The Polar Form of a Complex Number
Chapter Sixteen
Syllabus topic Module 1, "1.1 Complex Numbers"
Pages 38 to 39 of 303
In one line
Every complex number except zero can be written as r(cos t + i sin t), where r is its modulus and t is its argument.
Where it comes from
The argument chapter gave two equations relating the parts to the modulus and the argument.
x = r cos(t)
y = r sin(t)
Substitute both into x + iy and take out the common factor r.
r cos(t) + i r sin(t) = r(cos(t) + i sin(t))
That is the polar form. The name is from polar coordinates, which is exactly what r and t are: a distance and a direction.
The form x + iy is called the Cartesian form, or the rectangular form, or the standard form. The two forms describe the same number and you must be able to move between them in both directions.
Cartesian to polar
Two steps, and the second is the one with the trap.
Step one, the modulus.
r = sqrt(x^2 + y^2)
Step two, the argument, using the acute angle and then the quadrant, exactly as the previous chapter set out. Never straight from a calculator's inverse tangent.
Worked: put 1 + i root 3 into polar form.
abs(1 + i sqrt(3)) = sqrt(1 + 3) = 2
The acute angle has tangent root three over one, so it is 60 degrees, pi over three. Both parts are positive, so the point is in the first quadrant and the argument is pi over three. Therefore:
2(cos(pi/3) + i sin(pi/3)) = 1 + i sqrt(3)
The check is the last line itself: expanding it must return the number you started with, and it does.
Worked again, in an awkward quadrant: put minus 1 minus i into polar form. The modulus is root two, the acute angle is 45 degrees, and both parts are negative so the point is in the third quadrant, giving an argument of pi over four minus pi, which is minus three pi over four.
sqrt(2)(cos(-3 pi/4) + i sin(-3 pi/4)) = -1 - i
Polar to Cartesian
Easier: work out the cosine and the sine and multiply.
4(cos(pi/6) + i sin(pi/6)) = 2 sqrt(3) + 2i
3(cos(pi) + i sin(pi)) = -3
5(cos(pi/2) + i sin(pi/2)) = 5i
The two forms, side by side
| Number | Cartesian | Modulus | Argument | Polar |
|---|---|---|---|---|
| 1 | 1 | 1 | 0 | cos 0 + i sin 0 |
| i | i | 1 | pi/2 | cos(pi/2) + i sin(pi/2) |
| -1 | -1 | 1 | pi | cos(pi) + i sin(pi) |
| 1 + i | 1 + i | sqrt(2) | pi/4 | sqrt(2)(cos(pi/4) + i sin(pi/4)) |
| -2i | -2i | 2 | -pi/2 | 2(cos(-pi/2) + i sin(-pi/2)) |
The two rules about the form itself
r must be non-negative. The modulus is a distance. If a calculation leaves you with a negative number in front of the bracket, absorb the minus sign into the angle by adding or subtracting 180 degrees, since minus one has argument pi.
The Polar Form of a Complex Number
-2(cos(pi/6) + i sin(pi/6)) = 2(cos(pi/6 + pi) + i sin(pi/6 + pi))
Both sides are the same number, but only the right-hand side is in proper polar form.
The sign inside the bracket must be plus. A bracket written cos t minus i sin t is not in polar form. Use the fact that cosine is even and sine is odd to flip it.
cos(t) - i sin(t) = cos(-t) + i sin(-t)
So a bracket with a minus sign inside it is the polar form of a number whose argument is the negative of the angle shown. That is the conjugate, which is why conjugating negates the argument.
Why it is worth the trouble
Two reasons, and both are about to be used.
Multiplication becomes easy. The previous chapter showed that multiplying multiplies moduli and adds arguments. In polar form that is one line, where in Cartesian form it is four products and a collection.
Powers and roots become possible at all. Squaring a Cartesian complex number is manageable; raising one to the tenth power by expanding brackets is not. In polar form it is one multiplication of the argument, and that is De Moivre's theorem.
A third reason, less examined but worth knowing: the polar form is what makes the connection to the exponential visible. Two chapters from here, the bracket cos t + i sin t turns out to be e to the power it, and then the polar form becomes r e^(it), which is the form every application actually uses.
Check yourself
Put each into polar form with its principal argument.
| Number | Modulus | Argument |
|---|---|---|
| 3i | 3 | pi/2 |
| -4 | 4 | pi |
| 1 - i | sqrt(2) | -pi/4 |
| -sqrt(3) + i | 2 | 5 pi/6 |
| -2 - 2i sqrt(3) | 4 | -2 pi/3 |
And in the other direction:
6(cos(pi/3) + i sin(pi/3)) = 3 + 3i sqrt(3)
sqrt(2)(cos(-pi/4) + i sin(-pi/4)) = 1 - i
The fourth row of the table is the one to check carefully: the acute angle has tangent one over root three, which is 30 degrees, and the point is in the second quadrant, so the argument is 180 minus 30, that is 150 degrees, which is five pi over six.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.