INTRODUCTION

In this and the next section, we shall examine the exponential, logarithmic, trigonometric, and hyperbolic functions of a complex variable z. Although the definitions of these complex functions are motivated by their real variable analogues, the properties of these complex functions will yield some surprises.

Exponential Function

Recall that in real variables the exponential function f(x) = e^x has the properties

f′(x) = f(x)     and     f(x₁ + x₂) = f(x₁)f(x₂). (1)

We certainly want the definition of the complex function f(z) = e^z, where z = x + iy, to reduce e^x for y = 0 and to possess the same properties as in (1).

We have already used an exponential function with a pure imaginary exponent. Euler’s formula,

e^iy = cos y + i sin y,     y a real number, (2)

played an important role in Section 3.3. We can formally establish the result in (2) by using the Maclaurin series for e^x and replacing x by iy and rearranging terms:

For z = x + iy, it is natural to expect that

e^x+iy = e^xe^iy

and so by (2), e^x+iy = e^x(cos y + i sin y).

Inspired by this formal result, we make the following definition.

The exponential function e^z is also denoted by the symbol exp z. Note that (3) reduces to e^x when y = 0.

EXAMPLE 1 Complex Value of the Exponential Function

The real and imaginary parts of e^z, u(x, y) = e^x cos y and v(x, y) = e^x sin y, are continuous and have continuous first partial derivatives at every point z of the complex plane. Moreover, the Cauchy–Riemann equations are satisfied at all points of the complex plane:

It follows from Theorem 17.5.2 that f(z) = e^z is analytic for all z; in other words, f is an entire function.

Properties

We shall now demonstrate that e^z possesses the two desired properties given in (1). First, the derivative of f is given by (5) of Section 17.5:

f′(z) = e^x cos y + i(e^x sin y) = e^x(cos y + i sin y) = f(z).

As desired, we have established that

Second, if z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂, then by multiplication of complex numbers and the addition formulas of trigonometry, we obtain

In other words, (4)

It is left as an exercise to prove that

Periodicity

Unlike the real function e^x, the complex function f(z) = e^z is periodic with the complex period 2πi. Since e^2πi = cos 2π + i sin 2π = 1 and, in view of (4), e^z+^2πi = e^ze^2πi = e^z for all z, it follows that f(z + 2πi) = f(z). Because of this complex periodicity, all possible functional values of f(z) = e^z are assumed in any infinite horizontal strip of width 2π. Thus, if we divide the complex plane into horizontal strips defined by (2n − 1)π < y ≤ (2n + 1)π, n = 0, ±1, ±2, . . ., then, as shown in FIGURE 17.6.1, for any point z in the strip −π < y ≤ π, the values f(z), f(z + 2πi), f(z − 2πi), f(z + 4πi), and so on, are the same. The strip −π < y ≤ π is called the fundamental region for the exponential function f(z) = e^z. The corresponding flow over the fundamental region is shown in FIGURE 17.6.2.

Polar Form of a Complex Number

In Section 17.2, we saw that the complex number z could be written in polar form as z = r(cos θ + i sin θ). Since e^iθ = cos θ + i sin θ, we can now write the polar form of a complex number as

z = re^iθ.

For example, in polar form z = 1 + i is z = .

Circuits

In applying mathematics, mathematicians and engineers often approach the same problem in completely different ways. Consider, for example, the solution of Example 10 in Section 3.8. In this example we used strictly real analysis to find the steady-state current i_p(t) in an LRC-series circuit described by the differential equation

Electrical engineers often solve circuit problems such as this using complex analysis. To illustrate, let us first denote the imaginary unit by the symbol j to avoid confusion with the current i. Since current i is related to charge q by i = dq/dt, the differential equation is the same as

Moreover, the impressed voltage E₀ sin γt can be replaced by Im(E₀e^jγt), where Im means the “imaginary part of.” Because of this last form, the method of undetermined coefficients suggests that we assume a solution in the form of a constant multiple of complex exponential—that is, i_p(t) = Im(Ae^jγt). We substitute this expression into the last differential equation, use the fact that q is an antiderivative of i, and equate coefficients of e^jγt:

The quantity Z = R + j(Lγ − 1/Cγ) is called the complex impedance of the circuit. Note that the modulus of the complex impedance, |Z| = , was denoted in Example 10 of Section 3.8 by the letter Z and called the impedance.

Now, in polar form the complex impedance is

Hence, A = E₀/Z = E₀/(|Z|e^jθ), and so the steady-state current can be written as

The reader is encouraged to verify that this last expression is the same as (35) in Section 3.8.

