12.6 Trigonometric Functions

Next we calculate $\exp(it)$ for $t\in\mbox{{\bf R}}$.

$$\exp(it)=\sum_{j=0}^\infty {{(it)^j}\over {j!}},\quad t\in\mbox{{\bf R}}.$$

Now $\{i^n\}=\{1,i,-1,-i,1,i,-1,-i,\cdots\}$ and it is clear that $(i)^{2n}=(-1)^n\in\mbox{{\bf R}}$, $(i)^{2n+1}=i(-1)^n$ is pure imaginary. Hence,

$$\begin{eqnarray*} \mbox{\rm Re}\left(\exp(it)\right)&=&\sum_{j=0}^\infty{{(-1)^j... ...j=0}^\infty{{(-1)^jt^{2j+1}}\over {(2j+1)!}}=\sin t;\index{sine} \end{eqnarray*}$$

i.e.,

$$\exp(it)=\cos t+i\sin t\mbox{ for all } t\in\mbox{{\bf R}}.$$ (12.46)

For any complex number $(x,y)=x+iy$, we have

$$\begin{eqnarray*} \exp(x+iy)&=&\exp(x)\exp(iy)=\exp(x)[\cos (y)+i\sin (y)] \\ &=&\exp(x)\cos (y)+i\exp(x)\sin (y). \end{eqnarray*}$$

Since your calculator has buttons that calculate approximations to $\exp$, $\sin$ and $\cos$, you can approximately calculate the exponential of any complex number with a few key strokes.

The relation (12.46)

$$\exp(it)=\cos t+i\sin t$$

actually holds for all $t\in\mbox{{\bf C}}$, since

$$\begin{eqnarray*} \sum_{j=0}^n{(-1)^jz^{2j} \over (2j)! } +i\sum_{j=0}^n{(-1)^... ...{2j+1}\over {(2j+1)!}} \\ &=&\sum_{j=0}^{2n+1}{(iz)^j\over j!}. \end{eqnarray*}$$

Hence

$$e^{iz}=\cos z+i\sin z\mbox{ for all }z\in\mbox{{\bf C}},$$ (12.47)

so

$$e^{-iz}=\cos z-i\sin z\mbox{ for all }z\in\mbox{{\bf C}}.$$ (12.48)

We can solve (12.47) and (12.48) for $\sin(z)$ and $\cos(z)$ to obtain

$$\cos(z)={{e^{iz}+e^{-iz}}\over 2}\mbox{ for all }z\in\mbox{{\bf C}}.$$ (12.49)

$$\sin(z)={{e^{iz}-e^{-iz}}\over {2i}}\mbox{ for all }z\in\mbox{{\bf C}}.$$ (12.50)

From (12.47) it follows that

$$\vert e^{it}\vert = 1 \mbox{ for all }t \in \mbox{{\bf R}},$$

i.e., $e^{it}$ is in the unit circle for all $t\in\mbox{{\bf R}}$.

12.51   Exercise (Addition laws for $\sin$ and $\cos$.) A Prove that

$$\begin{eqnarray*} \cos(z+a)&=&\cos (z)\cos (a)-\sin (z)\sin (a)\\ \sin(z+a)&=&\sin (z)\cos (a)+\cos (z)\sin (a) \end{eqnarray*}$$

for all $z,a\in\mbox{{\bf C}}$.

By the addition laws, we have (for all $x,y\in\mbox{{\bf C}}$),

$$\cos(x+iy)=\cos (x)\cos(iy)-\sin (x)\sin(iy)$$ (12.52)

$$\sin(x+iy)=\sin x\cos(iy)+\cos x\sin(iy).$$ (12.53)

By (12.49) and (12.50)

$$\cos(iy)={{e^{i(iy)}+e^{-i(iy)}}\over 2}={{e^y+e^{-y}}\over 2}$$

and

$$\sin(iy)={{e^{i(iy)}-e^{-i(iy)}}\over {2i}}={{e^{-y}-e^y}\over {2i}}=i\left({{e^y-e^{-y}\over 2}}\right).$$

