Complex exponential and trigonometric functions · Selected prerequisite proofs

L11.4.2: Proposition and retained argument completed by P16.1–P16.3, including its actual exercises, first-zero supremum, return-time existence issue and omitted axes. The following arc-length assertion is excluded.

Proposition 11.4.2.

The sine and cosine functions have the following properties:

For all $z \in \C\text{,}$

$\begin{equation*} e^{iz} = \cos(z) + i\sin(z) \qquad \text{(Euler's formula)}. \end{equation*}$
$\cos(0) = 1\text{,}$ $\sin(0) = 0\text{.}$
For all $z \in \C\text{,}$

$\begin{equation*} \cos(-z) = \cos(z), \qquad \sin(-z) = -\sin(z). \end{equation*}$
For all $z \in \C\text{,}$

$\begin{equation*} \cos(z) = \sum_{k=0}^\infty \frac{{(-1)}^k}{(2k)!} z^{2k} , \qquad \sin(z) = \sum_{k=0}^\infty \frac{{(-1)}^k}{(2k+1)!} z^{2k+1} . \end{equation*}$
For all $x \in \R$

$\begin{equation*} \cos(x) = \Re (e^{ix}) \qquad\text{and}\qquad \sin(x) = \Im (e^{ix}) . \end{equation*}$
For all $z \in \C\text{,}$

$\begin{equation*} {\bigl( \cos(z) \bigr)}^2 + {\bigl( \sin(z) \bigr)}^2 = 1 . \end{equation*}$
For all $x \in \R\text{,}$

$\begin{equation*} \babs{\sin(x)} \leq 1, \qquad \babs{\cos(x)} \leq 1 . \end{equation*}$
For all $x \in \R\text{,}$

$\begin{equation*} \frac{d}{dx} \bigl[ \cos(x) \bigr] = -\sin(x) \qquad \text{and} \qquad \frac{d}{dx} \bigl[ \sin(x) \bigr] = \cos(x) . \end{equation*}$
For all $x \geq 0\text{,}$

$\begin{equation*} \sin(x) \leq x . \end{equation*}$
There exists an $x > 0$ such that $\cos(x) = 0\text{.}$ We define

$\begin{equation*} \pi \coloneqq 2 \, \inf \{ x > 0 : \cos(x) = 0 \} . \end{equation*}$
For all $z \in \C\text{,}$

$\begin{equation*} e^{2\pi i} = 1 \qquad \text{and} \qquad e^{z + i 2\pi} = e^z. \end{equation*}$
Sine and cosine are $2\pi$ -periodic and not periodic with any smaller period. That is, $2\pi$ is the smallest number such that for all $z \in \C\text{,}$

$\begin{equation*} \sin(z+2\pi) = \sin(z) \qquad \text{and} \qquad \cos(z+2\pi) = \cos(z) . \end{equation*}$
The function $x \mapsto e^{ix}$ is a bijective map from $[0,2\pi)$ onto the set of $z \in \C$ such that $\sabs{z} = 1\text{.}$

Proof.

The first three items follow directly from the definition. The computation of the power series for both is left as an exercise. As the complex conjugate is a continuous function, the definition of

e^z

implies

\overline{e^z} = e^{\bar{z}}\text{.}

x

is real,

\begin{equation*} \overline{e^{ix}} = e^{-ix} . \end{equation*}

Thus for real

x\text{,}

\cos(x) = \frac{e^{ix}+e^{-ix}}{2} = \frac{e^{ix}+\overline{e^{ix}}}{2} = \Re (e^{ix})

and similarly

\sin(x) = \Im (e^{ix})\text{.}

For real

x\text{,}

we compute

\begin{equation*} 1 = e^{ix} e^{-ix} = e^{ix} \, \overline{e^{ix}} = \sabs{e^{ix}}^2 = \babs{\cos(x) + i \sin(x)}^2 = {\bigl( \cos(x) \bigr)}^2 + {\bigl( \sin(x) \bigr)}^2 . \end{equation*}

A slightly more complicated computation shows this fact for complex numbers, see Exercise 11.4.6. In particular,

e^{ix}

is unimodular for real

x\text{;}

the values lie on the unit circle. A square of a real number is always nonnegative:

\begin{equation*} {\bigl(\sin(x)\bigr)}^2 = 1-{\bigl(\cos(x)\bigr)}^2 \leq 1 . \end{equation*}

\babs{\sin(x)} \leq 1

and similarly

\babs{\cos(x)} \leq 1\text{.}

We leave the computation of the derivatives to the reader as exercises. Let us prove that

