi/2)) = -i/8
The argument of 1 minus i is minus pi over four, and minus six times that is plus six pi over four, which is 270 degrees.
What it is for: expanding cos(nt) and sin(nt)
This is the second standard use and it is asked in its own right.
Expand the left-hand side of the theorem with the binomial theorem, then match real parts with real parts and imaginary with imaginary. For n = 3:
(cos(t) + i sin(t))^3 = cos(t)^3 + 3i cos(t)^2 sin(t) - 3 cos(t) sin(t)^2 - i sin(t)^3
The real part of that is cos cubed minus three cos sin squared, and the imaginary part is three cos squared sin minus sin cubed. By the theorem the whole thing is cos 3t + i sin 3t, so comparing parts gives two identities for the price of one.
cos(3t) = cos(t)^3 - 3 cos(t) sin(t)^2
sin(3t) = 3 cos(t)^2 sin(t) - sin(t)^3
And using sin squared equals one minus cos squared on the first, and cos squared equals one minus sin squared on the second, gives the forms usually quoted.
cos(3t) = 4 cos(t)^3 - 3 cos(t)
sin(3t) = 3 sin(t) - 4 sin(t)^3
Those two are standard results, and this is where they come from. The same method gives cos 4t and sin 4t, and cos 5t, and so on for ever, which no amount of trigonometric identity-juggling would do.
Check yourself
(cos(t) + i sin(t))^5 = cos(5t) + i sin(5t)
(1 + i)^8 = 16
(1 - i sqrt(3))^3 = -8
(2(cos(pi/6) + i sin(pi/6)))^6 = -64
cos(2t) = cos(t)^2 - sin(t)^2
sin(2t) = 2 sin(t) cos(t)
Take the third line slowly, because it is the one that catches people. The modulus of 1 minus i root three is 2 and its argument is minus pi over three. Cubing gives modulus 8 and argument minus pi, and a number of modulus 8 at an argument of minus pi is the real number minus 8, not plus 8. If you write plus 8 you have lost the half turn.
The expansion agrees, and doing it once is worth the minute. The square of 1 minus i root three is minus 2 minus 2i root three, and multiplying that by 1 minus i root three gives minus 2 plus 2i root three minus 2i root three plus 2i squared times three, in which the two root-three terms cancel and the last term is minus 6, leaving minus 8.
De Moivre's Theorem
(1 - i sqrt(3))^2 = -2 - 2i sqrt(3)
(-2 - 2i sqrt(3))(1 - i sqrt(3)) = -8
The fourth line is the same trap in a different costume: six times pi over six is pi, so the answer is a negative real number.
The last two lines are De Moivre with n = 2, which is how the double-angle formulae are usually first met, and it is worth realising that the two of them are one statement about a complex number rather than two facts about triangles.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.