(1 + sqrt(3) i)^5 with De Moivre's theorem, step by step

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The problem

Work out this power and give the answer in the form $a + bi$ .

\left( 1 + \sqrt{3} i \right)^{5}

The result

\left( 1 + \sqrt{3} i \right)^{5} = 16 - 16 \sqrt{3} i

In polar form the answer has length $32$ and direction $-\frac{\pi}{3}$ .

checked 14 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 8Step 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{\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.

Both parts are positive, so $z$ lies in the first quadrant: right of the vertical axis and above the horizontal one. Its angle will be between $0$ and $\frac{\pi}{2}$ .

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{\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{1^{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 \sin \theta

=

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

Sine is the up side over the long side:

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

. checked

\theta = \frac{\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

60^\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{\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{\pi}{3} \right) + i \sin\left( \frac{\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{\pi}{3}

See what multiplying does in polar form

Why this step

Multiplying out $(a + bi)^{5}$ directly means 4 rounds of expanding brackets, each one a chance to slip. Polar form replaces all of that with one idea: multiplying by $z$ turns the plane through the angle $\theta$ and stretches it by the factor $r$ . A power is the same turn and the same stretch repeated, so angles ADD and lengths MULTIPLY.

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

A first test of the idea with

n = 2

: squaring

z

should square the length and double the angle. Both sides are evaluated and they agree. checked

Multiplying by z once more. The second arrow is the first turned through the same angle theta again, and stretched by the same factor r again: that is what squaring does, seen rather than computed.

Apply De Moivre: raise the length, multiply the angle

Why this step

The theorem splits the job into two small real-number calculations that cannot interfere with each other: a power of the positive number $r$ , and $n$ times the angle. The only subtlety is that $n\theta$ may run past a full turn, so we bring it back into the principal range at the end.

\displaystyle 2^{5}

=

\displaystyle 32

Multiply

2

by itself

5

times. checked

\displaystyle 5 \cdot \frac{\pi}{3}

=

\displaystyle \frac{5 \pi}{3}

Multiply the angle by

5

. checked

\displaystyle \frac{5 \pi}{3} - 1 \cdot 2 \pi

=

\displaystyle -\frac{\pi}{3}

A full turn is

2\pi

and changes nothing about a direction. Take away

1

full turn to land in the principal range

-\pi < \theta \le \pi

. checked

\left( 1 + \sqrt{3} i \right)^{5} = 32 \left( \cos\left( -\frac{\pi}{3} \right) + i \sin\left( -\frac{\pi}{3} \right) \right)

De Moivre's theorem with the two results above. This is the answer in polar form. checked

Every power from z up to the answer. Each arrow is the one before it turned through theta and stretched by r, so reaching the 5th power is that same turn done 5 times. Count the arcs: there are 5 of them.

Turn the answer back into the form a + bi

Why this step

Polar form was the right tool for the power, but an answer is usually wanted in the form the question was asked in. Going back is the easy direction: work out one cosine and one sine, and multiply through by the length.

\displaystyle 32 \left( \cos\left( -\frac{\pi}{3} \right) + i \sin\left( -\frac{\pi}{3} \right) \right)

=

\displaystyle 32 \left( \frac{1}{2} - \frac{\sqrt{3}}{2} i \right)

The exact cosine and sine of

-\frac{\pi}{3}

, from the special triangles with the signs of its quadrant. checked

=

\displaystyle 16 - 16 \sqrt{3} i

Multiply each part by

32

. checked

The answer back on the plane, with its two coordinates read off the axes: across to the real part, down to the imaginary part. That pair of readings IS the a + bi form.

Check by multiplying the long way

Why this step

An answer reached by one road is a claim; the same answer reached by two unrelated roads is a fact. Here the second road is brute force: multiply by $z$ again and again, using $i^2 = -1$ each time. It also shows how much labour the polar form saved.

$k$	$z^k$	length	direction
$1$	$1 + \sqrt{3} i$	$2$	$60.0^\circ$
$2$	$-2 + 2 \sqrt{3} i$	$4$	$120.0^\circ$
$3$	$-8$	$8$	$180.0^\circ$
$4$	$-8 - 8 \sqrt{3} i$	$16$	$-120.0^\circ$
$5$	$16 - 16 \sqrt{3} i$	$32$	$-60.0^\circ$

Each row is the row above multiplied by z, exactly. Watch the length column multiply by r and the angle column step forward by the same amount each time.

The DIRECTIONS of z, z squared, z cubed and so on, all drawn at length 1 so they fit on one picture. Each multiplication by z turns the arrow through the same angle theta.

Power of a complex number with De Moivre's theorem