Polar form of -1 + sqrt(3) i: modulus and argument, step by step

Femtobytes

femtobytes.com/solve/complex-polar-form/-1_1_1_3

The problem

Write this complex number in polar form.

z = -1 + \sqrt{3} i

The result

z = -1 + \sqrt{3} i = 2 \left( \cos \left( \frac{2 \pi}{3} \right) + i \sin \left( \frac{2 \pi}{3} \right) \right)

The modulus is $r = 2$ and the principal argument is $\theta = \frac{2 \pi}{3}$ .

checked 8 lines of this solution were evaluated with exact arithmetic before it was shown to you. A solution with a false line is never served.

Picture 1 of 5Step 1: See where the number lives

The number z drawn as an arrow from the origin. Its length is the modulus r and the angle it makes with the positive real axis is the argument, written with the Greek letter theta. Here

r = 2

and

\theta = \frac{2 \pi}{3}

See where the number lives

Why this step

A complex number is two real numbers travelling together, so it cannot sit on a number line; it needs a plane. Rectangular form $a + bi$ says "go $a$ across and $b$ up". Polar form describes the SAME point as "face this direction, walk this far". The reason to bother: when complex numbers are multiplied, their lengths multiply and their directions add. Rectangular form hides that completely; polar form makes it visible.

Here $z = -1 + \sqrt{3} i$ , so the real part is $a = -1$ and the imaginary part is $b = \sqrt{3}$ . The symbol $i$ is the imaginary unit, the number with $i^2 = -1$ . On the plane, $z$ is the point $(a, b)$ : $a$ is read on the horizontal (real) axis and $b$ on the vertical (imaginary) axis.

The real part is negative and the imaginary part positive, so $z$ lies in the second quadrant: to the left and above. Its angle will be between $\frac{\pi}{2}$ and $\pi$ .

The number z drawn as an arrow from the origin. Its length is the modulus r and the angle it makes with the positive real axis is the argument, written with the Greek letter theta. Here

r = 2

and

\theta = \frac{2 \pi}{3}

Find the modulus: how far z is from the origin

Why this step

The arrow to $z$ is the long side of a right-angled triangle whose other sides are $\left| a \right|$ across and $\left| b \right|$ up. Pythagoras gives its length. We square $a$ and $b$ first, which also throws away their signs: a distance cannot care whether we went left or right.

\displaystyle r

=

\displaystyle \sqrt{\left( -1 \right)^{2} + \left( \sqrt{3} \right)^{2}}

The modulus

r = \left| z \right| = \sqrt{a^2 + b^2}

, by Pythagoras. checked

=

\displaystyle \sqrt{1 + 3}

Square each part. Squaring a square root undoes it, for example

\left( \sqrt{3} \right)^2 = 3

, and the square of a negative number is positive. checked

=

\displaystyle \sqrt{4}

Add. checked

=

\displaystyle 2

4

is a perfect square, so the root comes out exactly. checked

The same arrow, with the right-angled triangle it is the long side of. The side across is the real part and the side up is the imaginary part, so the modulus is the length Pythagoras gives from those two.

Find the argument: which way z points

Why this step

The direction is an angle $\theta$ measured from the positive real axis, anticlockwise counted as positive. One reading is not enough to fix it: the cosine alone cannot tell up from down, and the sine alone cannot tell left from right. Two readings together leave exactly one direction.

\displaystyle \cos \theta

=

\displaystyle \frac{-1}{2}

Cosine is the across side over the long side:

\cos \theta = \frac{a}{r}

. checked

=

\displaystyle -\frac{1}{2}

Simplify. If a root is left on the bottom, multiply top and bottom by that root to move it to the top. checked

\displaystyle \sin \theta

=

\displaystyle \frac{\sqrt{3}}{2}

Sine is the up side over the long side:

\sin \theta = \frac{b}{r}

. checked

\theta = \frac{2 \pi}{3}

This is the one angle in the range

-\pi < \theta \le \pi

whose cosine is

-\frac{1}{2}

and whose sine is

\frac{\sqrt{3}}{2}

: the sizes pick out the reference angle, and the two signs pick out the quadrant. In degrees it is

120^\circ

The argument read off the plane: the arc is the turn from the positive real axis round to the arrow, and that turn is what the argument names. The quadrant the arrow lands in is what decides whether the arctangent needs correcting. Here

r = 2

and

\theta = \frac{2 \pi}{3}

Write the polar form, and check it by turning it back

Why this step

From the triangle, $a = r \cos \theta$ and $b = r \sin \theta$ : cosine is "across over long side" and sine is "up over long side". Putting those into $a + bi$ and taking $r$ out as a common factor gives the polar form. We then multiply it back out, because a conversion that cannot be reversed has not been checked.

2 \left( \cos\left( \frac{2 \pi}{3} \right) + i \sin\left( \frac{2 \pi}{3} \right) \right) = -1 + \sqrt{3} i

Work out the cosine and the sine, multiply by

r

, and the original number returns exactly. checked

The polar form read off the picture: the circle is every number this same distance from the origin, and the angle turned from the positive real axis picks out which one. A length and a turn name the number just as well as a real part and an imaginary part. Here

r = 2

and

\theta = \frac{2 \pi}{3}

Every other angle that names the same direction

Why this step

Turning a full circle brings an arrow back to where it started, so a direction does not have one angle; it has endlessly many, all differing by whole turns. This looks like a nuisance. It is in fact the reason a complex number has $n$ different $n$ -th roots, so it is worth stating now.

\arg z = \frac{2 \pi}{3} + 2k\pi, \qquad k = 0, \pm 1, \pm 2, \ldots

Every one of these angles points the same way as

\theta

. The one we found, the unique choice with

-\pi < \theta \le \pi

, is called the principal argument and is written

\operatorname{Arg} z

with a capital A.

The same arrow, reached by the angle we found or by that angle plus any number of whole turns. Every one of them points here; the principal one is simply the name we agree to use. Here

r = 2

and

\theta = \frac{2 \pi}{3}

Polar form of a complex number