## Section11.4The complex exponential and the trigonometric functions

Note: 1 lecture

### Subsection11.4.1The complex exponential

Define

\begin{equation*} E(z) := \sum_{k=0}^\infty \frac{1}{k!} z^k . \end{equation*}

This series converges for all $$z \in \C$$ and so by Corollary 11.3.7, $$E$$ is analytic on $$\C\text{.}$$ We notice that $$E(0) = 1\text{,}$$ and that for $$z=x \in \R\text{,}$$ $$E(x) \in \R\text{.}$$ Keeping $$x$$ real, we find

\begin{equation*} \frac{d}{dx} \bigl( E(x) \bigr) = E(x) \end{equation*}

by direct calculation. In Section 5.4 (or by Picard's theorem), we proved that the unique function satisfying $$E' = E$$ and $$E(0) = 1$$ is the exponential. In other words, for $$x \in \R\text{,}$$ $$e^x = E(x)\text{.}$$

For complex numbers $$z\text{,}$$ we define

\begin{equation*} e^z := E(z) = \sum_{k=0}^\infty \frac{1}{k!} z^k . \end{equation*}

On the real line this new definition agrees with our previous one. See Figure 11.7. Notice that in the $$x$$ direction (the real direction) the graph behaves like the real exponential, and in the $$y$$ direction (the imaginary direction) the graph oscillates.

#### Proof.

We already know that the equality $$e^{x+y} = e^x e^y$$ holds for all real numbers $$x$$ and $$y\text{.}$$ For every fixed $$y \in \R\text{,}$$ consider the expressions as functions of $$x$$ and apply the identity theorem (Theorem 11.3.9) to get that $$e^{z+y} = e^ze^y$$ for all $$z \in \C\text{.}$$ Fixing an arbitrary $$z \in \C\text{,}$$ we get $$e^{z+y} = e^ze^y$$ for all $$y \in \R\text{.}$$ Again by the identity theorem $$e^{z+w} = e^z e^w$$ for all $$w \in \C\text{.}$$

A simple consequence is that $$e^z\not=0$$ for all $$z \in \C\text{,}$$ as $$e^z e^{-z} = e^{z-z} = 1\text{.}$$ A more complicated consequence is that we can easily compute the power series for the exponential at a point $$a \in \C\text{:}$$ $$e^z = e^a e^{z-a} = \sum \frac{e^a}{k!} {(z-a)}^k\text{.}$$

### Subsection11.4.2Trigonometric functions and $$\pi$$

We can now finally define sine and cosine by the equation

\begin{equation*} e^{x+iy} = e^x \bigl( \cos(y) + i \sin(y) \bigr) . \end{equation*}

In fact, we define sine and cosine for all complex $$z\text{:}$$

\begin{equation*} \cos(z) := \frac{e^{iz} + e^{-iz}}{2} \qquad\text{and}\qquad \sin(z) := \frac{e^{iz} - e^{-iz}}{2i} . \end{equation*}

Let us use our definition to prove the common properties we usually associate with sine and cosine. In the process we also define the number $$\pi\text{.}$$

The proposition immediately implies that $$\sin(x)$$ and $$\cos(x)$$ are real whenever $$x$$ is real.

#### Proof.

The first three items follow directly from the definition. The computation of the power series for both is left as an exercise.

As complex conjugate is a continuous function, the definition of $$e^z$$ implies $$\overline{(e^z)} = e^{\bar{z}}\text{.}$$ If $$x$$ is real,

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

Thus for real $$x\text{,}$$ $$\cos(x) = \Re (e^{ix})$$ and $$\sin(x) = \Im (e^{ix})\text{.}$$

For real $$x$$ we compute

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

In particular, is $$e^{ix}$$ is unimodular, the values lie on the unit circle. A square is always nonnegative:

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

So $$\sabs{\sin(x)} \leq 1$$ and similarly $$\sabs{\cos(x)} \leq 1\text{.}$$

We leave the computation of the derivatives to the reader as exercises.

