Module 22 - Complex Analysis 2 (self Study)

SCHOOL OF MATHEMATICS MATHEMATICS FOR PART I ENGINEERING Self Study Course

MODULE 22

COMPLEX NUMBERS II

Module Topics 1. Relationships between trigonometric and hyperbolic functions 2. Logarithm of a complex number 3. De Moivre’s theorem 4. The nth roots of a complex number 5. Simple loci in the complex plane

A:

Work Scheme based on JAMES (THIRD EDITION)

1. The work in this module follows on from Module 5, studied in semester 1. Refresh your memory about that material, if necessary. Turn to p.172 of J. and start section 3.2.7. Using Euler’s formula and the definitions of the hyperbolic functions cosh x and sinh x, which all appear on the Formula Sheet or in the Data Book, it is not difficult to derive equations (3.13) and (3.14) which connect the hyperbolic and trigonometric functions. You need not remember these relationships but you are expected to know how to derive them using results which appear on the Formula Sheet or in the Data Book. 2. The method for obtaining cos z, sin z etc. is shown on p.173. Study Example 3.14(a),(b),(d). In part (b), J. uses Osborne’s rule to write down the expansion of sinh(A + B) from knowledge of the corresponding trigonometric identity for sin(A + B). Another way to proceed is to use formula (3.14b) with x replaced by 3 + 4j, in which case it is the trigonometric identity for sin(C + D) which is required to derive the answer. When θ is real you know −1 ≤ cos θ ≤ 1. When θ is complex the situation is different, however, and it follows from equation (3.14a) and the range of cosh x that the values of cos θ can lie outside the interval [−1, 1]. Part (d) shows there are infinitely many solutions of the equation cos z = 2. 3. Study section 3.2.8 on the logarithm of a complex number. Writing w = u + jv, expressions for u and v are found on p.176, and hence the final result on p.176 should read w = ln |z| + j arg z + j 2nπ,

n = 0, ±1, ±2, . . .

For a particular z there an infinite number of values of w = ln z, one for each value of n, and this makes ln z very different from the standard functions of a real variable you are used to working with. When the argument of z takes its principal value, i.e. lies between −π and +π, then it is usual to call (3.15) the principal value of ln z, and this is sometimes denoted by Ln z. Work through Example 3.15. J. obtains only the principal value of ln(−3 + 4j). Clearly multiples of 2π can be added to arg z and so the general answer should be written ln(−3 + 4j) = ln 5 + j(2.214 + 2nπ), –1–

n = 0, ±1, ±2, . . . .

In obtaining solutions it is useful to observe that the exponential form of the complex number z can be used to calculate ln z. You know that a given complex number z can be written z = rejθ , where θ is the principal value of the argument. This can also be expressed z = rej(θ+2nπ) , n = 0, ±1, ±2, . . ., and taking logarithms of each side and using the usual properties of logarithms and exponentials, we obtain ln z = ln r + j(θ + 2nπ),

n = 0, ±1, ±2, . . . ,

the expression obtained in J. The latter provides the most straightforward way of obtaining the values of ln z. ***Do Exercises 20(a),(b), 21(a)(harder) on p.177*** ***Do Exercise A:

Obtain all values of

(a) ln(5 + 12j),

√ (b) ln(− 12 − j 12 3).

