Complex Number

269  Chapter 11 Complex Numbers 11.1 Introduction Earlier in the text we discussed about different types of numbers, namely, the natual numbers (the counting numbers), the integers, the rational numbers and the irrational numbers. All these numbers taken together form the set of real numbers. We have also seen how real numbers can be represented by points in a straight line, called the real line or the number line. The set of real numbers is everywhere dense—between every two real numbers there is a real number. Also we have noted that the system of real numbers is complete – every real number corresponds to a point in the real line and conversely. One important fact about real number is “The square of a real number is never negative.” Though the system of real numbers is complete in itself we find it inadequate to find a solution for equation of the type x2 + 4 = 0 which demands a value of x whose square is negative. We cannot find a solution for the equation in the system of real numbers. If we were to make equations like x2 + 4 = 0 sensible naturally we have to extend the system of numbers and we shall be dealing with this extended system of numbers in this chapter. This is done by introducing new set of numbers called complex numbers. It is believed that Cardon used complex numbers in 1546. 11.2 Complex Numbers An ordered pair of real numbers is defined to be a Complex Number. Thus, if a and b are real numbers, the ordered pair written as (a, b) is a complex number. The first number, a, is called the real part and the second number, b, the imaginary part of the complex number. The imaginary part (the second number) is as good a real number as the first part. We shall agree to call it the imaginary part since we have called the first number, the real part, and there is nothing imaginary about it. In line with this interesting terminology, the x-axis is frequently referred to as the real axis, and the y-axis as the imaginary axis. A complex number is usually denoted by a single letter such as z, w etc. If z = (a, b) is a complex number, the real and the imaginary parts of z are respectively denoted by Re(z) and Im(z). Thus Re(z) = a and Im(z) = b. Since a complex number (a, b) is an ordered pair of real numbers a and b, two complex numbers (a, b) and (c, d) are said to be equal if and only if a = c and b = d.

270  BASIC MATHEMATICS : GRADE XI  a) Three Definitions i) The sum of two complex numbers z = (a, b) and w = (c, d) is defined to be a complex number z + w such that z + w = (a + c, b + d). For example, (2, 1) + (3, 4) = (5, 5) ii) The product of two complex numbers z = (a, b) and w = (c, d) is defined to be a complex number zw such that zw = (ac – bd, ad + bc) For example, (2, 1) · (3, 4) = (2·3 – 1·4, 2·4 + 1·3) = (2, 11) iii) If z = (a, b) ≠ (0, 0) be a complex number then the reciprocal of z, denoted by 1 or z–1, is z defined as z–1 =  a b2 , –b  a2 + a2 + b2  This is the multiplicative inverse of z, as zz–1 = (1, 0) For example 1 = 130 , –1  (3, 1) 10  We have w = wz–1 where w and z are complex numbers. z Also if z = (a, b) and w = (c, d) then w = (c, d)  a b2 , –b  z a2 + a2 + b2  = ac + bd , ad – bc   a2 + b2 a2 + b2  b) Some Theorems Theorem I. The operations of addition and multiplication satisfy the commutative, associative and distributive laws. Proof. Let z1 = (a, b), z2 = (c, d) and z3 = (e, f), then i) z1 + z2 = (a, b) + (c, d) = (a + c, b + d) = (c + a, d + b) = z2 + z1  addition of complex numbers is commutative.

