The Form r(cos θ + i sin θ)

The two legs

Polar form names a complex number by its modulus r and its argument $\theta$θ, as r at $\theta$θ. To write it as a single expression, go back to the right-angled triangle under its arrow on the Argand diagram.

The hypotenuse is the arrow, of length r, and the angle at the origin is $\theta$θ. The leg along the real axis is adjacent to $\theta$θ, so it is $r\cos \theta$r cos θ. The leg parallel to the imaginary axis is opposite $\theta$θ, so it is $r\sin \theta$r sin θ.

r = 2 at 60°. The dashed lines drop to the legs: 2 cos 60° = 1 on the real axis and $2\sin 60^{\circ }=\sqrt{3}$2 sin 60° = √3, about 1.732, on the imaginary axis.

Factor out the r

The number is its real leg plus i times its imaginary leg: $z=r\cos \theta +i\text{r}\sin \theta$z = r cos θ + i r sin θ. Both terms carry the factor r, so take it out: $z=r\left(\cos \theta +i\sin \theta \right)$z = r(cos θ + i sin θ).

This is the full notation for r at $\theta$θ. The r stands outside the brackets and multiplies both terms. The angle appears twice inside, once in the cosine and once in the sine, and it is the same angle both times.

r = 2 at 60° is 2(cos 60° + i sin 60°). Multiplied out, it is $2\times \frac{1}{2}+2\times \left(\frac{\sqrt{3}}{2}\right)i=1+\sqrt{3}i$2 × 1/2 + 2 × (√3/2)i = 1 + √3 i, the point in the figure. Reading the other way, 5(cos 40° + i sin 40°) has r = 5 and $\theta =40^{\circ }$θ = 40°; its parts are $5\cos 40^{\circ }\approx 3.830$5 cos 40° ≈ 3.830 and $5\sin 40^{\circ }\approx 3.214$5 sin 40° ≈ 3.214, so it is about 3.830 + 3.214i.

The real part is $r\cos \theta$r cos θ, the term with no i. The imaginary part is $r\sin \theta$r sin θ, the number that multiplies i.

A negative argument

1 − i has modulus $\sqrt{1+1}=\sqrt{2}$√(1 + 1) = √2. It lies in the fourth quadrant, below the real axis, and its parts are equal in size, so its argument is −45°. In the form it is $\sqrt{2}\left(\cos \left(-45^{\circ }\right)+i\sin \left(-45^{\circ }\right)\right)$√2(cos(−45°) + i sin(−45°)).

Since cos(−45°) = cos 45° and sin(−45°) = −sin 45°, the same value can be written $\sqrt{2}\left(\cos 45^{\circ }-i\sin 45^{\circ }\right)$√2(cos 45° − i sin 45°). That has the right value but is not in the form: the form has a plus sign between the two terms, and the angle inside both of them is the argument. With a negative argument, the negative angle goes in both places.

1 − i is $\sqrt{2}$√2 long and turned 45° clockwise from the positive real axis, so its argument is −45° and it is $\sqrt{2}\left(\cos \left(-45^{\circ }\right)+i\sin \left(-45^{\circ }\right)\right)$√2(cos(−45°) + i sin(−45°)).

The r is never negative

−3(cos 30° + i sin 30°) looks like the notation but is not in it, because r is a length and a length is never negative. Its value is −3 cos 30° − 3i sin 30°, about −2.598 − 1.5i, a point in the third quadrant.

Its length is 3, and it points in the opposite direction to 30°, half a turn away: 30° − 180° = −150°. So −3(cos 30° + i sin 30°) = 3(cos(−150°) + i sin(−150°)). Check: $3\cos \left(-150^{\circ }\right)\approx -2.598$3 cos(−150°) ≈ −2.598 and 3 sin(−150°) = −1.5.

Real and imaginary numbers fit the form too. −2 points along the negative real axis, so it is 2(cos 180° + i sin 180°). −4i points straight down, so it is 4(cos(−90°) + i sin(−90°)).

Multiplying in the form

Multiplying multiplies the lengths and adds the angles, and the notation shows both at once: 2(cos 20° + i sin 20°) × 3(cos 25° + i sin 25°) = 6(cos 45° + i sin 45°). The factors in front multiply, 2 × 3 = 6, and the angle inside becomes 20° + 25° = 45°.

Shortened to cis

Some books shorten $\cos \theta +i\sin \theta$cos θ + i sin θ to cis $\theta$θ, from the letters c, i, s of cos, i, sin. Then 2(cos 60° + i sin 60°) is written 2 cis 60°, and 1 − i is $\sqrt{2}$√2 cis(−45°).

The shorthand changes nothing else: the real part of r cis $\theta$θ is $r\cos \theta$r cos θ and the imaginary part is $r\sin \theta$r sin θ.

The usual mistakes

Letting the r reach only the cosine. 2 cos 60° + i sin 60° is 1 + 0.866i, whose length is about 1.32, not 2. The brackets put the 2 on both terms.

A minus sign between the terms. 2(cos 60° − i sin 60°) is $1-\sqrt{3}i$1 − √3 i, below the real axis; the form always has a plus, and the sign of the angle carries the direction.

Swapping r and $\theta$θ. In 5(cos 40° + i sin 40°) the length is the factor in front, 5, and the angle is inside, 40°.

Doubling the angle. $\theta$θ is written twice, once with cos and once with sin, but there is one angle: in 3(cos 25° + i sin 25°), $\theta =25^{\circ }$θ = 25°, not 50°.

Taking $r\sin \theta$r sin θ as the real part. $r\sin \theta$r sin θ multiplies i, so it is the imaginary part; the real part is $r\cos \theta$r cos θ.