12.54   Definition (Hyperbolic functions.) For all $z\in\mbox{{\bf C}}$, we define the hyperbolic sine and hyperbolic cosine of $z$ by

$$\begin{eqnarray*} \sinh(z)&=&{{e^z-e^{-z}}\over 2} \glossary{$\sinh$, hyperbolic... ...sh(z)&=&{{e^z+e^{-z}}\over 2}.\glossary{cosh, hyperbolic cosine} \end{eqnarray*}$$

Note that if $z$ is real, $\sinh(z)$ and $\cosh(z)$ are real. Most calculators have buttons that calculate $\cosh$ and $\sinh$. We can now rewrite (12.52) and (12.53) as

$$\begin{eqnarray*} \cos(x+iy)&=&\cos (x)\cosh (y)-i\sin (x)\sinh (y) \\ \sin(x+iy)&=&\sin (x)\cosh (y)+i\cos (x)\sinh (y). \end{eqnarray*}$$

These formulas hold true for all complex $x$ and $y$.

Since

$\sin '=\cos$, $\cos '=-\sin$, $\sin(0)=0$ and $\cos(0)=1$,

it follows from our discussion in example 10.45 that

$$\sin(x)\geq x-{{x^3}\over 6} \mbox{ and } \cos x\leq 1-{{x^2}\over 2}+{{x^4}\over {24}}$$

for all $x>0$. In particular

$$\sin(x)\geq x\left(1-{{x^2}\over 6}\right)>0\mbox{ for } 0<x<\sqrt 6$$

and

$$\cos(2)<1-{4\over 2}+{{16}\over {24}}<0.$$

Hence $\cos'=-\sin<0$ on $(0,2)$, so $\cos$ is strictly decreasing on $[0,2]$. Moreover $\cos$ is continuous (since it is differentiable) so by the intermediate value theorem there is a number $c$ in $(0,2)$ such that $\cos(c)=0$. Since $\cos$ is strictly decreasing on $(0,2)$ this number $c$ is unique. (Cf. exercise 5.48.)

12.55   Definition ($\pi$.) We define the real number $\pi$ by the condition $${{\pi\over 2}}$$ is the unique number in $(0,2)$ satisfying $${\cos\left({\pi\over 2}\right)=0}$$.

12.56   Theorem. $\exp$ is periodic of period $2\pi i$; i.e.,

$$\exp(z+2\pi i)=\exp(z)\mbox{ for all }z\in\mbox{{\bf C}}.$$

Proof: Since $\sin^2t+\cos^2t=1$ for all $t\in\mbox{{\bf C}}$, we have $${\sin^2\left({\pi\over 2}\right)=1}$$, so $${\sin\left({\pi\over 2}\right)=\pm 1}$$. We have noted that $\sin t>0$ on $(0,2)$ so $${\sin\left({\pi\over 2}\right)=1}$$. Hence

$$e^{ {{i\pi}\over 2}}=\cos\left({\pi\over 2}\right) +i\sin\left({\pi\over 2}\right)=i,$$

and

$$e^{2i\pi}=\left(e^{{{i\pi}\over 2}}\right)^4 =i^4=1.$$ (12.57)

It follows that $e^{2\pi i+z}=e^{2\pi i}e^z=1e^z=e^z$ for all $z\in\mbox{{\bf C}}$. $\mid\!\mid\!\mid$

12.58   Entertainment. If Maple or Mathematica are asked for the numerical values of $(-1)^{3.14}$ and $i^i$, they agree that

$$(-1)^{3.14}=-.9048\cdots-i\cdot.4257\cdots$$

and

$$i^i=.2078\cdots.$$

Can you propose a reasonable definition for $(-1)^z$ and $i^z$ when $z$ is an arbitrary complex number, that is consistent with these results? To be reasonable you would require that when $z\in\mbox{{\bf Z}}$, $(-1)^z$ and $i^z$ give the expected values, and

$$\begin{eqnarray*} (-1)^{z+w}&=&(-1)^z(-1)^w\mbox{ for all } z,w\in\mbox{{\bf C}}, \\ (i)^{z+w}&=&i^zi^w\mbox{ for all }z,w\in\mbox{{\bf C}}. \end{eqnarray*}$$

12.59   Exercise. A Prove that:

a) $\cos\pi=-1$, and $\sin\pi=0$.