  Complex Numbers  271  ii) (z1 + z2) + z3 = (a + c, b + d) + (e, f) = (a + c + e, b + d + f) = (a, b) + (c + e, d + f) = z1 + (z2 + z3)  addition of complex numbers is associative. iii) z1z2 = (a, b) (c, d) = (ac – bd, ad + bc) = (ca – db, da + cb) = (c, d) (a, b) = z2z1 Hence, multiplication of complex numbers is commutative. iv) (z1z2)z3 = (ac – bd, ad + bc) (e, f) = (ace – bde – adf – bcf, acf – bdf + ade + bce) Also, z1(z2z3) = (a, b) (ce – df, cf + de) = (ace – adf – bcf – bde, acf + ade + bce – bdf)  (z1z2)z3 = z1(z2z3) Hence, multiplication of complex numbers is associative. v) z1(z2 + z3) = (a, b) [(c, d) + (e, f)] = (a, b) (c + e, d + f) = (ac + ae – bd – bf, ad + af + bc + be) z1z2 + z1z3 = (a, b) (c, d) + (a, b) (e, f) = (ac – bd, ad + bc) + (ae – bf, af + be) = (ac + ae – bd – bf, ad + af + bc + be)  z1(z2 + z3) = z1z2 + z1z3 Thus, addition and multiplication of complex numbers are distributive. Theorem II. If z = (a, b) be a complex number ii) (a, b) (0, 0) = (0, 0) i) (a, b) + (0, 0) = (a, b) iv) (a, b) = (a, b) iii) (a, b) (1, 0) = (a, b) (1, 0) v) (a, b) + (–a, –b) = (0, 0) Proofs are immediate from definitions, for example i) (a, b) + (0, 0) = (a + 0, b + 0) = (a, b) iv) (a, b) = (a, b) 1 = (a, b) (1, 0) = (a, b) (1, 0) (1, 0)

272  BASIC MATHEMATICS : GRADE XI  We observe from the above theorems that the complex numbers (0, 0) and (1, 0) have properties similar to those of the real numbers 0 and 1, and so we shall take them as the real numbers 0 and 1; and the complex number (–a, –b) acts as the additive inverse of z = (a, b). These complex numbers are therefore denoted by the symbols 0, 1 and –z. Thus by writing (0, 0) = 0, (1, 0) = 1 and (–a, –b) = –z the theorems will be restated as i) z + 0 = z ii) z·0 = 0 iii) z·1 = z iv) z = z 1 v) z + (–z) = z – z = 0 Here, –z is said to be the additive inverse of z. It can be easily seen that for any non-zero real numbers a, b i) (a, 0) + (b, 0) = (a + b, 0) ii) (a, 0) (b, 0) = (ab, 0) iii) (a, 0) = a , 0  (b, 0) b  Hence a complex number of the form (a, 0) has properties similar to the real number a. Hence we shall henceforth identify the complex number (a, 0) with the real number a. In other words, a complex number (a, 0) with the imaginary part 0 is the real number ‘a’ itself. Hence the real number system is a special case of the complex number system. 11.3 The Imaginary Unit We have seen that the complex number (a, 0) plays the same role as the real number a. Hence the real number a can be treated as the complex number (a, 0) with the imaginary part 0. Thus the complex number (1, 0) is same as the real number 1. Naturally, the question arises – “What if the real part is zero ?” The complex number (0, 1) with the real part zero and the imaginary part 1 is denoted by the letter ‘i’ (the iota in the Greek alphabet and the first letter of the word imaginary in the English alphabet) and it is called the imaginary unit. The symbol i was introduced by Euler in 1777. The complex number (0, 1) is denoted by i and is called the imaginary unit. a) Theorems Theorem I. i2 = –1 Proof i2 = i·i = (0, 1) (0, 1) = (0 – 1, 0 + 0) = (–1, 0) = – (1, 0) = –1 Theorem II. If a and b are real numbers then the complex number (a, b) can be written as a + ib where i = (0, 1), and i2 = –1.

  Complex Numbers  273  Proof a + ib = (a, 0) + (0, 1) (b, 0) = (a, 0) + (0, b) = (a, b) So, if a and b are real numbers then a + ib is said to be a complex number where i2 = –1. Now we are in a position to say that there is a solution to an equation like x2 + 4 = 0 x2 + 4 = 0    x2 = –4    x2 = 4i2    x = ± 2i Henceforth the complex number (a, 0) with the imaginary part zero will be written as the real number a, and it is said to be a purely real number (or simply a real number); the complex number (0, b) with the real part zero will be written as bi and it will be said to be a purely imaginary number. The complex number (a, b) will be written as a + ib. Taking i as the imaginary unit such that i2 = –1 and the complex number z = a + ib as a complex number, it is not necessary to memorize the definitions of the sum and the product of two complex numbers. We can use the simple rules of algebra for computing the sum, difference or products of complex numbers. For example, i) (2 + 3i) + (–1 + 4i) = 2 + 3i – 1 + 4i = 2 – 1 + 3i + 4i = 1 + 7i ii) (2 + 3i) – (–1 + 4i) = 2 + 3i + 1 – 4i =3–i iii) (2 + 3i)  (–1 + 4i) = –2 + 8i – 3i + 12i2 = –2 + 5i – 12 = –14 + 5i The set of complex numbers, denoted by the letter C, obviously contains the set of real numbers R, i.e. R  C. b) Powers of i We have defined i as that number whose square is –1, and it is the imaginary unit. Any real multiple of i is an imaginary number. Since i2 = –1, we have i3 = i2·i = (–1)·i = –i i4 = (i2)2 = (–1)2 = 1 i5 = (i4)i = (1)i = i i6 = (i2)3 = (–1)3 = –1 i10 = (i2)5 = (–1)5 = –1

