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The polar form and Euler’s formula

Name a complex number by its length and its angle and multiplying becomes easy: lengths multiply and angles add. For two points of the unit circle that proves the angle-addition formulas for every angle, and Euler’s formula gives each point of the circle a short name.
M01.9.3L0 Foundation25 minIn review

Before this lesson

M01.9.2 Adding and multiplying: moving, turning and scaling M01.4.2 Exponential growth and decay M01.5.4 The law of cosines and the angle-addition formulas

After this lesson you can

  • Write a complex number in polar form, r(cos θ + i sin θ), and convert between the polar and the rectangular form.
  • Derive the angle-addition formulas for every pair of angles by multiplying two numbers on the unit circle.
  • Use Euler’s formula, e to the power iθ equals cos θ + i sin θ, as a name whose rule of powers says that arguments add, and support it with a compounding table.
  • Compute moduli, arguments and powers with the cmath module, and compare complex results with a tolerance.

In review. The maths and code on this page are checked by automated tests on every change, but an independent reader has not signed it off yet. Spot a mistake or a confusing step? Tell us.

The question

Raise \({3 + 4i}\) to the tenth power by multiplying out brackets and you face nine long multiplications. Yet the answer can be read off at once: its length is \(5^{10}\) and its angle is ten times that of \({3 + 4i}\). And the odd-looking line \({e^{i\pi} = -1}\) ties together three numbers that seem to have nothing in common: \(e\) from growth (M01.4.2), \(i\) from this module and \(\pi\) from circles (M01.5.3). This lesson finds the language in which both facts are plain. Signal processing speaks it every day: the amplitude and the phase of a wave are one complex number.

From the last lesson

1. Where does multiplying by \(i\) send \({2 + 5i}\)?

Show answer

To \({-5 + 2i}\): the point \({(a, b)}\) goes to \({(-b, a)}\), a quarter turn anticlockwise about 0.

2. \(u\) has length 2 and angle \(30^\circ\), and \(z\) has length 5 and angle \(53.13^\circ\). Give the length and the angle of \(uz\).

Show answer

Lengths multiply and angles add: length 10, angle \(83.13^\circ\).

Intuition

Every arrow from 0 is a length and a direction. The direction is a point of the unit circle: at angle \(\theta\) it is \({(\cos\theta, \sin\theta)}\), by M01.5.2, which as a complex number is \({\cos\theta + i\sin\theta}\). The length is a positive number \(r\). Multiply the direction by \(r\) and you have the arrow:

\[ z = r(\cos\theta + i\sin\theta) . \]

Two panels. Left, titled A direction, on the unit circle: a unit circle with a blue arrow to the point cos theta plus i sin theta, equals 0.6 plus 0.8 i, with an orange angle theta at 0 and a right triangle whose green sides are labelled cos theta equals 0.6 across and sin theta equals 0.8 up. Right, titled A length times the direction: a complex plane with the unit circle and a dashed circle of radius 5. A blue arrow runs from 0 through the point 0.6 plus 0.8 i on the unit circle and on to z equals 3 plus 4 i on the large circle. It is labelled r equals 5 in orange, an orange arc marks theta equals 0.9273, which is 53.13 degrees, and a label reads z equals 5 times (cos theta plus i sin theta).
M01.9.3-F1

The polar form. The direction of \({3 + 4i}\) is the point \({0.6 + 0.8i}\) of the unit circle; five times that direction is \({3 + 4i}\).

For \({z = 3 + 4i}\) the length is 5 and the direction is \({0.6 + 0.8i}\), the 3-4-5 triangle shrunk to fit the unit circle. In this form the rule of M01.9.2 becomes a sentence a child could follow: to multiply two numbers, multiply their lengths and add their angles. A power then needs no brackets at all, and two points of the unit circle, multiplied, prove the angle-addition formulas of M01.5.4 for every angle there is.

Build it up

Step 1. The polar form. By M01.5.2, the point at distance \(r\) and angle \(\theta\) is \({(r\cos\theta, r\sin\theta)}\). As a complex number that is \({r\cos\theta + ir\sin\theta}\), which is \({r(\cos\theta + i\sin\theta)}\). The length \({r = \abs{z}}\) is the modulus, and the angle \(\theta\) is called the argument, written \(\arg z\). As everywhere from M01.5.3 on, the angle is in radians unless it carries a degree sign, and \(\arg z\) is taken between \({-\pi}\) and \(\pi\), the way Python’s cmath.phase returns it. For \({3 + 4i}\): \({r = 5}\), and \(\theta\) is atan2(4, 3), about \(0.9273\), which is \(53.13^\circ\).

