Dividing in Polar Form

Dividing undoes multiplying

Multiplying by $r_2$r₂ at ${\theta }_2$θ₂ multiplies a length by $r_2$r₂ and adds ${\theta }_2$θ₂ to an angle. Dividing by $r_2$r₂ at ${\theta }_2$θ₂ has to undo both: divide the length by $r_2$r₂ and take ${\theta }_2$θ₂ off the angle. So $r_1$r₁ at ${\theta }_1$θ₁ divided by $r_2$r₂ at ${\theta }_2$θ₂ is $\frac{r_1}{r_2}$r₁/r₂ at ${\theta }_1-{\theta }_2$θ₁ − θ₂.

To see that this is the only answer, call the quotient w. Then $w\times (r_2$w × (r₂ at ${\theta }_2)=r_1$θ₂) = r₁ at ${\theta }_1$θ₁. Multiplying multiplies the lengths, so $|w|\times r_2=r_1$|w| × r₂ = r₁ and $|w|=\frac{r_1}{r_2}$|w| = r₁/r₂. It adds the angles, so arg $w+{\theta }_2={\theta }_1$w + θ₂ = θ₁ and arg $w={\theta }_1-{\theta }_2$w = θ₁ − θ₂.

The order matters in both parts: the top’s modulus is divided by the bottom’s, and the bottom’s argument is taken from the top’s.

One division, two ways

Divide 6i by $\sqrt{3}+i$√3 + i. 6i is 6 at 90°. $\sqrt{3}+i$√3 + i has modulus $\sqrt{3+1}=2$√(3 + 1) = 2 and lies in the first quadrant with $\tan \theta =\frac{1}{\sqrt{3}}$tan θ = 1/√3, so it is 2 at 30°. The quotient is 6 ÷ 2 = 3 at 90° − 30° = 60°.

In parts, 3 at 60° is $3\cos 60^{\circ }+3i\sin 60^{\circ }=\frac{3}{2}+\left(\frac{3\sqrt{3}}{2}\right)i$3 cos 60° + 3i sin 60° = 3/2 + (3√3/2)i, about 1.5 + 2.598i.

Dividing Complex Numbers reaches the same number by the conjugate: $6i\left(\sqrt{3}-i\right)/\left(\left(\sqrt{3}+i\right)\left(\sqrt{3}-i\right)\right)=\left(6\sqrt{3}i-6i^2\right)/\left(3+1\right)=\frac{6+6\sqrt{3}i}{4}=1.5+1.5\sqrt{3}i$6i(√3 − i) / ((√3 + i)(√3 − i)) = (6√3 i − 6i²) / (3 + 1) = (6 + 6√3 i)/4 = 1.5 + 1.5√3 i, and $1.5\sqrt{3}=\frac{3\sqrt{3}}{2}$1.5√3 = 3√3/2.

6i is 6 long at 90°, and $\sqrt{3}+i$√3 + i is 2 long at 30°. Their quotient is 6 ÷ 2 = 3 long, at 90° − 30° = 60°.

Check by multiplying back

A quotient times the number divided by gives back the number divided into. 3 at 60° × 2 at 30° = 3 × 2 at 60° + 30° = 6 at 90°, which is 6i.

In parts: $\left(\frac{3}{2}+\left(\frac{3\sqrt{3}}{2}\right)i\right)\left(\sqrt{3}+i\right)=\frac{3\sqrt{3}}{2}+\left(\frac{3}{2}\right)i+\left(\frac{9}{2}\right)i+\left(\frac{3\sqrt{3}}{2}\right)i^2=\frac{3\sqrt{3}}{2}-\frac{3\sqrt{3}}{2}+6i=6i$(3/2 + (3√3/2)i)(√3 + i) = (3√3)/2 + (3/2)i + (9/2)i + (3√3/2)i² = (3√3)/2 − (3√3)/2 + 6i = 6i.

Bring the angle back into range

Subtracting can give a negative argument, which is fine as long as it is not below −180°. 4 at 30° ÷ 2 at 150° = 2 at 30° − 150° = 2 at −120°, which is $-1-\sqrt{3}i$−1 − √3 i, in the third quadrant.

8 at −150° ÷ 2 at 60° = 4 at −150° − 60° = 4 at −210°. −210° is below −180°, so add a full turn: −210° + 360° = 150°. The quotient is 4 at 150°, which is $4\cos 150^{\circ }+4i\sin 150^{\circ }=-2\sqrt{3}+2i$4 cos 150° + 4i sin 150° = −2√3 + 2i, about −3.464 + 2i.

One over a complex number

1 is 1 at 0°, so $1/(r$1 / (r at $\theta )=\frac{1}{r}$θ) = 1/r at $0^{\circ }-\theta =\frac{1}{r}$0° − θ = 1/r at $-\theta$−θ. The reciprocal of 2 at 60° is $\frac{1}{2}$1/2 at −60°, which is $\left(\frac{1}{2}\right)\cos \left(-60^{\circ }\right)+\left(\frac{1}{2}\right)i\sin \left(-60^{\circ }\right)=\frac{1}{4}-\left(\frac{\sqrt{3}}{4}\right)i$(1/2)cos(−60°) + (1/2)i sin(−60°) = 1/4 − (√3/4)i, about 0.25 − 0.433i.

For a number on the unit circle, r = 1, so $1/(1$1 / (1 at $\theta )=1$θ) = 1 at $-\theta$−θ: the reciprocal is the mirror image in the real axis. $\left(1+i\right)/\left(1-i\right)$(1 + i) / (1 − i) is $\sqrt{2}$√2 at $45^{\circ }÷\sqrt{2}$45° ÷ √2 at −45° = 1 at 90° = i.

The usual mistakes

Adding the angles. 6 at 90° ÷ 2 at 30° is not 3 at 120°; adding is what multiplying does, and dividing subtracts.

Subtracting the lengths. The modulus is 6 ÷ 2 = 3, not 6 − 2 = 4: dividing scales the length down by the factor 2.

Dividing the angles. 90° ÷ 30° = 3 is a number of times, not an angle; the angles subtract, 90° − 30° = 60°.

Subtracting the wrong way round. ${\theta }_1-{\theta }_2$θ₁ − θ₂ takes the bottom’s angle from the top’s. 30° − 90° = −60° belongs to (2 at 30°) ÷ (6 at 90°), which is $\frac{1}{3}$1/3 at −60°.

Leaving the argument out of range. 4 at −210° points the same way as 4 at 150°, but the argument is 150°.

A motor on a 260-volt supply

In the application below, the voltage 260 lies along the real axis, so it is 260 at 0°. The impedance 12 + 5i is put into polar form first. Dividing takes its argument away from 0°, so the current’s argument is negative: the current lags behind the voltage.