\sin(x) \leq x

for

x \geq 0\text{.}

Consider

f(x) \coloneqq x-\sin(x)

and differentiate:

\begin{equation*} f'(x) = \frac{d}{dx} \bigl[ x - \sin(x) \bigr] = 1 -\cos(x) \geq 0 , \end{equation*}

for all

x \in \R

\babs{\cos(x)} \leq 1\text{.}

In other words,

f

is increasing and

f(0) = 0\text{.}

f

must be nonnegative when

x \geq 0

and hence,

\sin(x) \leq x\text{.}

Next, we claim there exists a positive

x

such that

\cos(x) = 0\text{.}

\cos(0) = 1 > 0\text{,}

\cos(x) > 0

for

x

near

0\text{.}

Namely, there is some

y > 0

such that

\cos(x) > 0

[0,y)\text{.}

Then

\sin(x)

is strictly increasing on

[0,y)\text{.}

\sin(0) = 0\text{,}

we have

\sin(x) > 0

for

x \in (0,y)\text{.}

Take

a \in (0,y)\text{.}

By the mean value theorem, there is a

c \in (a,y)

such that

\begin{equation*} 2 \geq \cos(a)-\cos(y) = \sin(c)(y-a) \geq \sin(a)(y-a) . \end{equation*}

a \in (0,y)\text{,}

we have

\sin(a) > 0

and so

\begin{equation*} y \leq \frac{2}{\sin(a)} + a . \end{equation*}

Hence there is some largest

y

such that

\cos(x) > 0

[0,y)\text{,}

and let

y

be the largest such number. By continuity,

\cos(y) = 0\text{.}

In fact,

y

is the smallest positive

y

such that

\cos(y) = 0\text{.}

As mentioned,

\pi

is defined to be

2y\text{.}

\cos(\nicefrac{\pi}{2}) = 0\text{,}

we find

{\bigl(\sin(\nicefrac{\pi}{2})\bigr)}^2 = 1\text{.}

\sin

is positive on

(0,\nicefrac{\pi}{2})\text{,}

we have

\sin(\nicefrac{\pi}{2}) = 1\text{.}

Hence,

\begin{equation*} e^{i \pi /2} = i , \end{equation*}

and by the law of exponents,

\begin{equation*} e^{i \pi} = -1 , \qquad e^{i 2\pi} = 1 . \end{equation*}

e^{i2\pi} = 1 = e^0\text{.}

The law of exponents also says

\begin{equation*} e^{z+i2\pi} = e^z e^{i2\pi} = e^z \end{equation*}

for all

z \in \C\text{.}

Immediately, we also obtain

\cos(z+2\pi) = \cos(z)

and

\sin(z+2\pi) = \sin(z)\text{.}

\sin

and

\cos

are

2\pi

-periodic.

We claim that

\sin

and

\cos

are not periodic with a smaller period. It suffices to show that if

e^{ix} = 1

for the smallest positive

x\text{,}

then

x = 2\pi\text{.}

Let

x

be the smallest positive

x

such that

e^{ix} = 1\text{.}

Of course,

x \leq 2\pi\text{.}

By the law of exponents,

\begin{equation*} {\bigl(e^{ix/4}\bigr)}^4 = 1 . \end{equation*}

e^{ix/4} = a+ib\text{,}

then

\begin{equation*} {(a+ib)}^4 =a^4-6a^2b^2+b^4 + i\bigl(4ab(a^2-b^2)\bigr) =1 . \end{equation*}

Then either

a = 0

a^2 = b^2\text{.}

\nicefrac{x}{4} \leq \nicefrac{\pi}{2}\text{,}

we have

a = \cos(\nicefrac{x}{4}) \geq 0

and

b = \sin(\nicefrac{x}{4}) > 0\text{.}

a^2=b^2\text{,}

then

a^4-6a^2b^2+b^4 = -4a^4 < 0

and in particular not equal to 1. Therefore

a=0\text{,}

in which case

\nicefrac{x}{4} = \nicefrac{\pi}{2}\text{.}

Hence

2\pi

is the smallest period we could choose for

e^{ix}

and so also for

\cos

and

\sin\text{.}

Finally, we wish to show that

e^{ix}

is one-to-one and onto from the set

[0,2\pi)

to the set of

z \in \C

such that

\sabs{z} = 1\text{.}

Suppose

e^{ix} = e^{iy}

and

x > y\text{.}

Then

e^{i(x-y)} = 1\text{,}

meaning

x-y

is a multiple of

2\pi

and hence only one of them can live in

[0,2\pi)\text{.}

To show

e^{ix}

is onto, pick

(a,b) \in \R^2

such that

a^2+b^2 = 1\text{.}

Suppose first that

a,b \geq 0\text{.}

By the intermediate value theorem, there must exist an

x \in [0,\nicefrac{\pi}{2}]

such that

\cos(x) = a\text{,}

and hence

b^2 = \bigl(\sin(x)\bigr)^2\text{.}

b

and

\sin(x)

are nonnegative,

b = \sin(x)\text{.}

Since

-\sin(x)

is the derivative of

\cos(x)

and

\cos(-x) = \cos(x)\text{,}

we have that

\sin(x) < 0

for

x \in [\nicefrac{-\pi}{2},0)\text{.}

Using the same reasoning, we obtain that if

a > 0

and

b \leq 0\text{,}

we can find an

x

[\nicefrac{-\pi}{2},0)\text{,}

and by periodicity,

x \in [\nicefrac{3\pi}{2},2\pi)

such that

\cos(x) = a

and

\sin(x)=b\text{.}

Multiplying by

-1

is the same as multiplying by

e^{i\pi}

e^{-i\pi}\text{.}

So we can always assume that

a \geq 0

(details are left as an exercise).