Step 2. Multiplying in polar form. If \(z\) has length \(r_1\) and angle \(\alpha\), and \(u\) has length \(r_2\) and angle \(\beta\), then M01.9.2 says that \(uz\) has length \(r_1 r_2\) and angle \({\alpha + \beta}\):

\[ \begin{aligned} uz &= r_1 r_2\big(\cos(\alpha + \beta) \\ &\qquad + i\sin(\alpha + \beta)\big) . \end{aligned} \]

Step 3. Powers. Multiplying by \(z\) again and again multiplies the length by \(r\) each time and adds \(\theta\) each time. So \(z^n\) has length \(r^n\) and angle \(n\theta\). For \({z = 3 + 4i}\), the square \(z^2\) has length 25 and angle \(2\theta\), about 1.8546, which is \(106.26^\circ\).

Step 4. The angle-addition formulas, for every angle. Take two points of the unit circle, \({\cos\alpha + i\sin\alpha}\) and \({\cos\beta + i\sin\beta}\). Both lengths are 1, so by Step 2 their product is \({\cos(\alpha + \beta)} + {i\sin(\alpha + \beta)}\). Now multiply the same two brackets out, as in M01.9.1. The four products are

\[ \begin{aligned} \cos\alpha \times \cos\beta &= \cos\alpha\cos\beta , \\ \cos\alpha \times i\sin\beta &= i\cos\alpha\sin\beta , \\ i\sin\alpha \times \cos\beta &= i\sin\alpha\cos\beta , \\ i\sin\alpha \times i\sin\beta &= -\sin\alpha\sin\beta , \end{aligned} \]

where the last uses \({i^2 = -1}\). The first and the last are real; the middle two are multiples of \(i\). So the product is

\[ \begin{aligned} &(\cos\alpha\cos\beta - \sin\alpha\sin\beta) \\ &\quad + i(\sin\alpha\cos\beta + \cos\alpha\sin\beta) . \end{aligned} \]

Two complex numbers are equal exactly when their real parts are equal and their imaginary parts are equal (M01.9.1). Matching the two answers part by part gives

\[ \begin{aligned} \cos(\alpha + \beta) &= \cos\alpha\cos\beta \\ &\quad - \sin\alpha\sin\beta , \\ \sin(\alpha + \beta) &= \sin\alpha\cos\beta \\ &\quad + \cos\alpha\sin\beta . \end{aligned} \]

These are the formulas of M01.5.4. That lesson’s picture needed two positive angles with a sum below \(90^\circ\). The rule of M01.9.2 holds for every pair of angles and never used these formulas, so now they are proved for every pair.

Left: a unit circle with a blue arrow to the point at angle alpha equals 120 degrees, a sky-blue arrow to the point at angle beta equals 150 degrees, and a vermillion arrow straight down to minus i. A small blue arc marks alpha, a sky-blue arc marks beta, and a long green arc, labelled turn by beta, runs from the blue arrow round to the vermillion one. Right, the working: multiply the two points; lengths: 1 times 1 equals 1; angles: 120 degrees plus 150 degrees equals 270 degrees; product: minus i, the point at 270 degrees.
M01.9.3-F2

Two points of the unit circle multiply to the point at the sum of their angles. Here the sum is \(270^\circ\), far beyond the reach of the triangle picture of M01.5.4.

The formal statement

NoteDefinition: the polar form and Euler’s formula

A complex number \({z \neq 0}\) with modulus \({r = \abs{z}}\) and argument \({\theta = \arg z}\) (in radians, between \({-\pi}\) and \(\pi\)) is

\[ z = r(\cos\theta + i\sin\theta) . \]

Euler’s formula names the point of the unit circle at angle \(\theta\):

\[ e^{i\theta} = \cos\theta + i\sin\theta , \]

so that the polar form is written \({z = re^{i\theta}}\).