274  BASIC MATHEMATICS : GRADE XI  (i–1) = 1 = i = –i i i2 i–10 = 1 = 1 = –1 etc. i10 –1 Thus any integral power of i is one of the four numbers 1, –1, i or –i. c) Some Examples to Illustrate the Use of i i) –3 = i2 3 = i 3 ii) –5  –9 = 5 i2  9 i2 = i 5  i 9 = i2 5  9 = – 45 = –3 5 Note : It is wrong to write –5  –9 = –5  –9 = 45 = 3 5 as the square root of a negative real number is undefined. iii) a2 + b2 = a2 – (–1)b2 = a2 – i2b2 = (a + ib) (a – ib) d) Properties of Complex Numbers i) A real number cannot be equal to an imaginary number unless each is zero. Let a and b be real numbers such that a = ib. We have a = ib  a2 = i2b2  a2 = –b2  a2 + b2 = 0  a = 0, b = 0 Hence the statement. ii) If a + ib = 0, then a = 0 and b = 0. Here, a + ib = 0  a = –ib  a2 = i2b2  a2 = –b2  a2 + b2 = 0  a = 0 and b = 0 iii) a + ib = c + id iff a = c, b = d. Here, a + ib = c + id  a – c = i(d – b)  (a – c)2 = i2(d – b)2  (a – c)2 = –(d – b)2

  Complex Numbers  275   (a – c)2 + (d – b)2 = 0  a – c = 0, d – b = 0  a = c, b = d Worked Out Examples Example 1 b) (0, 1)7 Evaluate : a) (1, 0)6 Solution: b) (x, y) = (3, 1) . (2, 3) a) (1, 0)6 = (1)6 = 1 b) (0, 1)7 = (i)7 = i4.i2.i = 1.(–1).i = –i d) (x, y) = (2, –1) (–1, 3) Example 2 1. Find the values of x and y if a) (x, y) = (1, 2) + (2, 3) c) (3, 1) = (x, y) + (5, –1) Solutions: (a) (x, y) = (1, 2) + (2, 3) = (3, 5)  x = 3, y = 5 (b) (x, y) = (3, 1) . (2, 3) = (6 – 3, 9 + 2) = (3, 11)  x = 3, y = 11 (c) (3, 1) = (x, y) + (5, –1) = (x + 5, y – 1) Hence x + 5 = 3, y – 1 = 1  x = –2, y = 2 (d) (x, y) = (2, –1) = –12+–93 , –6 + 1  = – 1 , – 1  (–1, 3) 1+9  2 2   x = – 1 , y = – 1 2 2 Example 3 Simplify : 3 –4 + 5 –9 – 4 –25 Solution : 3 –4 + 5 –9 – 4 –25 = 3 4i2 + 5 9i2 – 4 25i2 = 6i + 15i – 20i = i