b) $${\cos{{3\pi}\over 2}=0}$$, and $${\sin{{3\pi}\over 2}=-1}$$.

c) $\cos 2\pi=1$, and $\sin 2\pi=0$.

d) $\sin(2\pi-t)=-\sin t$ for all $t\in\mbox{{\bf C}}$.

e) $\cos(2\pi-t)=\cos t$ for all $t\in\mbox{{\bf C}}$.

f) $\sin(\pi-t)=\sin t$ for all $t\in\mbox{{\bf C}}$.

g) $\cos(\pi-t)=-\cos t$ for all $t\in\mbox{{\bf C}}$.

h) $\sin(2\pi+t)=\sin t$ for all $t\in\mbox{{\bf C}}$.

i) $\cos(2\pi+t)=\cos t$ for all $t\in\mbox{{\bf C}}$.

12.60   Theorem. $\cos(2\pi)=1$ and $\cos t<1$ for $0<t<2\pi$.

Proof: From the previous exercise, $\cos (2\pi)=\cos (0)=1$. We've noted that $\sin t>0$ for ${t\in \left(0,{\pi\over 2}\right]}$,

$$\begin{eqnarray*} t\in\left(\textstyle{{\pi\over 2},\pi}\right)&\mbox{$\Longrigh... ...htarrow$}&\sin(\pi-t)>0 \\ &\mbox{$\Longrightarrow$}&\sin(t)>0. \end{eqnarray*}$$

Hence $\sin(t)>0$ for $t\in(0,\pi)$. Hence $\cos'(t)=-\sin(t)<0$ for $t\in(0,\pi)$. Hence $\cos$ is strictly decreasing on $(0,\pi)$. Hence $\cos(x)<\cos (0)=1$ for all $x \in (0,\pi)$.

Now

$$\begin{eqnarray*} t\in(\pi,2\pi)&\mbox{$\Longrightarrow$}&\pi<t<2\pi\mbox{$\hspa... ...htarrow$}&\cos(2\pi-t)<1 \\ &\mbox{$\Longrightarrow$}&\cos t<1, \end{eqnarray*}$$

and since $\cos(\pi)=-1<1$, we've shown that $\cos t<1$ for all $t\in(0,2\pi)$. $\mid\!\mid\!\mid$

12.61   Theorem. Every point $(x,y)$ in the unit circle can be written as
$(x,y)=e^{it}$ for a unique $t\in[0,2\pi)$.

Proof: We first show uniqueness.

Suppose $(x,y)=x+iy=e^{it}=e^{is}$ where $s,t\in [0,2\pi)$. Without loss of generality, say $s\leq t$. Then

$$1={{e^{it}}\over {e^{is}}}=e^{i(t-s)}=\cos(t-s)+i\sin(t-s),$$

and $t-s\in[0,2\pi)$. By the previous theorem, $0$ is the only number in $[0,2\pi)$ whose cosine is $1$, so $t-s = 0$, and hence $t=s$.

Let $(x,y)$ be a point in the unit circle, so $x^2+y^2=1$, and hence $-1\leq x\leq 1$. Since $\cos(0)=1$ and $\cos (\pi)=-1$, it follows from the intermediate value theorem that $x=\cos t$ for some $t\in[0,\pi]$. Then

$$y^2=1-x^2=1-\cos^2(t)=\sin^2(t),$$

so $y=\pm\sin (t)$.

$$\begin{eqnarray*} y=\sin t&\mbox{$\Longrightarrow$}&(x,y)=(\cos t,\sin t)=e^{it}... ... t,-\sin t)=\left(\cos(2\pi-t),\sin(2\pi-t)\right)=e^{i(2\pi-t)} \end{eqnarray*}$$