At this stage \(e^{i\theta}\) is a name, as Beck and others say of it, “purely a notation”. The rule of Step 2, for two numbers of length 1, now reads

\[ e^{i\alpha}\,e^{i\beta} = e^{i(\alpha + \beta)} . \]

That is the shape of the rule \({b^m\,b^n = b^{m + n}}\) of M01.4.1, which is one reason the name deserves its \(e\). Its first consequences: \({e^{i0} = 1}\); \({e^{i\pi} = \cos\pi + i\sin\pi = -1}\); the length of \(e^{i\theta}\) is always 1; and since \({e^{i\theta}e^{-i\theta} = e^{i0} = 1}\), dividing by \(e^{i\theta}\) is multiplying by \({e^{-i\theta}}\), a turn back.

The compounding table, with an imaginary rate. Lesson M01.4.2 found \(e\) as the number that \({(1 + \frac{1}{n})^n}\) settles towards. The same table with a rate of 2, \({(1 + \frac{2}{n})^n}\), climbs to 7.3889 by \({n = 100{,}000}\), close to \({e^2 \approx 7.3891}\): a rate of \(x\) leads towards \(e^x\). Now run it with the imaginary rate \(i\pi\). If the name \(e^{i\theta}\) has earned its letter, \({(1 + \frac{i\pi}{n})^n}\) should settle towards \({e^{i\pi} = -1}\).

\(n\) \({(1 + \frac{i\pi}{n})^n}\)
1 \({1 + 3.1416i}\)
4 \({-2.3206 + 1.2037i}\)
10 \({-1.5934 + 0.1561i}\)
100 \({-1.0506 + 0.0011i}\)
1,000 \({-1.0049 + 0.0000i}\)
100,000 \({-1.0000 + 0.0000i}\)

It does. The picture shows why. Each factor \({1 + \frac{i\pi}{n}}\) is a point just above 1: its angle is a little less than \(\frac{\pi}{n}\) and its length a little more than 1. Multiplying \(n\) of them adds the \(n\) small angles, to a little less than \(\pi\), and multiplies the \(n\) lengths, to a little more than 1. The more steps, the nearer both come to a half turn of length 1, which is the point \({-1}\).

A complex plane with the dashed unit circle, from 1 on the right to minus 1 on the left. Three vermillion paths start at 1 and climb anticlockwise towards the left. The dotted path for n equals 4 takes four big steps and ends far out at about minus 2.32 plus 1.20 i; a pointer at its first step reads times (1 plus i pi over 4): one small turn. The dashed path for n equals 10 ends at about minus 1.59 plus 0.16 i. The solid path for n equals 100 hugs the upper half of the unit circle and ends just beside the black point minus 1, labelled minus 1 equals e to the i pi. Title: (1 plus i pi over n) to the n: n small turns that add up to a half turn.
M01.9.3-F3

The compounding table as a picture. With more and smaller turns the path hugs the unit circle and ends ever closer to \({-1}\).

The table is evidence, not proof. What “settles towards” means exactly is a limit (M04.1), and the proof that \(e^{i\theta}\) and \({\cos\theta + i\sin\theta}\) agree, once \(e\) to a complex power has its own definition, uses the series of M04.8. Until then \(e^{i\theta}\) is a name with the right rule of powers and the right table behind it.

Go deeper: the exact length and angle of each row of the table

The factor \({1 + \frac{i\pi}{n}}\) is the point \({(1, \frac{\pi}{n})}\). Its length is \({\sqrt{1 + \pi^2/n^2}}\) and its angle is \({\operatorname{atan}(\pi/n)}\), since the point lies in the right half of the plane. By M01.9.2, the \(n\)-th power has

\[ \begin{aligned} \text{length} &= \big(1 + \pi^2/n^2\big)^{n/2} , \\ \text{angle} &= n\,\operatorname{atan}(\pi/n) . \end{aligned} \]

At \({n = 100}\) the length is about 1.0506 and the angle about 3.1406, which is \(179.94^\circ\): the point \({-1.0506 + 0.0011i}\) of the table. At \({n = 100{,}000}\) the length is about 1.00005 and the angle is within a ten-millionth of a degree of \(180^\circ\).

Worked example

Write \({z = 3 + 4i}\) in polar form, and use it to square \(z\).

