12.6 Trigonometric Functions

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

i.e.,

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

so

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

From (12.47) it follows that

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

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

By (12.49) and (12.50)

and

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.)

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

and

It follows that for all .

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

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 .

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 .

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., .

or

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

But

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.

maps every strip onto all of .

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