276  BASIC MATHEMATICS : GRADE XI  Example 4 Find the values of the real numbers x and y if (x + 2) + yi = (3 + i) (1 – 2i) Solution : x + 2 + yi = (3 + i) (1 – 2i) or, x + 2 + yi = 3 – 6i + i – 2i2 or, x + 2 + yi = 5 – 5i Equating real and imaginary parts, x + 2 = 5 and y = –5  x = 3and y = –5 EXERCISE 11.1 1. Evaluate : b) (1, 0)5 c) (0, 1)5 d) (0, 1)11 a) (1, 0)2 2. Find the values of x and y in each of the following a) (x, y) = (2, 3) + (3, 2) b) (x, y) = (2, 1) + (–2, –1) c) (x, y) = (2, 3) – (3, 2) d) (2, 3) = (1, 1) + (x, y) e) (x, y) = (1, 1).(2, 3) f) (x, y) = (1, 1) (3, 4) 3. Simplify : a) –9 + –25 – –36 b) (3 – –4) (2 + –9) c) –16 . –1 d) 3i2 + i3 + 9i4 – i7 e) 1 – 1 + 1 – 1 f) 1 + 1 + 1 + 1 i i2 i3 i4 i i2 i3 i4 4. Prove that : (1 + i)4.1 + 1 4 = 16 i 5. Find the real numbers x and y if b) (x – 1)i + (y + 1) = (1 + i) (4 – 3i) a) x + iy = (2 – 3i) (3 – 2i) 6. Show that 3 + 2i + 3 – 2i is purely a real number 2 – 5i 2 + 5i Answers 1. a) 1 b) 1 c) i d) –i 2. a) x = 5, y = 5 b) x = 0, y = 0 c) x = –1, y = 1 d) x = 1, y = 2 e) x = –1, y = 5 f) x = 7 , y = – 1 25 25 3. a) 2i b) 12 + 5i c) –4 d) 6 e) 0 f) 0 5. a) x = 0, y = –13 b) x = 2, y = 6

  Complex Numbers  277  11.4 Conjugate of a Complex Number Let a and b be real numbers and let z = a + ib be a complex number. The complex number a – ib is called the conjugate of z, and it is denoted by –z (read ‘bar z’) . We note here that two complex numbers are said to be conjugate of one another if their real parts are same and the imaginary parts differ in sign only. Thus 3 + 4i and 3 – 4i are conjugate complex numbers. Each is the conjugate of the other. a) Properties of Conjugates Let z = a + ib and w = c + id be two complex numbers, then we have the following properties i) Re (z) = 1 (z + –z) ii) Im (z) = 1 (z – –z) iii) —z +—w— = –z + w– 2 2i iv) —zw— = –z w– ( )v) —z 2 = —z 2 vi) =z = z vii) z—z = a real number Proofs : i) z + –z = a + ib + a – ib = 2a  1 (z + –z) = a = Re (z) 2 ii) z – –z = a + ib – a + ib = 2ib  1 (z – –z) = b = Im (z) 2i iii) z + w = a + ib + c + id = a + c + i(b + d) = a + c – i(b + d) = a – ib + c – id = –z + w– iv) zw = (a + ib) (c + id) = (ac – bd) + i(ad + bc) = ac – bd – i(ad + bc) = (a – ib) (c – id) = –z · —w v) In the result of (iv) if z = w, we get ( )—z 2 = —z 2 vi) =z = =a==+=i=b= = a – ib = a + ib = z vii) z —z = (a + ib) (a – ib) = a2 – i2b2 = a2 + b2, a real number.

278  BASIC MATHEMATICS : GRADE XI  11.5 Absolute Value of a Complex Number The absolute value of a complex number z = a + ib is defined as the non-negative real number a2 + b2 . It is denoted by | z | or | a + ib |. The absolute value of a complex number is also known as the modulus of the complex number. Examples: | 3 + 4i | = 32 + 42 = 5 | 4 – 3i | = 42 + (–3)2 = 5 a) Properties of Moduli of Complex Numbers i) | z | = | –z | ii) | z | = 0 iff z = 0 z —z = | z |2 iii) | zw | = | z | . | w | for two complex numbers iv)  z  = | z | if | w | ≠ 0 v) w | w | vi) Re ( z ) ≤ | z | and Im ( z ) ≤ | z | Proofs i) Let z = a + ib, then —z = a – ib By definition | z | = a2 + b2 and | —z | = a2 + (–b)2 = a2 + b2  | z | = | —z | ii) Let z = a + ib So, | z | = a2 + b2 | z | = 0  a2 + b2 = 0 or, a2 + b2 = 0 or, a = 0 and b = 0  z=0 Also, z = 0  | z | = 0. iii) Let z = a + ib and w = c + id so that | z | = a2 + b2 and | w | = c2 + d2 and zw = ac – bd + i(ad + bc) hence | zw | = (ac – bd)2 + (ad + bc)2 = a2c2 + b2d2 + a2d2 + b2c2 = (a2 + b2) (c2 + d2) = a2 + b2 c2 + d2 =|z||w|