The polar form. The length is \({r = \sqrt{9 + 16} = 5}\). The angle \(\theta\) is atan2(4, 3), about \(0.9273\) radians, which is \(53.13^\circ\); its cosine and sine are \({3/5 = 0.6}\) and \({4/5 = 0.8}\). So

\[ \begin{aligned} 3 + 4i &= 5e^{i\theta} \\ &\approx 5e^{0.9273i} . \end{aligned} \]

Back again: \({5\cos\theta = 5 \times 0.6 = 3}\) and \({5\sin\theta = 5 \times 0.8 = 4}\).

The square, by the polar form. Square the length and double the angle: \({5^2 = 25}\) and \({2\theta \approx 1.8546}\), which is \(106.26^\circ\). The cosine and sine of that angle are \({-0.28}\) and \(0.96\), so

\[ \begin{aligned} z^2 &= 25(\cos 2\theta + i\sin 2\theta) \\ &= 25(-0.28 + 0.96i) \\ &= -7 + 24i . \end{aligned} \]

Two routes drawn as boxes and arrows. Route 1, multiply out the brackets: a blue box (3 plus 4 i)(3 plus 4 i), a green arrow to a box 9 plus 24 i plus 16 i squared, and an arrow marked with i squared equals minus 1 to a yellow box minus 7 plus 24 i. Route 2, the polar form: an orange box 5 at 53.13 degrees, a green arrow marked 5 squared, 2 times 53.13 degrees, to an orange box 25 at 106.26 degrees, and an arrow to the same yellow box. Under it: 25 cos 106.26 degrees is about minus 7 and 25 sin 106.26 degrees is about 24; both routes land on minus 7 plus 24 i.
M01.9.3-F4

Two routes to \(z^2\). Multiplying out uses \({i^2 = -1}\); the polar form squares the length and doubles the angle. Both land on \({-7 + 24i}\).

The square, by brackets. \({(3 + 4i)^2 = 9 + 12i + 12i + 16i^2}\), which is \({9 + 24i - 16}\), or \({-7 + 24i}\). The two routes agree, and the polar route did no multiplying out at all.

The tenth power, from the hook. The length is \({5^{10} = 9{,}765{,}625}\), and the angle is \({10 \times 53.13^\circ = 531.30^\circ}\), which is one whole turn and \(171.30^\circ\) more. So \(z^{10}\) points just above the negative real axis, and the code below finds it to be \({-9{,}653{,}287 + 1{,}476{,}984i}\).

Code

First the two conversions from scratch, with math.hypot for the length and math.atan2 for the angle (in radians, \(y\) first), and math.cos and math.sin for the way back.

import math

def to_polar(a, b):
    return math.hypot(a, b), math.atan2(b, a)          # (length r, angle theta in radians)

def from_polar(r, theta):
    return r * math.cos(theta), r * math.sin(theta)    # (real part, imaginary part)

r, theta = to_polar(3, 4)
print(r, round(theta, 4), round(math.degrees(theta), 2))
x, y = from_polar(r, theta)
print(round(x, 10), round(y, 10))
print("z squared:", [round(v, 10) for v in from_polar(r ** 2, 2 * theta)])
5.0 0.9273 53.13
3.0 4.0
z squared: [-7.0, 24.0]

Python’s complex numbers, where \(i\) is written 1j, give powers directly with **. The polar rule agrees with them, up to rounding:

z = 3 + 4j
print(z ** 2, z ** 10)
print(from_polar(r ** 10, 10 * theta))                 # length 5 to the 10, angle 10 theta
(-7+24j) (-9653287+1476984j)
(-9653286.999999998, 1476984.000000011)

The compounding table by a loop. Each pass multiplies the running total by one small turn, \({1 + \frac{i\pi}{n}}\), and the f"..." string prints four decimal places with the sign shown:

for n in [1, 4, 10, 100, 1000, 100000]:
    step = complex(1, math.pi / n)        # one small turn, 1 + i pi / n
    total = 1
    for _ in range(n):
        total = total * step              # n small turns, one after another
    print(f"{n:>6}   {total.real:.4f} {total.imag:+.4f}i")
     1   1.0000 +3.1416i
     4   -2.3206 +1.2037i
    10   -1.5934 +0.1561i
   100   -1.0506 +0.0011i
  1000   -1.0049 +0.0000i
100000   -1.0000 +0.0000i

The cmath module, the complex twin of math, does each job in one call: polar gives the length and the argument, phase the argument alone, rect goes back, and exp raises \(e\) to a complex power.

