Polar Form

A length and a direction

A complex number a + bi is fixed by its two parts. It is fixed just as well by the arrow from the origin to it on the Argand diagram: how long the arrow is, and which way it points.

The length is the modulus, r = |z|. The direction is the argument, $\theta =$θ = arg z, the angle from the positive real axis to the arrow, counted positive counterclockwise. The argument is taken in the range $-180^{\circ }<\theta \le 180^{\circ }$−180° < θ ≤ 180°, so a number below the real axis has a negative argument.

For 3 + 4i, $r=\sqrt{3^2+4^2}=\sqrt{25}=5$r = √(3² + 4²) = √25 = 5. The point is in the first quadrant and $\tan \theta =\frac{4}{3}$tan θ = 4/3, so $\theta =53.13^{\circ }$θ = 53.13° to 2 decimal places. In polar form, 3 + 4i is r = 5 at 53.13°.

The arrow to 3 + 4i is 5 long and makes 53.13° with the positive real axis. The dashed lines drop to 3 on the real axis and 4 on the imaginary axis.

From polar form back to the parts

The arrow is the hypotenuse of a right-angled triangle whose legs are the two parts. The leg along the real axis is adjacent to $\theta$θ and the other leg is opposite it. So the real part is $a=r\cos \theta$a = r cos θ and the imaginary part is $b=r\sin \theta$b = r sin θ.

Check on 3 + 4i: 5 cos 53.13° = 3.00 and 5 sin 53.13° = 4.00, the parts the number started with.

r = 2 at 30° is $2\cos 30^{\circ }+2i\sin 30^{\circ }=\sqrt{3}+i$2 cos 30° + 2i sin 30° = √3 + i, about 1.732 + i. r = 4 at 135° is $4\cos 135^{\circ }+4i\sin 135^{\circ }=-2\sqrt{2}+2\sqrt{2}i$4 cos 135° + 4i sin 135° = −2√2 + 2√2 i, about −2.828 + 2.828i: cos 135° is negative, so the real part is negative and the point is in the second quadrant. r = 3 at −90° is 3 cos(−90°) + 3i sin(−90°) = 0 − 3i = −3i, straight down the imaginary axis.

From the parts to polar form

−2 − 2i has modulus $\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$√(4 + 4) = √8 = 2√2. Its parts are equal in size, so the angle the arrow makes with the real axis is 45°. The point is in the third quadrant, below the real axis and to the left, so the argument is −(180° − 45°) = −135°. In polar form, −2 − 2i is $2\sqrt{2}$2√2 at −135°.

Turning 225° counterclockwise points the arrow the same way, but 225° is outside the range, so the argument is −135°.

The lengths multiply and the angles add

$\left(1+i\right)\left(1+i\right)=1+2i+i^2=2i$(1 + i)(1 + i) = 1 + 2i + i² = 2i. In polar form, 1 + i is $\sqrt{2}$√2 at 45°, because $\sqrt{1+1}=\sqrt{2}$√(1 + 1) = √2 and its parts are equal. The product 2i is 2 at 90°. The lengths multiplied, $\sqrt{2}\times \sqrt{2}=2$√2 × √2 = 2, and the angles added, 45° + 45° = 90°.

Another pair: $\left(1+i\right)\left(\sqrt{3}+i\right)=\sqrt{3}+i+\sqrt{3}i+i^2=\left(\sqrt{3}-1\right)+\left(\sqrt{3}+1\right)i$(1 + i)(√3 + i) = √3 + i + √3 i + i² = (√3 − 1) + (√3 + 1)i, about 0.732 + 2.732i. Here $\sqrt{3}+i$√3 + i is 2 at 30°, since $\sqrt{3+1}=2$√(3 + 1) = 2 and $\tan \theta =\frac{1}{\sqrt{3}}$tan θ = 1/√3. The product’s modulus is $\sqrt{{\left(\sqrt{3}-1\right)}^2+{\left(\sqrt{3}+1\right)}^2}=\sqrt{4-2\sqrt{3}+4+2\sqrt{3}}=\sqrt{8}=2\sqrt{2}$√((√3 − 1)² + (√3 + 1)²) = √(4 − 2√3 + 4 + 2√3) = √8 = 2√2, which is $\sqrt{2}\times 2$√2 × 2. Its argument is 75°, which is 45° + 30°.