Logarithmic Function

The logarithm of a complex number z = x + iy, z ≠ 0, is defined as the inverse of the exponential function—that is,

(5)

In (5) we note that ln z is not defined for z = 0, since there is no value of w for which e^w = 0. To find the real and imaginary parts of ln z , we write w = u + iv and use (3) and (5):

x + iy = e^u+iv = e^u(cos v + i sin v) = e^u cos v + ie^u sin v.

The last equality implies x = e^u cos v and y = e^u sin v. We can solve these two equations for u and v. First, by squaring and adding the equations, we find

where log_e|z| denotes the real natural logarithm of the modulus of z. Second, to solve for v, we divide the two equations to obtain

This last equation means that v is an argument of z; that is, v = θ = arg z. But since there is no unique argument of a given complex number z = x + iy, if θ is an argument of z, then so is θ + 2nπ, n = 0, ±1, ±2, . . . .

As is clearly indicated in (6), there are infinitely many values of the logarithm of a complex number z. This should not be any great surprise since the exponential function is periodic.

In real calculus, logarithms of negative numbers are not defined. As the next example will show, this is not the case in complex calculus.

EXAMPLE 2 Complex Values of the Logarithmic Function

Find the values of (a) ln(−2), (b) ln i, and (c) ln(−1 − i).

SOLUTION

(a) With θ = arg(−2) = π and log_e|−2| = 0.6932, we have from (6)

ln(−2) = 0.6932 + i(π + 2nπ).

(b) With θ = arg(i) = π/2 and log_e|i| = log_e 1 = 0, we have from (6)

In other words, ln i = πi/2, −3πi/2, 5πi/2, −7πi/2, and so on.

(c) With θ = arg(−1 − i) = 5π/4 and log_e|−1 − i| = log_e = 0.3466, we have from (6)

≡

EXAMPLE 3 Solving an Exponential Equation

Principal Value

It is interesting to note that as a consequence of (6), the logarithm of a positive real number has many values. For example, in real calculus, log_e5 has only one value: log_e 5 = 1.6094, whereas in complex calculus, ln 5 = 1.6094 + 2nπi. The value of ln 5 corresponding to n = 0 is the same as the real logarithm log_e 5 and is called the principal value of ln 5. Recall that in Section 17.2 we stipulated that the principal argument of a complex number, written Arg z, lies in the interval (−π, π]. In general, we define the principal value of ln z as that complex logarithm corresponding to n = 0 and θ = Arg z. To emphasize the principal value of the logarithm, we shall adopt the notation Ln z. In other words,

(7)

Since Arg z is unique, there is only one value of Ln z for each z ≠ 0.

EXAMPLE 4 Principal Values

The principal values of the logarithms in Example 2 are as follows:

(a) Since Arg(−2) = π, we need only set n = 0 in the result given in part (a) of Example 2:

Ln(−2) = 0.6932 + πi.

(b) Similarly, since Arg(i) = π/2, we set n = 0 in the result in part (b) of Example 2 to obtain

(c) In part (c) of Example 2, arg(−1 − i) = 5π/4 is not the principal argument of z = −1 − i. The argument of z that lies in the interval (−π, π] is Arg(−1 − i) = −3π/4. Hence, it follows from (7) that

≡

Up to this point we have avoided the use of the word function for the obvious reason that ln z defined in (6) is not a function in the strictest interpretation of that word. Nonetheless, it is customary to write f(z) = ln z and to refer to f(z) = ln z by the seemingly contradictory phrase multiple-valued function. Although we shall not pursue the details, (6) can be interpreted as an infinite collection of logarithmic functions (standard meaning of the word). Each function in the collection is called a branch of ln z. The function f(z) = Ln z is then called the principal branch of ln z, or the principal logarithmic function. To minimize the confusion, we shall hereafter simply use the words logarithmic function when referring to either f(z) = ln z or f(z) = Ln z.

Some familiar properties of the logarithmic function hold in the complex case:

(8)

Equations (8) and (9) are to be interpreted in the sense that if values are assigned to two of the terms, then a correct value is assigned to the third term.

EXAMPLE 5 Properties of Logarithms

Just as (7) of Section 17.2 was not valid when arg z was replaced with Arg z , so too (8) is not true, in general, when ln z is replaced by Ln z. See Problems 45 and 46 in Exercises 17.6.