Let us now prove that $$\sin(x) \leq x$$ for $$x \geq 0\text{.}$$ Consider $$f(x) := x-\sin(x)$$ and differentiate:

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

for all $$x$$ as $$\sabs{\cos(x)} \leq 1\text{.}$$ In other words, $$f$$ is increasing and $$f(0) = 0\text{.}$$ So $$f$$ must be nonnegative when $$x \geq 0\text{.}$$

We claim there exists a positive $$x$$ such that $$\cos(x) = 0\text{.}$$ As $$\cos(0) = 1 > 0\text{,}$$ $$\cos(x) > 0$$ for $$x$$ near $$0\text{.}$$ Namely, there is some $$y > 0\text{,}$$ such that $$\cos(x) > 0$$ on $$[0,y)\text{.}$$ Then $$\sin(x)$$ is strictly increasing on $$[0,y)\text{.}$$ As $$\sin(0) = 0\text{,}$$ then $$\sin(x) > 0$$ 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*}

As $$a \in (0,y)\text{,}$$ then $$\sin(a) > 0$$ and so

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

Hence there is some largest $$y$$ such that $$\cos(x) > 0$$ in $$[0,y)\text{,}$$ 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 $$2y\text{.}$$

As $$\cos(\nicefrac{\pi}{2}) = 0\text{,}$$ then $${\bigl(\sin(\nicefrac{\pi}{2})\bigr)}^2 = 1\text{.}$$ As $$\sin$$ is positive on $$(0,y)\text{,}$$ we have $$\sin(\nicefrac{\pi}{2}) = 1\text{.}$$ Hence,

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

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

So $$e^{i2\pi} = 1 = e^0\text{.}$$ The addition formula says

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

for all $$z \in \C\text{.}$$ Immediately we also obtain $$\cos(z+2\pi) = \cos(z)$$ and $$\sin(z+2\pi) = \sin(z)\text{.}$$ So $$\sin$$ and $$\cos$$ are $$2\pi$$-periodic.

We claim that $$\sin$$ and $$\cos$$ are not periodic with a smaller period. It would suffice to show that if $$e^{ix} = 1$$ for the smallest positive $$x\text{,}$$ then $$x = 2\pi\text{.}$$ So let $$x$$ be the smallest positive $$x$$ such that $$e^{ix} = 1\text{.}$$ Of course, $$x \leq 2\pi\text{.}$$ By the addition formula,

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

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

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

As $$\nicefrac{x}{4} \leq \nicefrac{\pi}{2}\text{,}$$ then $$a = \cos(\nicefrac{x}{4}) \geq 0$$ and $$0 < b = \sin(\nicefrac{x}{4})\text{.}$$ Then either $$a = 0$$ or $$a^2 = b^2\text{.}$$ If $$a^2=b^2\text{,}$$ then $$a^4-6a^2b^2+b^4 = -4a^4 < 0$$ and in particular not equal to 1. Therefore $$a=0$$ 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 also 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 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 $$b^2 = \bigl(\sin(x)\bigr)^2\text{.}$$ As $$b$$ and $$\sin(x)$$ are nonnegative, we have $$b = \sin(x)\text{.}$$ Since $$-\sin(x)$$ is the derivative of $$\cos(x)$$ and $$\cos(-x) = \cos(x)\text{,}$$ then $$\sin(x) < 0$$ 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 $$x$$ in $$[\nicefrac{-\pi}{2},0)\text{,}$$ 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 $$e^{i\pi}$$ or $$e^{-i\pi}\text{.}$$ So we can always assume that $$a \geq 0$$ (details are left as exercise).

### Subsection11.4.3The unit circle and polar coordinates

The arclength of a curve parametrized by $$\gamma \colon [a,b] \to \C$$ is given by

\begin{equation*} \int_a^b \sabs{\gamma^{\:\prime}(t)} \, dt . \end{equation*}

We have that $$e^{it}$$ parametrizes the circle for $$t$$ in $$[0,2\pi)\text{.}$$ As $$\frac{d}{dt} \bigl( e^{it} \bigr) = ie^{it}\text{,}$$ the circumference of the circle (the arclength) is

\begin{equation*} \int_0^{2\pi} \sabs{i e^{it}} \, dt = \int_0^{2\pi} 1 \, dt = 2\pi . \end{equation*}

More generally we notice that $$e^{it}$$ parametrizes the circle by arclength. That is, $$t$$ measures the arclength, and hence a circle of radius 1 by the angle in radians. So the definitions of $$\sin$$ and $$\cos$$ we have used above agree with the standard geometric definitions.

All the points on the unit circle can be achieved by $$e^{it}$$ for some $$t\text{.}$$ Therefore, we can write a complex number $$z \in \C$$ (in so-called polar coordinates) as

\begin{equation*} z = r e^{i\theta} \end{equation*}

for some $$r \geq 0$$ and $$\theta \in \R\text{.}$$ The $$\theta$$ is, of course, not unique as $$\theta$$ or $$\theta+2\pi$$ gives the same number. The formula $$e^{a+b} = e^a e^b$$ leads to a useful formula for powers and products of complex numbers in polar coordinates:

\begin{equation*} {(r e^{i\theta})}^n = r^n e^{i n \theta} , \qquad (r e^{i\theta}) (s e^{i\gamma}) = rs e^{i(\theta+\gamma)} . \end{equation*}

### Subsection11.4.4Exercises

#### Exercise11.4.1.

Derive the power series for $$\sin(z)$$ and $$\cos(z)$$ at the origin.