import cmath

print(cmath.polar(z))
print(cmath.phase(z))
print(cmath.rect(5, cmath.phase(z)))
print(cmath.exp(1j * math.pi))
print(cmath.exp(1j * math.pi) == -1, cmath.isclose(cmath.exp(1j * math.pi), -1))
(5.0, 0.9272952180016122)
0.9272952180016122
(3+3.9999999999999996j)
(-1+1.2246467991473532e-16j)
False True

The third line comes back to \({3 + 4i}\) up to a rounding error in the last digit. The last line deserves a closer look. cmath.exp(1j * math.pi) misses \({-1}\) by a tiny imaginary part, about \({1.2 \times 10^{-16}}\), because math.pi stops after sixteen or so digits and the sine of it is not exactly 0. So == says False, while cmath.isclose says True. Compare complex results with a tolerance, just like floats.

Finally the angle-addition formulas checked on a grid of 625 pairs of angles, every \(15^\circ\) from \({-180^\circ}\) to \(180^\circ\). This is evidence for what Step 4 proved:

agree = 0
for a in range(-180, 181, 15):
    for b in range(-180, 181, 15):
        alpha, beta = math.radians(a), math.radians(b)
        product = cmath.exp(1j * alpha) * cmath.exp(1j * beta)
        if cmath.isclose(product, cmath.exp(1j * (alpha + beta))):
            agree += 1
print(agree, "pairs agree")
625 pairs agree

The same idea runs through signal processing. The wave \({A\sin(t + \phi)}\) of M01.5.3 is the imaginary part of \(ce^{it}\) with \({c = Ae^{i\phi}}\), because \({ce^{it} = Ae^{i(t + \phi)}}\). So a single complex number \(c\) holds both the amplitude (its modulus) and the phase (its argument). A Fourier transform (M10.9) describes a signal by one such number for each frequency, and characteristic functions (M07.2) and spectral analysis (S07.6) rest on the same idea.

c = cmath.rect(2, 0.5)                    # amplitude 2, phase 0.5
for t in [0.0, 1.0, 2.5]:
    print(round((c * cmath.exp(1j * t)).imag, 4), round(2 * math.sin(t + 0.5), 4))
0.9589 0.9589
1.995 1.995
0.2822 0.2822

Explore

The turn-and-scale explorer of lesson M01.9.2 is a polar-form machine. Set the length of \(u\) to 1 and its angle to \(45^\circ\): \(u\) is then \(e^{i\pi/4}\). Press Multiply again eight times. The powers of \(u\) step round the circle of radius 5, an eighth of a turn at a time, and come home, because eight turns of \(\frac{\pi}{4}\) make \(2\pi\), a whole turn. Now set the length to 0.9: each press multiplies the length by 0.9, so after eight presses it is \({0.9^8 \approx 0.4305}\) of what it was, and the walk spirals in.

Pitfalls

Two complex planes, each with a dashed half circle of radius 5. Left, marked with a cross and titled rect(5, 53.13): a reddish-purple arc runs from the positive real axis round to 164.12 degrees, with the note 8 whole turns, then 164.12 degrees, and a reddish-purple arrow ends at the point minus 4.81 plus 1.37 i, far from 3 plus 4 i. Right, marked with a tick and titled rect(5, radians(53.13)): an orange arc marks 53.13 degrees and a vermillion arrow ends at 3 plus 4 i.
M01.9.3-F5

Wrong beside right. cmath.rect reads its angle in radians, so 53.13 means 53.13 radians: eight whole turns and \(164.12^\circ\) more.

“cmath.rect(5, 53.13) gives \({3 + 4i}\).”

Every cmath function measures angles in radians, as math does. Read as radians, 53.13 is eight whole turns and \(164.12^\circ\) more, and the result is about \({-4.81 + 1.37i}\). Convert first: cmath.rect(5, math.radians(53.13)) is about \({3 + 4i}\), and cmath.rect(5, cmath.phase(3 + 4j)) is \({3 + 4i}\) up to rounding.

“The argument of a product is exactly the sum of the arguments.”

