The Argument of a Complex Number

The direction of the arrow

The modulus says how long the arrow from the origin is. The argument says which way it points: it is the angle from the positive real axis to the arrow, measured counterclockwise.

For 4 + 4i the arrow goes 4 across and 4 up, along the diagonal, so its argument is 45°.

The arrow for 4 + 4i. The arc turns counterclockwise from the positive real axis to the arrow, through 45°: the argument of 4 + 4i.

$\tan \theta =\frac{b}{a}$tan θ = b/a

The arrow for a + bi is the hypotenuse of a right-angled triangle with the real part a across and the imaginary part b up. Seen from the origin, b is opposite the angle $\theta$θ and a is adjacent to it, so $\tan \theta =\frac{b}{a}$tan θ = b/a.

For 4 + 4i, $\tan \theta =\frac{4}{4}=1$tan θ = 4/4 = 1, so $\theta =45^{\circ }$θ = 45°. For 3 + 4i, $\tan \theta =\frac{4}{3}$tan θ = 4/3 and $\theta ={\tan }^{-1}\left(\frac{4}{3}\right)\approx 53.13^{\circ }$θ = tan⁻¹(4/3) ≈ 53.13°. For $1+\sqrt{3}i$1 + √3i, $\tan \theta =\sqrt{3}$tan θ = √3 and $\theta =60^{\circ }$θ = 60°.

The triangle for 3 + 4i, with legs 3 and 4. The side opposite $\theta$θ is the imaginary part 4 and the side next to it is the real part 3, so $\tan \theta =\frac{4}{3}$tan θ = 4/3 and $\theta \approx 53.13^{\circ }$θ ≈ 53.13°.

On the axes

A positive real number points along the positive real axis, so the argument of 3 is 0°. The arrow for 5i points straight up, a quarter turn, so its argument is 90°; there $\tan \theta =\frac{5}{0}$tan θ = 5/0 has no value, and the picture gives the answer.

A negative real number points the other way along the real axis: the argument of −2 is 180°. The arrow for −3i points straight down, and its argument is −90°.

One angle for each direction

Turning a full 360° comes back to the same direction, so 45° and 405° name the same arrow. To give each direction one argument, it is taken between −180° and 180°: $-180^{\circ }<\theta \le 180^{\circ }$−180° < θ ≤ 180°.

Arrows above the real axis then have positive arguments, turned counterclockwise. Arrows below it have negative arguments, turned clockwise: the argument of 2 − 2i is −45°, not 315°. In radians the same range is $-\pi <\theta \le \pi$−π < θ ≤ π, and 135° is $\frac{3\pi }{4}$3π/4.

In the second quadrant ${\tan }^{-1}\left(\frac{b}{a}\right)$tan⁻¹(b/a) points the wrong way

Take −3 + 3i, 3 to the left and 3 up. Here $\frac{b}{a}=\frac{3}{-3}=-1$b/a = 3/(−3) = −1, and a calculator gives ${\tan }^{-1}\left(-1\right)=-45^{\circ }$tan⁻¹(−1) = −45°. An arrow at −45° points down and to the right, toward 3 − 3i, exactly opposite to −3 + 3i. Both arrows have $\tan \theta =-1$tan θ = −1, and a calculator’s ${\tan }^{-1}$tan⁻¹ only returns angles between −90° and 90°.

Use the sizes of the parts instead. The arrow makes an angle $\alpha$α with the negative real axis, where $\tan \alpha =\frac{3}{3}=1$tan α = 3/3 = 1, so $\alpha =45^{\circ }$α = 45°. The arrow is in the second quadrant, so $\theta =180^{\circ }-45^{\circ }=135^{\circ }$θ = 180° − 45° = 135°.

The gold arrow is −3 + 3i, at 135°. The dashed arrow is the direction ${\tan }^{-1}\left(\frac{b}{a}\right)=-45^{\circ }$tan⁻¹(b/a) = −45° gives, which points the opposite way, into the fourth quadrant.

And in the third

Take −4 − 3i, 4 to the left and 3 down. Here $\frac{b}{a}=\frac{-3}{-4}=0.75$b/a = (−3)/(−4) = 0.75, and ${\tan }^{-1}\left(0.75\right)\approx 36.87^{\circ }$tan⁻¹(0.75) ≈ 36.87°, which points up and to the right, toward 4 + 3i: again exactly opposite.

From the sizes, $\tan \alpha =\frac{3}{4}$tan α = 3/4 and $\alpha \approx 36.87^{\circ }$α ≈ 36.87°, the angle below the negative real axis. The arrow is below the real axis, so the argument is negative: $\theta =-\left(180^{\circ }-36.87^{\circ }\right)=-143.13^{\circ }$θ = −(180° − 36.87°) = −143.13°.

The gold arrow is −4 − 3i. Its argument turns clockwise from the positive real axis, through −143.13°. The dashed arrow is the direction ${\tan }^{-1}\left(\frac{b}{a}\right)\approx 36.87^{\circ }$tan⁻¹(b/a) ≈ 36.87° gives, the opposite one.

One method for every quadrant

Sketch the point first. Find the angle $\alpha$α between the arrow and the real axis from the sizes of the parts, $\tan \alpha =|b|$tan α = |b|/|a|. Then:

in the first quadrant, $\theta =\alpha$θ = α; in the second, $\theta =180^{\circ }-\alpha$θ = 180° − α; in the third, $\theta =-\left(180^{\circ }-\alpha \right)$θ = −(180° − α); in the fourth, $\theta =-\alpha$θ = −α.

For the four numbers $1+\sqrt{3}i$1 + √3i, $-1+\sqrt{3}i$−1 + √3i, $-1-\sqrt{3}i$−1 − √3i and $1-\sqrt{3}i$1 − √3i, $\tan \alpha =\sqrt{3}$tan α = √3 each time, so $\alpha =60^{\circ }$α = 60°. Their arguments are 60°, 120°, −120° and −60°.

Four arrows of length 2, each at 60° to the real axis. Their arguments are 60° and 120° above the axis, −60° and −120° below it. The gold arrows, in the second and third quadrants, are the ones ${\tan }^{-1}\left(\frac{b}{a}\right)$tan⁻¹(b/a) gets wrong.

The conjugate and the negative

The conjugate is the reflection in the real axis, so its argument is the negative of the original: 3 + 4i has argument about 53.13°, and 3 − 4i has about −53.13°.

The negative is a half turn: −3 − 4i has argument $53.13^{\circ }-180^{\circ }\approx -126.87^{\circ }$53.13° − 180° ≈ −126.87°. A modulus and an argument together fix a point, just as a real part and an imaginary part do: modulus 5 and argument 53.13° give 3 + 4i.

The usual mistakes

Trusting tan alone. The argument of −1 + i is not 45°: the arrow is in the second quadrant, so it is 180° − 45° = 135°.

Taking the calculator’s ${\tan }^{-1}\left(\frac{b}{a}\right)$tan⁻¹(b/a) as the answer. For −1 + i it gives −45°, which is in the fourth quadrant, the opposite corner.

Going outside the range. The argument of −1 − i is −135°, not 225°.

Measuring from the imaginary axis. $\tan \theta =\frac{b}{a}$tan θ = b/a, imaginary over real; for $1+\sqrt{3}i$1 + √3i, $\frac{a}{b}$a/b would give 30° instead of 60°.

Turning the wrong way for a real number. The argument of 3 is 0°, not 180°: 180° is the direction of a negative real number.