Taking a Power with De Moivre's Theorem
Chapter Twenty
Syllabus topic Module 1, "1.1 Complex Numbers"
Pages 47 to 48 of 303
In one line
To raise a complex number to a power, put it in polar form, raise the modulus, multiply the angle, and convert back.
The procedure
Four steps, and the fourth is where the marks are.
- Find the modulus r and the principal argument t of the number.
- Raise r to the power n.
- Multiply t by n.
- Reduce the resulting angle by whole turns until it is in the principal range, then read off the cosine and the sine.
Step four is not optional. An angle of 450 degrees is a correct answer and a careless one; reducing it to 90 degrees is what lets you write the Cartesian form without a calculator.
Worked, with every step
Find the value of the expression below.
(sqrt(3) + i)^7
Modulus. The parts are root three and one, so the modulus is 2.
abs(sqrt(3) + i) = sqrt(3 + 1) = 2
Argument. The tangent of the acute angle is one over root three, so the acute angle is 30 degrees; both parts are positive, so the argument is pi over six.
Raise and multiply. Two to the seventh is 128, and seven times pi over six is seven pi over six.
Reduce. Seven pi over six is 210 degrees, which is outside the principal range, so subtract a full turn to get minus five pi over six, which is minus 150 degrees.
(sqrt(3) + i)^7 = 128(cos(7 pi/6) + i sin(7 pi/6))
128(cos(7 pi/6) + i sin(7 pi/6)) = -64 sqrt(3) - 64i
So the answer is minus 64 root three minus 64i. The cosine of 210 degrees is minus root three over two and the sine is minus one half, which gives those two terms.
Worked, a negative index
(1 - i)^(-8)
The modulus is root two and the argument is minus pi over four. Raising: root two to the minus eight is one over sixteen. Multiplying: minus eight times minus pi over four is two pi, a whole turn, so the angle reduces to zero.
(1 - i)^(-8) = (1/16)(cos(2 pi) + i sin(2 pi)) = 1/16
The answer is real, which the whole-turn angle told you before you computed anything.
Why this beats the binomial theorem
Expanding (root three plus i) to the seventh with the binomial theorem means eight terms, each with a power of i to reduce, and eight chances to lose a sign. The polar route has one modulus, one angle, and two trigonometric values. On a paper that gives you ten minutes an answer, that difference decides whether you finish.
There is one case where the binomial theorem is the better tool: a small power of a number whose argument is not a standard angle. (2 + 3i) squared is quicker expanded than converted, because the argument of 2 + 3i is not a nice angle and you would end up with an inverse tangent you could not evaluate.
Taking a Power with De Moivre's Theorem
So the rule of thumb: a standard angle, or a large power, means polar; an awkward angle and a small power means expand.
Powers that come out real or purely imaginary
Worth recognising, because a question is often built around one.
If the argument times n lands on a multiple of 180 degrees, the answer is real. If it lands on an odd multiple of 90 degrees, the answer is purely imaginary.
(1 + i)^4 = -4
(1 + i)^8 = 16
(1 + i)^2 = 2i
(1 + i)^6 = -8i
The argument of 1 + i is 45 degrees. Times two is 90, so the square is purely imaginary. Times four is 180, so the fourth power is a negative real. Times six is 270, purely imaginary again. Times eight is 360, a positive real.
That pattern is worth having: the powers of 1 + i walk round the four axes in order.
Using the theorem to prove an identity
A common form of question: prove something about cosines and sines, using De Moivre.
Prove that cos 4t equals eight cos to the fourth t minus eight cos squared t plus one.
Expand (cos t + i sin t) to the fourth with the binomial theorem and take the real part, which by the theorem must be cos 4t.
cos(4t) = cos(t)^4 - 6 cos(t)^2 sin(t)^2 + sin(t)^4
Now replace every sin squared by one minus cos squared.
cos(t)^4 - 6 cos(t)^2 (1 - cos(t)^2) + (1 - cos(t)^2)^2 = 8 cos(t)^4 - 8 cos(t)^2 + 1
cos(4t) = 8 cos(t)^4 - 8 cos(t)^2 + 1
The same expansion's imaginary part gives sin 4t.
sin(4t) = 4 cos(t)^3 sin(t) - 4 cos(t) sin(t)^3
Check yourself
(1 + i sqrt(3))^4 = -8 - 8i sqrt(3)
(2 + 2i)^5 = -128 - 128i
(cos(pi/9) + i sin(pi/9))^9 = -1
(1 - i)^12 = -64
The third is the neatest: an angle of 20 degrees taken nine times is 180 degrees, so the answer is exactly minus one, whatever you expected. That is the kind of question De Moivre is set for.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.