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# 12.6 Trigonometric Functions

Next we calculate for .

Now and it is clear that , is pure imaginary. Hence,

i.e.,
 (12.46)

For any complex number , we have

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

The relation (12.46)

actually holds for all , since

Hence
 (12.47)

so
 (12.48)

We can solve (12.47) and (12.48) for and to obtain
 (12.49)

 (12.50)

From (12.47) it follows that

i.e., is in the unit circle for all .

12.51   Exercise (Addition laws for and .) A Prove that

for all .

By the addition laws, we have (for all ),

 (12.52)

 (12.53)

By (12.49) and (12.50)

and

12.54   Definition (Hyperbolic functions.) For all , we define the hyperbolic sine and hyperbolic cosine of by

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

These formulas hold true for all complex and .

Since

, , and ,
it follows from our discussion in example 10.45 that

for all . In particular

and

Hence on , so is strictly decreasing on . Moreover is continuous (since it is differentiable) so by the intermediate value theorem there is a number in such that . Since is strictly decreasing on this number is unique. (Cf. exercise 5.48.)

12.55   Definition (.) We define the real number by the condition is the unique number in satisfying .

12.56   Theorem. is periodic of period ; i.e.,

Proof: Since for all , we have , so . We have noted that on so . Hence

and
 (12.57)

It follows that for all .

12.58   Entertainment. If Maple or Mathematica are asked for the numerical values of and , they agree that

and

Can you propose a reasonable definition for and when is an arbitrary complex number, that is consistent with these results? To be reasonable you would require that when , and give the expected values, and

12.59   Exercise. A Prove that:
a) , and .
b) , and .
c) , and .
d) for all .
e) for all .
f) for all .
g) for all .
h) for all .
i) for all .

12.60   Theorem. and for .

Proof: From the previous exercise, . We've noted that for ,

Hence for . Hence for . Hence is strictly decreasing on . Hence for all .

Now

and since , we've shown that for all .

12.61   Theorem. Every point in the unit circle can be written as
for a unique .

Proof: We first show uniqueness.

Suppose where . Without loss of generality, say . Then

and . By the previous theorem, is the only number in whose cosine is , so , and hence .

Let be a point in the unit circle, so , and hence . Since and , it follows from the intermediate value theorem that for some . Then

so .

and since , we have .

12.62   Lemma. The set of all complex solutions to is .

Proof: By exercise 12.59A

so

Let be any solution to ; i.e.,

By uniqueness of polar decomposition,

so (since for real , ). We can write where and by theorem 5.30 , so where . Now

By theorem 12.61, , so , and ; i.e., .

12.63   Definition (Argument.) Let and write in its polar decomposition , where . We know for a unique . I will call the argument of and write . Hence

12.64   Remark. Our definition of is rather arbitrary. Other natural definitions are

or

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

But

12.65   Theorem. Let . Then the complex solutions to the equation are exactly the numbers of the form

In particular, every non-zero is the exponential of some .

Proof: Since

the numbers given are solutions to . Let be any solution to . Then . Hence, by the lemma 12.62,

We will now look at geometrically as a function from to .

Claim: maps the vertical line into the circle .

Proof: If , then

Claim: maps the horizontal line into the ray through with direction .

Proof: If , then

Since is periodic of period , maps an infinite horizontal strip of width into an infinite circular segment making  angle " at the origin.

The Exponentials of Some Cats

maps every strip onto all of .

12.66   Theorem (Roots of complex numbers.) Let and let . Then the solutions to in are exactly the numbers

(These numbers are distinct.)

Proof: These numbers are clearly solutions to . Let be any solution to . Then

By uniqueness of polar decomposition,

i.e., and . Hence, for some and

Thus

For each , the number

is a solution to . For , the numbers are distinct numbers in , so the numbers are distinct. For every , where and , so where and ; i.e.,

Then , so

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