and since $t\in[0,\pi]$, we have $2\pi-t\in[\pi,2\pi]$. $\mid\!\mid\!\mid$

12.62   Lemma. The set of all complex solutions to $e^z=1$ is $\{2\pi in:n\in\mbox{{\bf Z}}\}$.

Proof: By exercise 12.59A

$$e^{2\pi i}=\cos 2\pi+i\sin 2\pi=1+i0=1,$$

so

$$e^{2\pi in}=\left(e^{2\pi i}\right)^n=1^n=1.$$

Let $w=(a,b)=a+ib$ be any solution to $e^z=1$; i.e.,

$$1=e^{a+bi}=e^ae^{ib}.$$

By uniqueness of polar decomposition,

$$e^a=1\mbox{ and } e^{ib}=1,$$

so $a=0$ (since for real $a$, $e^a=1\iff a=0$). We can write $${ {b\over {2\pi}}=n+\varepsilon}$$ where $n\in\mbox{{\bf Z}}$ and $\varepsilon\in[0,1)$ by theorem 5.30 , so $b=2\pi n+2\pi\varepsilon$ where $2\pi\varepsilon\in[0,2\pi)$. Now

$$1=e^{ib}=e^{2\pi in+i2\pi\varepsilon}=e^{2\pi i\varepsilon}.$$

By theorem 12.61, $2\pi i\varepsilon=0$, so $\varepsilon=0$, and $b=2\pi n$; i.e., $w=2\pi in$. $\mid\!\mid\!\mid$

12.63   Definition (Argument.) Let $a\in\mbox{{\bf C}}\backslash\{0\}$ and write $a$ in its polar decomposition $a=\vert a\vert u$, where $\vert u\vert=1$. We know $u=e^{iA}$ for a unique $A\in[0,2\pi)$. I will call $A$ the argument of $a$ and write $A=\mbox{{\rm Arg}}(a)$. Hence

$$a=\vert a\vert e^{i{\mathrm{\rm Arg}}(a)} \hspace{1em} A \in [0,2\pi).$$

12.64   Remark. Our definition of $\mbox{{\rm Arg}}$ is rather arbitrary. Other natural definitions are

$$\mbox{{\rm Arg}}_1(z)\mbox{ is the unique number } a\mbox{ in }[-\pi,\pi) \mbox{ such that }z=\vert z\vert e^{ia},$$

or

$$\mbox{{\rm Arg}}_2(z)\mbox{ is the unique number } b\mbox{ in }(-\pi,\pi] \mbox{ such that }z=\vert z\vert e^{ib}.$$

None of these argument functions is continuous; e.g.,

$$\left\{e^{{{-i\pi}\over n}}\right\}_{n\geq 1}\to 1.$$

But

$$\left\{\mbox{{\rm Arg}}\left(e^{{{-i\pi}\over n}}\right)\righ... ...ver n}\right)\right\}_{n\geq 1}\to 2\pi\neq\mbox{{\rm Arg}}(1).$$

12.65   Theorem. Let $a\in\mbox{{\bf C}}\backslash\{0\}$. Then the complex solutions to the equation $e^z=a$ are exactly the numbers of the form

$$z=\ln\vert a\vert+i\mbox{{\rm Arg}}(a)+2\pi in\mbox{ where } n\in\mbox{{\bf Z}}.$$

In particular, every non-zero $a\in\mbox{{\bf C}}$ is the exponential of some $z\in\mbox{{\bf C}}$.

Proof: Since

$$\begin{eqnarray*} e^{(\ln\vert a\vert+i{\mathrm{\rm Arg}}(a)+2\pi in)}&=&e^{\ln\... ...a)}e^{2\pi in} \\ &=&\vert a\vert e^{i{\mathrm{\rm Arg}}(a)}=a, \end{eqnarray*}$$

the numbers given are solutions to $e^z=a$. Let $w$ be any solution to $e^w=a$. Then $e^{w-\ln\vert a\vert-i{\mathrm{\rm Arg}}(a)}={a \over a}=1$. Hence, by the lemma 12.62,

$$w-\ln\vert a\vert+i\mbox{{\rm Arg}}(w)=2\pi in\mbox{ for some }n\in\mbox{{\bf Z}}.\mbox{ $\mid\!\mid\!\mid$}$$

We will now look at $\exp$ geometrically as a function from $\mbox{{\bf C}}$ to $\mbox{{\bf C}}$.

Claim: $\exp$ maps the vertical line $x=x_0$ into the circle $C(0,e^{x_0})$.