  Complex Numbers  279  iv) z = a + ib = a + ib  c – id w c + id c + id c – id = (ac + bd) + i(bc – ad) c2 + d2 = ac + bd + bc – ad i c2 + d2 c2 + d2 Now,  z  = ac + bd2 + bc – ad2  w   c2 + d2   c2 + d2  = a2c2 + b2d2 + b2c2 + a2d2 (c2 + d2)2 = (a2 + b2) (c2 + d2) (c2 + d2)2 = a2 + b2 = a2 + b2 = | z | c2 + d2 c2 + d2 | w | v) z —z = (a + ib) (a – ib) = a2 – i2b2 = a2 + b2 = | z |2 vi) Re (z) ≤ | z | and Im(z) ≤ | z | Let z = a + ib Then Re(z) = a, Im(z) = b and | z | = a2 + b2 Since, a2 ≤ a2 + b2   a ≤ a2 + b2   Re(z) ≤ | z | Similarly, b ≤ a2 + b2   Im(z) ≤ | z | b) The Triangle Inequality |z|+|w|≥|z+w| If z and w are complex numbers Proof. Let z = a + ib and w = c + id so that z + w = a + c + i(b + d), then | z | = a2 + b2 , | w | = c2 + d2 Also | z + w | = (a + c)2 + (b + d)2 Now, | z | + | w | ≥ | z + w | will be true if a2 + b2 + c2 + d2 ≥ (a + c)2 + (b + d)2 i.e. a2 + b2 + c2 + d2 + 2 (a2 + b2) (c2 + d2) ≥ (a + c)2 + (b + d)2

280  BASIC MATHEMATICS : GRADE XI  i.e. (a2 + b2) (c2 + d2) ≥ ac + bd i.e. (a2 + b2) (c2 + d2) ≥ a2c2 + b2d2 + 2abcd i.e. a2d2 + b2c2 ≥ 2abcd i.e. a2d2 + b2c2 – 2abcd ≥ 0 i.e. (ad – bc)2 ≥ 0 which is true for all real numbers a, b, c, d. Hence | z | + | w | ≥ | z + w | Second Method | z + w |2 = (z + w)2 ——— — = (z + w) (z + w) (˙.˙ | z |2 = z z ) —— ——— — — = (z + w) ( z + w ) (˙.˙ z + w = z + w ) — —— — = =z z +zw + z w+ww (˙.˙ w = w) ( )= | z |2 + | w |2 + — —— (˙.˙ z + z = 2 Re(z)) zw + z w (˙.˙ Re(z) ≤ | z |) (˙.˙ | zw | = | z | | w | ) ( )— —= — = | z |2 + | w |2 + z w + z w (˙.˙ | w | = | w |) — = | z |2 + | w |2 + 2Re(z w ) — ≤ | z |2 + | w |2 + 2 | z w | — = | z |2 + | w |2 + 2 | z | | w | = | z |2 + | w |2 + 2 | z | | w | = (| z | + | w |) 2  |z+w|≤|z|+|w| i.e. | z | + | w | ≥ | z + w | 11.6 Square Roots of Complex Numbers Let z = a + ib be a complex number whose square roots we want to compute. Let a square root of z = a + ib be x + iy, so that (x + iy)2 = a + ib or, x2 – y2 + 2xyi = a + ib  x2 – y2 = aand 2xy = b Solving these two equations we get two sets of values of x and y, thus giving us two square roots of a + ib.

