Proposition 11.4.1. Law of exponents.
Let \(z,w \in \C\) be complex numbers. Then
\begin{equation*}
e^{z+w} = e^z e^w.
\end{equation*}
Section 11.4 Complex exponential and trigonometric functions
Note: 1 lecture
Subsection 11.4.1 The complex exponential
Let
\begin{equation*}
E(z) \coloneqq \sum_{k=0}^\infty \frac{1}{k!} z^k .
\end{equation*}
This series converges for all \(z \in \C\text{,}\) 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, direct computation shows
\begin{equation*}
\frac{d}{dx} \bigl( E(x) \bigr) = E(x) .
\end{equation*}
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 \coloneqq 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 of the proposition is that \(e^z\not=0\) for all \(z \in \C\text{,}\) as \(e^z e^{-z} = e^{z-z} = 1\text{.}\) This computation means that \({(e^z)}^{-1} = e^{-z}\text{.}\) Combining that fact with the law of exponents gives
\begin{equation*}
{(e^z)}^{n} = e^{nz} \qquad \text{for all } n \in \Z.
\end{equation*}
A yet more complicated consequence is that we can compute the power series for the exponential at any point \(a \in
\C\text{:}\)
\begin{equation*}
e^z = e^a e^{z-a} = \sum_{k=0}^\infty \frac{e^a}{k!} {(z-a)}^k .
\end{equation*}
Subsection 11.4.2 Trigonometric 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) \coloneqq \frac{e^{iz} + e^{-iz}}{2}
\qquad\text{and}\qquad
\sin(z) \coloneqq \frac{e^{iz} - e^{-iz}}{2i} .
\end{equation*}
Let us use our definition to prove common properties of sine and cosine. In the process, we also define the number \(\pi\text{.}\)
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*} \sabs{\sin(x)} \leq 1, \qquad \sabs{\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 \(\sabs{z} = 1\text{.}\)
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 \(x\text{,}\) \(\cos(x) =
\frac{e^{ix}-e^{-ix}}{2} =
\frac{e^{ix}-\overline{e^{ix}}}{2} =
\Re (e^{ix})\) 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, is \(e^{ix}\) 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*}
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 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 \(x \in \R\) as \(\sabs{\cos(x)} \leq 1\text{.}\) In other words, \(f\) is increasing and \(f(0) = 0\text{.}\) So \(f\) must be nonnegative when \(x \geq 0\) and hence, \(\sin(x) \geq x\text{.}\)
Next, 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 \((0,\nicefrac{\pi}{2})\text{,}\) 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*}
So \(e^{i2\pi} = 1 = e^0\text{.}\) 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 \(\sin(z+2\pi) = \sin(z)\text{.}\) 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 \(x = 2\pi\text{.}\) 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*}
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*}
Then either \(a = 0\) or \(a^2 = b^2\text{.}\) As \(\nicefrac{x}{4} \leq \nicefrac{\pi}{2}\text{,}\) then \(a = \cos(\nicefrac{x}{4}) \geq 0\) and \(b = \sin(\nicefrac{x}{4}) > 0\text{.}\) 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 \(a=0\text{,}\) 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 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, \(b = \sin(x)\text{.}\) 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).
Subsection 11.4.3 The 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, \(e^{it}\) parametrizes the circle by arclength. That is, \(t\) measures the arclength on a circle of radius 1 by the angle in radians. So the definitions of \(\sin\) and \(\cos\) given 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 law of exponents \(e^{a+b} = e^a e^b\) 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*}
Exercises 11.4.4 Exercises
11.4.1.
Derive the power series for \(\sin(z)\) and \(\cos(z)\) at the origin.
11.4.2.
Using the power series, show that for real \(x\text{,}\) we have \(\frac{d}{dx} \bigl[ \sin(x)\bigr] = \cos(x)\) and \(\frac{d}{dx} \bigl[ \cos(x)\bigr] = -\sin(x)\text{.}\)
11.4.3.
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 \(a^2+b^2=1\) and \(a \geq 0\text{.}\) By multiplying by \(e^{i\pi}\) or \(e^{-i\pi}\text{,}\) show that we get everything, that is, even points where \(a < 0\text{.}\)
11.4.4.
Show that the exponential is onto \(\C \setminus \{ 0 \}\text{,}\) and in fact, that for every nonzero \(w\text{,}\) there are infinitely many \(z \in \C\) such that \(e^z=w\text{.}\)
11.4.5.
Prove that for every \(w \not= 0\) and every \(\epsilon > 0\text{,}\) there exists a \(z \in \C\) with \(\sabs{z} < \epsilon\) such that \(e^{1/z} = w\text{.}\)
11.4.6.
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 \(z \in \C\text{.}\)
11.4.7.
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 \(z,w \in \C\text{.}\)
11.4.8.
Define \(\operatorname{sinc}(z) \coloneqq \frac{\sin(z)}{z}\) for \(z \not=0\) and \(\operatorname{sinc}(0) \coloneqq 1\text{.}\) Show that sinc is analytic and compute its power series at zero.
Define the hyperbolic sine and hyperbolic cosine by
\begin{equation*}
\sinh(z) \coloneqq \frac{e^z-e^{-z}}{2}, \qquad
\cosh(z) \coloneqq \frac{e^z+e^{-z}}{2}.
\end{equation*}
11.4.9.
Derive the power series at the origin for the hyperbolic sine and cosine.
11.4.10.
Show
- \(\sinh(0) = 0\text{,}\) \(\cosh(0) = 1\text{.}\)
- For \(x \in \R\text{,}\) \(\frac{d}{dx} \bigl[ \sinh(x) \bigr] = \cosh(x)\) and \(\frac{d}{dx} \bigl[ \cosh(x) \bigr] = \sinh(x)\text{.}\)
- \(\cosh(x) > 0\) for all \(x \in \R\) and show that \(\sinh(x)\) is strictly increasing and bijective from \(\R\) to \(\R\text{.}\)
- \({\bigl(\cosh(x)\bigr)}^2 = 1 + {\bigl(\sinh(x)\bigr)}^2\) for all \(x\text{.}\)
11.4.11.
Define \(\tan(x) \coloneqq \frac{\sin(x)}{\cos(x)}\) as usual.
- 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.
- Show that \(\sin^{-1}\) and \(\tan^{-1}\) are differentiable and that \(\frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}}\) and \(\frac{d}{dx} \tan^{-1}(x) = \frac{1}{1+x^2}\text{.}\)
- 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.
- 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*}
