# 2 Complex numbers

## 2.10 The complex exponential function

Consider the real exponential function *f* (*x*) = *e*^{x} (that is, *f* (*x*) = exp *x*). We now extend the definition of this function to define a function *f*(*z*) = *e*^{z} whose domain and codomain are .

We expect complex powers of *e* to satisfy the familiar properties of real powers of *e*. For example, we expect that

If this is to be achieved, then the definition of *e*^{z} has to be as follows.

### Definition

If *z* = *x* + *iy*, then *e*^{z} = *e*^{x}*e*^{iy} = *e*^{x}(cos *y* + *i* sin *y*).

### Example 8

Show that

for all complex numbers *z*_{1} and *z*_{2}.

### Solution

We use Strategy 1.

Suppose that *z*_{1} = *x*_{1} + *iy*_{1} and *z*_{2} = *x*_{2} + *iy*_{2}. Then

We assume that if *x*_{1}, *x*_{2} are *real* numbers, then

### Exercise 26

- (a) Using the above definition for
*e*^{z}and de Moivre's Theorem, show that - (b) Show that
*e*^{z1}/*e*^{z2}=*e*^{z1−z2}, for all*z*_{1},*z*_{2}.

### Solution

- (a) Let
*z*=*x*+*iy*; then - (by de Moivre's Theorem with
*n*= − 1) - (b)

So the rules for multiplication and division of complex powers of *e* are exactly the same as those for real powers. Furthermore, when the exponent *z* is real, that is when *z* = *x* + 0*i*, where *x* , the definitions of a real and a complex power of *e* coincide, since

On the other hand, if *z* = 0 + *iy*, where *y* , then the definition gives

This is known as **Euler's formula**. Putting *y* = , we obtain

or

This is a remarkable relationship between five important numbers: 0, 1, *i*, and *e*.

The formula *e*^{iy} = cos *y* + *i* sin *y* gives us an alternative form for the expression of a complex number in polar form. If

then we can write cos *θ* + *i* sin *θ *as *e*^{iθ}, so

Some texts refer to *re*^{iθ} as *polar* form.

A complex number expressed in this way is said to be in **exponential form**. Using this notation, de Moivre's Theorem becomes the simple result

We can use the complex exponential function to find some further useful trigonometric identities. At the end of Section 2.9 we showed how de Moivre's Theorem can be used to express the sine or cosine of a multiple of* θ* in terms of powers of sin *θ *and cos *θ*. Here we do the opposite, expressing a power of sin *θ* or cos *θ* as a combination of sines or cosines of multiples of *θ*.

First, we deduce two useful equations. We know that, for all *θ* ,

Also, since cos(−*θ*) + *i* sin(−*θ*)=cos *θ *− *i* sin *θ*, we have

Adding equations 2.6 and 2.7 gives

and subtracting them gives

Equations 2.8 and 2.9 enable us to express cos *θ* and sin *θ* as combinations of complex exponentials, and it is these equations we use to obtain the identities we are looking for.

### Example 9

- (a) Show that
- .
- (b) Expand the expression (
*e*^{iθ}+*e*^{−iθ})^{4}using the Binomial Theorem, and hence show that

### Solution

- (a) 2cos
*θ*=*e*^{iθ}+*e*^{−iθ}, so (2 cos*θ*)^{4}= (*e*^{iθ}+*e*^{iθ})^{4}. - Hence
- .
- (b)
- Hence
- Using Equation 2.8 first with 4
*θ*in place of*θ*and then with 2*θ*in place of*θ*, we have - as required.

### Exercise 27

- (a) Show that
- .
- (b) Expand (
*e*^{iθ}−*e*^{−iθ})^{5}using the Binomial Theorem, and hence show that - .

### Solution

- (a) 2
*i*sin*θ*=*e*^{iθ}−*e*^{− iθ}, so (2*i*sin*θ*)^{5}=(*e*^{i θ}−*e*^{−i θ})^{5}. - Hence
- (b)
- Hence