#### Exercise11.4.2.

Using the power series, show that for $$x$$ real, we have $$\frac{d}{dx} \bigl[ \sin(x)\bigr] = \cos(x)$$ and $$\frac{d}{dx} \bigl[ \cos(x)\bigr] = -\sin(x)\text{.}$$

#### Exercise11.4.3.

Finish the proof of the argument that $$x \mapsto e^{ix}$$ from $$[0,2\pi)$$ is onto the unit circle. In particular, assume that we get all points of the form $$(a,b)$$ where $$a^2+b^2=1$$ for $$a \geq 0\text{.}$$ By multiplying by $$e^{i\pi}$$ or $$e^{-i\pi}$$ show that we get everything.

#### Exercise11.4.4.

Prove that there is no $$z \in \C$$ such that $$e^z = 0\text{.}$$

#### Exercise11.4.5.

Prove that for every $$w \not= 0$$ and every $$\epsilon > 0\text{,}$$ there exists a $$z \in \C\text{,}$$ $$\sabs{z} < \epsilon$$ such that $$e^{1/z} = w\text{.}$$

#### Exercise11.4.6.

We showed $${\bigl( \cos(x) \bigr)}^2 + {\bigl( \sin(x) \bigr)}^2 = 1$$ for all $$x \in \R\text{.}$$ Prove that $${\bigl( \cos(z) \bigr)}^2 + {\bigl( \sin(z) \bigr)}^2 = 1$$ for all $$z \in \C\text{.}$$

#### Exercise11.4.7.

Prove the trigonometric identities $$\sin(z + w) = \sin(z) \cos(w) + \cos(z) \sin(w)$$ and $$\cos(z + w) = \cos(z) \cos(w) - \sin(z) \sin(w)$$ for all $$z,w \in \C\text{.}$$

#### Exercise11.4.8.

Define $$\operatorname{sinc}(z) := \frac{\sin(z)}{z}$$ for $$z \not=0$$ and $$\operatorname{sinc}(0) := 1\text{.}$$ Show that sinc is analytic and compute its power series at zero.

Define the hyperbolic sine and hyperbolic cosine by

\begin{equation*} \sinh(z) := \frac{e^z-e^{-z}}{2}, \qquad \cosh(z) := \frac{e^z+e^{-z}}{2}. \end{equation*}

#### Exercise11.4.9.

Derive the power series at the origin for the hyperbolic sine and cosine.

#### Exercise11.4.10.

Show

1. $$\sinh(0) = 0\text{,}$$ $$\cosh(0) = 1\text{.}$$

2. $$\frac{d}{dx} \bigl[ \sinh(x) \bigr] = \cosh(x)$$ and $$\frac{d}{dx} \bigl[ \cosh(x) \bigr] = \sinh(x)\text{.}$$

3. $$\cosh(x) > 0$$ for all $$x \in \R$$ and show that $$\sinh(x)$$ is strictly increasing and bijective from $$\R$$ to $$\R\text{.}$$

4. $${\bigl(\cosh(x)\bigr)}^2 = 1 + {\bigl(\sinh(x)\bigr)}^2$$ for all $$x\text{.}$$

#### Exercise11.4.11.

Define $$\tan(x) := \frac{\sin(x)}{\cos(x)}$$ as usual.

1. Show that for $$x \in (\nicefrac{-\pi}{2},\nicefrac{\pi}{2})$$ both $$\sin$$ and $$\tan$$ are strictly increasing, and hence $$\sin^{-1}$$ and $$\tan^{-1}$$ exist when we restrict to that interval.

2. Show that $$\sin^{-1}$$ and $$\tan^{-1}$$ are differentiable and that $$\frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}}$$ and $$\frac{d}{dx} \tan^{-1}(x) = \frac{1}{1+x^2}\text{.}$$

3. Using the finite geometric sum formula show

\begin{equation*} \tan^{-1}(x) = \int_0^x \frac{1}{1+t^s} dt = \sum_{k=0}^\infty \frac{{(-1)}^k}{2k+1} x^{2k+1} \end{equation*}

converges for all $$-1 \leq x \leq 1$$ (including the end points). Hint: Integrate the finite sum, not the series.

4. Use this to show that

\begin{equation*} 1 - \frac{1}{3} + \frac{1}{5} - \cdots = \sum_{k=0}^\infty \frac{{(-1)}^k}{2k+1} = \frac{\pi}{4} . \end{equation*}
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