Proof: If $z=x_0+iy$, then

$$\vert e^z\vert=\vert e^{x_0+iy}\vert=\vert e^{x_0}e^{iy}\vert=\vert e^{x_0}\vert \vert e^{iy}\vert=e^{x_0}.$$

Claim: $\exp$ maps the horizontal line $y=y_0$ into the ray through $0$ with direction $e^{iy_0}$.

Proof: If $z=x+iy_0$, then

$$e^z=e^{x+iy_0}=e^x\cdot e^{iy_0}\mbox{ and }e^x>0.$$

Since $\exp$ is periodic of period $2\pi i$, $\exp$ maps an infinite horizontal strip of width $w$ into an infinite circular segment making `` angle $w$" at the origin.

The Exponentials of Some Cats

$\exp$ maps every strip $\{(x,y)\colon y_0\leq y < y_0+2\pi\}$ onto all of $\mbox{{\bf C}}\backslash\{0\}$.

12.66   Theorem (Roots of complex numbers.) Let $a\in\mbox{{\bf C}}\backslash\{0\}$ and let $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$. Then the solutions to $z^n=a$ in $\mbox{{\bf C}}$ are exactly the numbers

$$z=\vert a\vert^{{1\over n}}e^{i\left({{{\mathrm{\rm Arg}}(a)+... ...}\right)}\mbox{ where }k\in\mbox{{\bf Z}}\mbox{ and }0\leq k<n.$$

(These numbers are distinct.)

Proof: These numbers are clearly solutions to $z^n=a$. Let $w=\vert w\vert e^{i{\mathrm{\rm Arg}}(w)}$ be any solution to $z^n=a$. Then

$$\vert w\vert^ne^{in{\mathrm{\rm Arg}}(w)}=w^n=a=\vert a\vert e^{i{\mathrm{\rm Arg}}(a)}.$$

By uniqueness of polar decomposition,

$$\vert w\vert^n=\vert a\vert\mbox{ and }e^{in{\mathrm{\rm Arg}}(w)}=e^{i{\mathrm{\rm Arg}}(a)},$$

i.e., $\vert w\vert=\vert a\vert^{1/n}$ and $e^{i[n{\mathrm{\rm Arg}}(w)-{\mathrm{\rm Arg}}(a)]}=1$. Hence, $n\mbox{{\rm Arg}}(w)-\mbox{{\rm Arg}}(a)=2\pi k$ for some $k \in\mbox{{\bf N}}$ and

$$\mbox{{\rm Arg}}(w)={{\mbox{{\rm Arg}}(a)+2\pi k}\over n}\mbox{ for some }k\in\mbox{{\bf N}}.$$

Thus

$$e^{i{\mathrm{\rm Arg}}w}=e^{i\left({{{\mathrm{\rm Arg}}(a)+2\pi k}\over n}\right)} \mbox{ for some }k\in\mbox{{\bf N}}.$$

For each $k\in\mbox{{\bf Z}}$, the number

$$w_k=\vert w\vert^{{1\over n}}e^{{{i{\mathrm{\rm Arg}}(a)}\over n}}\cdot e^{{{2\pi i k}\over n}}$$

is a solution to $w^n=a$. For $0\leq k<n$, the numbers $${ {{2\pi i k}\over n}}$$ are distinct numbers in $[0,2\pi)$, so the numbers $${e^{{2\pi i k}\over n}}$$ are distinct. For every $K\in\mbox{{\bf Z}}$, $${{K\over n}=M+\varepsilon}$$ where $M\in\mbox{{\bf Z}}$ and $\varepsilon\in[0,1)$, so $K=nM+\varepsilon n$ where $\varepsilon n\in[0,n)$ and $\varepsilon n=K-nM\in\mbox{{\bf Z}}$; i.e.,

$$K=nM+k\mbox{ where }k\in\mbox{{\bf Z}}\mbox{ and }0\leq k<n.$$

Then ${K\over n}=M+{k\over n}$, so

$$e^{2\pi i{K\over n}}=e^{2\pi iM}e^{{{2\pi ik}\over n}}=e^{{{2... ...n\mbox{{\bf Z}}\mbox{ and }0\leq k<n.\mbox{ $\mid\!\mid\!\mid$}$$