Analyticity

The logarithmic function f(z) = Ln z is not continuous at z = 0 since f(0) is not defined. Moreover, f(z) = Ln z is discontinuous at all points of the negative real axis. This is because the imaginary part of the function, v = Arg z, is discontinuous only at these points. To see this, suppose x₀ is a point on the negative real axis. As z → x₀ from the upper half-plane, Arg z → π, whereas if z → x₀ from the lower half-plane, then Arg z → −π. This means that f(z) = Ln z is not analytic on the nonpositive real axis. However, f(z) = Ln z is analytic throughout the domain D consisting of all the points in the complex plane except those on the nonpositive real axis. It is convenient to think of D as the complex plane from which the nonpositive real axis has been cut out. Since f(z) = Ln z is the principal branch of ln z, the nonpositive real axis is referred to as a branch cut for the function. See FIGURE 17.6.3. It is left as exercises to show that the Cauchy–Riemann equations are satisfied throughout this cut plane and that the derivative of Ln z is given by

(9)

for all z in D.

FIGURE 17.6.4 shows w = Ln z as a flow. Note that the vector field is not continuous along the branch cut.

Complex Powers

Inspired by the identity x^a = e^{a ln x} in real variables, we can define complex powers of a complex number. If α is a complex number and z = x + iy, then z^α is defined by

(10)

In general, z^α is multiple-valued since ln z is multiple-valued. However, in the special case when α = n, n = 0, ±1, ±2, . . ., (10) is single-valued since there is only one value for z², z³, z⁻¹, and so on. To see that this is so, suppose α = 2 and z = re^iθ, where θ is any argument of z. Then

If we use Ln z in place of ln z, then (10) gives the principal value of z^α.

EXAMPLE 6 Complex Power

Find the value of i²ⁱ.

SOLUTION

With z = i, arg z = π/2, and α = 2i , it follows from (10) that

where n = 0, ±1, ±2, . . . . Inspection of the equation shows that i²ⁱ is real for every value of n. Since π/2 is the principal argument of z = i, we obtain the principal value of i²ⁱ for n = 0. To four rounded decimal places, this principal value is i²ⁱ = e^−π = 0.0432. ≡

17.6 Exercises Answers to selected odd-numbered problems begin on page ANS-46.

In Problems 1–10, express e^z in the form a + ib.

1. 
2. 
3. 
4. 
5. z = π + πi
6. 
7. z = 1.5 + 2i
8. z = −0.3 + 0.5i
9. z = 5i
10. z = −0.23 − i

In Problems 11 and 12, express the given number in the form a + ib.

11. e^1+5πi/4e^−1−πi/3
12. 

In Problems 13–16, use Definition 17.6.1 to express the given function in the form f(z) = u + iv.

13. f(z) = e^−iz
14. f(z) =
15. f(z) =
16. f(z) = e^1/z

In Problems 17–20, verify the given result.

17. |e^z| = e^x
18. 
19. e^z+πi = e^z−πi
20. (e^z)ⁿ = e^nz, n an integer
21. Show that is nowhere analytic.
22. (a) Use the result in Problem 15 to show that f(z) = is an entire function.

    (b) Verify that u(x, y) = Re() is a harmonic function.

In Problems 23–28, express ln z in the form a + ib.

23. z = −5
24. z = −ei
25. z = −2 + 2i
26. z = 1 + i
27. 
28. 

In Problems 29–34, express Ln z in the form a + ib.

29. z = 6 − 6i
30. z = −e³
31. z = −12 + 5i
32. z = 3 − 4i
33. z = (1 + i)⁵
34. z = (1 + i)⁴

In Problems 35–38, find all values of z satisfying the given equation.

35. e^z = 4i
36. e^1/z = −1
37. e^z−1 = −ie²
38. e^2z + e^z + 1 = 0

In Problems 39–42, find all values of the given quantity.

39. (−i)⁴ⁱ
40. 3^i/π
41. (1 + i)^{(1 + i)}
42. (1 + i)³ⁱ

In Problems 43 and 44, find the principal value of the given quantity. Express answers in the form a + ib.

43. (−1)^(−2i/π)
44. (1 − i)²ⁱ
45. If z₁ = i and z₂ = −1 + i, verify that

    Ln(z₁z₂) ≠ Ln z₁ + Ln z₂.

46. Find two complex numbers z₁ and z₂ such that

    Ln(z₁/z₂) ≠ Ln z₁ − Ln z₂.

47. Determine whether the given statement is true.

    (a) Ln(−1 + i)² = 2 Ln(−1 + i)

    (b) Ln i³ = 3 Ln i

    (c) ln i³ = 3 ln i

48. The laws of exponents hold for complex numbers α and β:

    n an integer.

    However, the last law is not valid if n is a complex number. Verify that (iⁱ)² = i²ⁱ, but (i²)ⁱ ≠ i²ⁱ.

49. For complex numbers z satisfying Re(z) > 0, show that (7) can be written as

    

50. The function given in Problem 49 is analytic.

    (a) Verify that u(x, y) = log_e(x² + y²) is a harmonic function.

    (b) Verify that v(x, y) = tan⁻¹(y/x) is a harmonic function.