Only up to a whole turn, because an argument is kept between \({-\pi}\) and \(\pi\). The argument of \({-1 + i}\) is \(\frac{3\pi}{4}\) (2.3562) and that of \(i\) is \(\frac{\pi}{2}\) (1.5708); they add to 3.9270, which is \(225^\circ\). But cmath.phase of the product \({-1 - i}\) is \({-2.3562}\), which is \({-135^\circ}\): the same direction, \(2\pi\) lower. To compare two directions, compare the points cmath.rect(1, a) and cmath.rect(1, b), not the raw angles.

“\(e^{i\theta}\) grows as \(\theta\) grows, like \(e^x\).”

For a real \(\theta\) the length of \(e^{i\theta}\) is always 1: it goes round the unit circle and never grows. Growth needs a real part. NumPy’s documentation of exp puts it this way: for \({x = a + ib}\), \({e^x = e^a\,e^{ib}}\), where \(e^a\) stretches and \(e^{ib}\) only turns.

Check yourself

Recall

1. What is the polar form of a complex number with modulus \(r\) and argument \(\theta\)?

Show answer

\({z = r(\cos\theta + i\sin\theta)}\), which Euler’s formula writes \({z = re^{i\theta}}\).

2. What are the modulus and the argument of \({-2}\)?

Show answer

The modulus is 2 and the argument is \(\pi\): the point lies on the negative real axis, a half turn from the positive one.

3. What is \(e^{i\pi/2}\)?

Show answer

\({\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}} = {0 + 1i} = i\): the point a quarter turn round the unit circle.

4. Multiply \(2e^{0.5i}\) by \(3e^{1.2i}\).

Show answer

Multiply the lengths and add the angles: \(6e^{1.7i}\).

5. Why does \({(1 + \frac{i\pi}{n})^n}\) close on \({-1}\) as \(n\) grows?

Show answer

It is \(n\) small turns of a little less than \(\frac{\pi}{n}\) each, which add up to nearly \(\pi\), a half turn; the \(n\) lengths, each a hair over 1, multiply to nearly 1. The point at angle \(\pi\) and length 1 is \({-1}\).

Exercises

1. ★ Write \({-1 + i}\) in polar form, and use it to find \({(-1 + i)^4}\). Check by algebra.

Show solution

The length is \({\sqrt{1 + 1} = \sqrt{2}}\), and the point is in the second quarter, at \(\frac{3\pi}{4}\), which is \(135^\circ\). So \({-1 + i = \sqrt{2}\,e^{3i\pi/4}}\). The fourth power has length \({(\sqrt{2})^4 = 4}\) and angle \(3\pi\), which is one whole turn and a half turn more. So \({(-1 + i)^4 = -4}\). By algebra: \({(-1 + i)^2 = 1 - 2i + i^2 = -2i}\), and \({(-2i)^2 = 4i^2 = -4}\).

2. ★★ Use the formulas proved in Step 4 with \({\alpha = 120^\circ}\) and \({\beta = 150^\circ}\) to find \(\cos 270^\circ\) and \(\sin 270^\circ\). Why could lesson M01.5.4’s picture not do this?

Show solution

The values needed are \({\cos 120^\circ = -\frac{1}{2}}\), \({\sin 120^\circ = \frac{\sqrt{3}}{2}}\), \({\cos 150^\circ = -\frac{\sqrt{3}}{2}}\) and \({\sin 150^\circ = \frac{1}{2}}\). Then

\[ \begin{aligned} \cos 270^\circ &= \cos 120^\circ\cos 150^\circ \\ &\quad - \sin 120^\circ\sin 150^\circ \\ &= \tfrac{\sqrt{3}}{4} - \tfrac{\sqrt{3}}{4} = 0 , \\ \sin 270^\circ &= \sin 120^\circ\cos 150^\circ \\ &\quad + \cos 120^\circ\sin 150^\circ \\ &= -\tfrac{3}{4} - \tfrac{1}{4} = -1 . \end{aligned} \]

The point at \(270^\circ\) is \({(0, -1)}\), which is \({-i}\), as the figure in Step 4 shows. M01.5.4’s picture stacks two right triangles, so it needs both angles positive and their sum below \(90^\circ\).

3. ★★★ Show with the angle-addition formula that \({A\sin(t + \phi) = p\sin t + q\cos t}\), where \({p + qi = Ae^{i\phi}}\). Then write the wave \({3\sin t + 4\cos t}\) as a single sine wave.