4. Read the introduction to section 3.3 on p.177. Then study the start of section 3.3.1 on de Moivre’s theorem. Since z = rejθ = r(cos θ + j sin θ) it follows immediately that theorem (3.16) can be written {r(cos θ + j sin θ)}n = rn (cos nθ + j sin nθ), which is the usual form in which the theorem is stated. On replacing θ in the above expression by −θ, it can be shown that the theorem also holds when the addition signs on both sides of the result are replaced by subtraction signs. The latter comment is required to understand Example 3.16, which you should work through. Another way of obtaining the answer, which does not use de Moivre’s theorem, is (1 − j)12 = {(1 − j)2 }6 = {1 − 2j + j 2 }6 = (1 − 2j − 1)6 = (−2j)6 = (−2)6 (j 2 )3 = 64(−1)3 = −64. It should be pointed out that J. shows that, starting with the exponential form of z, the usual properties of indices and exponentials imply theorem (3.16) holds for all n. In many texts, however, de Moivre’s theorem is often considered first for integer n and in that situation the theorem is proved by direct expansion and the use of trigonometric identities. 5. Study the theory below the horizontal line on p.178 and at the top of p.179. Using de Moivre’s theorem to calculate the roots of a complex number is very important. Equations (3.17) and (3.18) are two equivalent expressions for the same result. Given a complex number z you first determine its modulus, r, and principal argument, θ. To find the roots of z it is crucial to add to this value of θ integral multiples of 2π. The nth roots of z are then found using formula (3.17) or (3.18). Study Example 3.17. The given z is expressed in polar form and then z 1/2 and z 1/3 are determined in parts (a) and (b) respectively. The solution for z 1/2 when k = 1 contains an argument lying outside the interval (−π, π]. However, this solution can be rewritten      5 5 . 2−1/4 cos − π + j sin − π 8 8 In a similar way the solution in part (b) when k = 2 should be expressed      5 5 . 2−1/6 cos − π + j sin − π 12 12 6. J. states after Example 3.17 on p.180 that the formula 3.17 is easily extended to deal with z p , where p = m/n is a rational. The result is used in Example 3.18 which you should work through. The three values here all involve arguments lying in (−π, π]. –2–

7. Study Example 3.19, which shows how to find the solution of a quadratic equation with complex coefficients. ***Do Exercises 27, 28(a),(d), 30 on p.184*** 8. Study section 3.3.2 on p.182, leading to the formulae (3.20a) and (3.20b). 9. Read the introduction to section 3.4 on loci in the complex plane. Move to section 3.4.1 and study Example 3.22. The solution to part (a) is straightforward. In part (b) the argument of a complex quantity is stated, so defining u = z − 1 − j we are given arg u = π/4. Thus we can write u = rejπ/4 where r is unknown. Introducing the definition of u, and rearranging, then leads to z = 1 + j + rejπ/4 . Part (c) is solved algebraically, and then looked at geometrically (using the fact that |z − a| denotes the distance of the complex number z from a). Finally part (d) is solved algebraically. 10. Study the introduction to section 3.4.2 at the bottom of p.186, and then work through Examples 3.23, 3.24 and 3.25. ***Do Exercises 35(a),(c), 37(b) on p.189***

B:

Work Scheme based on STROUD (FIFTH EDITION)

Turn to p.494 of S. and work through frames 39–50 of Programme 3. Go back to p.440 and work through frames 54–59 in Programme 1. In expressing z in exponential form the principal value of the argument should be used. However, any integral multiple of 2π can be added without altering z and so, in general, z = rej(θ+2nπ) , n = 0, ±1, ±2, . . .. With this expression it follows that ln z = ln r + j(θ + 2nπ), n = 0, ±1, ±2, . . ., and so for a given value of z there are an infinite number of values of ln z, one for each value of n. Finally turn to p.456 in Programme 2 and work through frames 23–39 and 49–56. Note that if you complete the square in the quadratic equations which appear as the answers in frames 53 and 56 you can show these loci are circles, with centres and radii which can be identified. For example, in frame 56 the solution can be written (x−1)2 +y 2 = 25 = 52 , which denotes a circle of radius 5 and centre at x = 1, y = 0 .

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Specimen Test 22 1.

2.

Find the real and imaginary parts of sin

π

 +j .

2 [You may use without proof the identities cos(jx) = cosh x and sin(jx) = j sinh x. ]

Write −3 in exponential form and hence, or otherwise, deduce the real and imaginary parts of ln(−3).

√ 3. (i) Express z = 4 2(1 − j) in exponential form and hence find the cube roots of z, leaving your answers in exponential form. (ii) Write your roots in the form x + jy . (iii) Display the roots on an Argand diagram.

4.

Given that z = cos θ + j sin θ, show by considering z n and z −n and using de Moivre’s theorem, that 2 cos nθ = z n + z −n

5.

and

2j sin nθ = z n − z −n

for any integer n .

z + 2j = 2. Find the locus of the point z in the Argand diagram which satisfies the equation z−j

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