The arrows to 1 + i and $\sqrt{3}+i$√3 + i, and to their product $\left(\sqrt{3}-1\right)+\left(\sqrt{3}+1\right)i$(√3 − 1) + (√3 + 1)i. The product is $\sqrt{2}\times 2=2\sqrt{2}$√2 × 2 = 2√2 long and points at 45° + 30° = 75°.

Why the angles add

Take $r_1$r₁ at $\alpha$α and $r_2$r₂ at $\beta$β. In parts they are $r_1\cos \alpha +i{r}_1\sin \alpha$r₁ cos α + i r₁ sin α and $r_2\cos \beta +i{r}_2\sin \beta$r₂ cos β + i r₂ sin β. Each r is a factor of both of its number’s parts, so the product is $r_1{r}_2$r₁r₂ times $\left(\cos \alpha +i\sin \alpha \right)\left(\cos \beta +i\sin \beta \right)$(cos α + i sin α)(cos β + i sin β).

Multiply out, with $i^2=-1$i² = −1: $\left(\cos \alpha +i\sin \alpha \right)\left(\cos \beta +i\sin \beta \right)=\left(\cos \alpha \cos \beta -\sin \alpha \sin \beta \right)+i\left(\sin \alpha \cos \beta +\cos \alpha \sin \beta \right)$(cos α + i sin α)(cos β + i sin β) = (cos α cos β − sin α sin β) + i(sin α cos β + cos α sin β). The real part is the expansion of $\cos \left(\alpha +\beta \right)$cos(α + β), and the imaginary part is the expansion of $\sin \left(\alpha +\beta \right)$sin(α + β).

So the product is $r_1{r}_2\cos \left(\alpha +\beta \right)+i{r}_1{r}_2\sin \left(\alpha +\beta \right)$r₁r₂ cos(α + β) + i r₁r₂ sin(α + β): length $r_1{r}_2$r₁r₂ at angle $\alpha +\beta$α + β. That holds for any two complex numbers, which is why multiplying is quicker in polar form than in parts.

An angle past 180°

Adding angles can carry the argument out of range. r = 2 at 120° times r = 3 at 150° is 2 × 3 = 6 at 120° + 150° = 270°. 270° is past 180°, so take off a full turn: 270° − 360° = −90°. The product is 6 at −90°, which is −6i.

Check in parts. 2 at 120° is $-1+\sqrt{3}i$−1 + √3 i, and 3 at 150° is $-\frac{3\sqrt{3}}{2}+\left(\frac{3}{2}\right)i$−(3√3)/2 + (3/2)i. The real part of their product is $\left(-1\right)\left(-\frac{3\sqrt{3}}{2}\right)-\left(\sqrt{3}\right)\left(\frac{3}{2}\right)=\frac{3\sqrt{3}}{2}-\frac{3\sqrt{3}}{2}=0$(−1)(−(3√3)/2) − (√3)(3/2) = (3√3)/2 − (3√3)/2 = 0. The imaginary part is $\left(-1\right)\left(\frac{3}{2}\right)+\left(\sqrt{3}\right)\left(-\frac{3\sqrt{3}}{2}\right)=-\frac{3}{2}-\frac{9}{2}=-6$(−1)(3/2) + (√3)(−(3√3)/2) = −3/2 − 9/2 = −6. So the product is −6i.

The usual mistakes

Adding the lengths. 2 at 30° times 3 at 60° is 6 at 90°, not 5 at 90°: multiplying by 3 at 60° scales the length 2 by 3.

Multiplying the angles. The angle is not 30° × 60° = 1800°: each factor adds its own turn, so it is 30° + 60° = 90°.

Leaving the argument out of range. 2 at 120° times 3 at 150° is 6 at −90°; 270° names the same direction, but the argument is −90°.

Reading the angle off tan alone. For −2 − 2i, $\tan \theta =1$tan θ = 1, but the point is in the third quadrant, so the argument is −135°, not 45°.

Two contacts on a radar

In the application below, contact A is given as 10(cos 120° + i sin 120°), which is r = 10 at 120° in the notation of the next lesson: its real part is 10 cos 120° and its imaginary part is 10 sin 120°. Contact B = 6 − 8i goes the other way, from parts to a length and an angle. B lies below the real axis, so its argument is negative.