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Section 11.4 The complex exponential and the trigonometric functions

Note: 1 lecture

Subsection 11.4.1 The complex exponential


\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.

Figure 11.7. Graphs of the real part (left) and imaginary part (right) of the complex exponential \(e^z = e^{x+iy}\text{.}\) The \(x\)-axis goes from \(-4\) to \(4\text{,}\) the \(y\)-axis goes from \(-6\) to \(6\text{,}\) and the vertical axis goes from \(-e^{4} \approx -54.6\) to \(e^{4} \approx 54.6\text{.}\) The plot of the real exponential (\(y=0\)) is marked in a bold line.


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 \(a \in \C\text{:}\) \(e^z = e^a e^{z-a} = \sum \frac{e^a}{k!} {(z-a)}^k\text{.}\)

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) := \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.


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*}

and by the addition formula

\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).

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 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*}

Subsection 11.4.4 Exercises

Exercise 11.4.1.

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

Exercise 11.4.2.

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

Exercise 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 \(a^2+b^2=1\) for \(a \geq 0\text{.}\) By multiplying by \(e^{i\pi}\) or \(e^{-i\pi}\) show that we get everything.

Exercise 11.4.4.

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

Exercise 11.4.5.

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

Exercise 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 \(z \in \C\text{.}\)

Exercise 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 \(z,w \in \C\text{.}\)

Exercise 11.4.8.

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*}

Exercise 11.4.9.

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

Exercise 11.4.10.


  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 \(x\text{.}\)

Exercise 11.4.11.

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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