Show solution

By the formula for \({\sin(\alpha + \beta)}\) with \({\alpha = t}\) and \({\beta = \phi}\),

\[ \begin{aligned} A\sin(t + \phi) &= A\sin t\cos\phi \\ &\quad + A\cos t\sin\phi \\ &= (A\cos\phi)\sin t \\ &\quad + (A\sin\phi)\cos t . \end{aligned} \]

And \({Ae^{i\phi} = A\cos\phi + iA\sin\phi}\), so \({p = A\cos\phi}\) and \({q = A\sin\phi}\), as claimed. For \({3\sin t + 4\cos t}\) we need \({p + qi = 3 + 4i}\), which is \(5e^{i\theta}\) with \(\theta\) the angle of \({3 + 4i}\), about 0.9273. So the wave is \({5\sin(t + \theta)}\), about \({5\sin(t + 0.9273)}\): amplitude 5 and phase \(\theta\), the modulus and the argument of \({3 + 4i}\). Check at \({t = 0}\): \({5\sin\theta = 5 \times 0.8 = 4}\), and \({3\sin 0 + 4\cos 0 = 4}\).

Summary card

A complex number is a length times a direction: \({z = re^{i\theta}}\).

The anchor figure again: the direction cos theta plus i sin theta, equal to 0.6 plus 0.8 i, on the unit circle, and five times that direction, z equals 3 plus 4 i, at the angle theta equals 0.9273.
M01.9.3-F1

Length times direction.

  • Polar form: \({z = r(\cos\theta + i\sin\theta)} = {re^{i\theta}}\), with \({r = \abs{z}}\) and \({\theta = \arg z}\).
  • Multiplying: lengths multiply, arguments add. Powers: \({z^n = r^n e^{in\theta}}\).
  • Euler’s formula: \({e^{i\theta} = \cos\theta + i\sin\theta}\), so \({e^{i\alpha}e^{i\beta}} = {e^{i(\alpha + \beta)}}\) and \({e^{i\pi} = -1}\).
  • The angle-addition formulas are the two parts of \(e^{i\alpha}e^{i\beta}\), multiplied out.

Sources and next steps

Sources. Each was opened and checked against this lesson on 12 November 2026.

  • Jay Abramson and others, Algebra and Trigonometry 2e, Section 10.5 Polar Form of Complex Numbers, OpenStax. The polar form \({r(\cos\theta + i\sin\theta)}\) with \(r\) the modulus and \(\theta\) the argument; the product “calls for multiplying the moduli and adding the angles”; and De Moivre’s theorem, \({z^n = r^n(\cos n\theta + i\sin n\theta)}\) for a positive whole number \(n\). The section stops before Euler’s formula.
  • Matthias Beck, Gerald Marchesi, Dennis Pixton and Lucas Sabalka, A First Course in Complex Analysis, edition 1.6, Section 1.2. Defines \(e^{i\phi}\) as \({\cos\phi + i\sin\phi}\) and calls it, at that point, “purely a notation”; Proposition 1.2.7 lists \({e^{i\phi_1}e^{i\phi_2}} = {e^{i(\phi_1 + \phi_2)}}\), \({e^{i0} = 1}\), \({1/e^{i\phi} = e^{-i\phi}}\) and that the length of \(e^{i\phi}\) is 1; the polar form \({x + iy = re^{i\phi}}\); and arguments that differ by whole multiples of \(2\pi\).
  • Python Software Foundation, cmath, Python 3.12 documentation. phase(z) is math.atan2(z.imag, z.real), between \({-\pi}\) and \(\pi\); polar(z) is (abs(z), phase(z)); rect(r, phi) is complex(r * math.cos(phi), r * math.sin(phi)); exp raises \(e\) to a complex power; isclose uses a relative tolerance of \({10^{-9}}\) by default.
  • NumPy reference, numpy.exp, version 2.5, Notes: for \({x = a + ib}\), \({e^x = e^a\,e^{ib}}\), where \(e^{ib}\) is \({\cos b + i\sin b}\), “a function with magnitude 1 and a periodic phase”.

Next. M01.9.4 Quadratics with a negative discriminant returns to the quadratic formula where M01.3.3 had to stop.

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M01.9.4 Quadratics with a negative discriminant M10.9 Fast transforms: FFT and convolution (planned)

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