  Complex Numbers  281  Worked Out Examples (1 – i)2 Example 1 1 + 2i Express the following complex number into a + ib form : Solution : (1 – i)2 = 1 – 2i + i2 1+ 2i 1 + 2i = 1 – 2i – 1 = –2i 1 + 2i 1 + 2i = –2i  1 – 2i = –2i + 4i2 1 + 2i 1 – 2i 1 – 4i2 = –2i – 4 = – 4 – 2 i 1+4 5 5 which is in the form of a + ib where a = – 4 , b = – 2 5 5 Example 2 Find the absolute value of 1 – 2i 2+i Solution :  1 – 2i  = | 1 – 2i |  2+i  |2+i| = (1)2 + (–2)2 = 1 (2)2 + (1)2 Example 3 If x – iy = 3 – 2i , prove that x2 + y2 = 1. 3 + 2i Solution : x – iy = 3 – 2i 3 + 2i or, x – iy = 3 – 2i  3 – 2i = 5 – 12i 3 + 2i 3 – 2i 9+4 or, x – iy = 5 – 12 i 13 13 Equating real and imaginary parts, x = 5 , y = 12 13 13

282  BASIC MATHEMATICS : GRADE XI  Now, x2 + y2 = 153 2 + 1123 2 = 25 + 144 = 1 169 Example 4 If (x + iy) (3 + 2i) = 1 + i, show that x2 + y2 = 2 13 Solution : (x + iy) (3 + 2i) = 1 + i Taking modulus on both sides, | (x + iy) (3 + 2i) | = | 1 + i | or, | x + iy | | 3 + 2i | = | 1 + i | or, x2 + y2 9 + 4 = 1 + 1 or, x2 + y2 = 2 13 Example 5 Find the multiplicative inverse of 2 + 3i 3–i Solution : Let z be the multiplicative inverse of 2 + 3i . Then, 3–i z . 2 + 3i = 1 3–i z = 3–i = 3–i  2 – 3i 2 + 3i 2 + 3i 2 – 3i = 6 – 9i – 2i – 3 = 3 – 11 i 4 + 9 13 13 Example 6 Find the square roots of 3 – 4i. Solution. Let x + iy be the square roots of 3 – 4i so that (x + iy)2 = 3 – 4i or, x2 – y2 + 2xyi = 3 – 4i so x2 – y2 = 3and 2xy = –4 We have (x2 + y2)2 = (x2 – y2)2 + 4x2y2 = 9 + 16 = 25  x2 + y2 = 5 , since x2 + y2 cannot be negative.

  Complex Numbers  283  So we have x2 – y2 = 3 and x2 + y2 = 5 Solving the two equations, we have x = ± 2, y = ± 1 Since xy = –2, x and y are of opposite signs  x = 2, y = –1 or, x = –2, y = 1 Hence the square roots are 2 – i and –2 + i i.e. ± (2 – i) Note 1. If we want the square roots of 3 + 4i, i.e. x and y have the same sign, proceeding as above we see that xy = 2, so x = 2, y = 1 and x = –2, y = –1 are the permissible sets of values of x and y and so the square roots will be ± (2 + i) Note 2. Sometimes the square roots of a complex number can be found by inspection. For example, a) 3 – 4i = 4 – 1 – 4i = 22 + i2 – 2.2.i = (2 – i)2   the square roots of 3 – 4i are ±(2 – i) b) 5 + 12i = 9 – 4 + 12i = 32 + (2i)2 + 2.3.2i = (3 + 2i)2   the square roots of 5 + 12i are ±(3 + 2i) Example 7 Find the square roots of the complex number (8, –15) (0, 1) Solution : (8, –15) = 8 – 15i = 8 – 15i = –15 – 8i (0, 1) 0+i i Let x + iy be the square roots of –15 – 8i. Then, (x + iy)2 = –15 – 8i or, (x2 – y2) + i.2xy = –15 – 8i Equating real and imaginary parts, x2 – y2 = –15 …… (i) 2xy = –8(–ve) …… (ii) Again, (x2 + y2)2 = (x2 – y2)2 + (2xy)2 = (–15)2 + (–8)2 = (17)2   x2 + y2 = 17 …… (iii) Adding (i) and (iii) 2x2 = 2 x=±1 Substituting the value of x in (i) y2 = 16  y = ±4

284  BASIC MATHEMATICS : GRADE XI  Since xy is negative (from (ii)), so the possible values of x and y are as follows: x = 1, y = –4 and x = –1, y = 4  the required square roots of (8, –15) i.e. of –15 – 8i are ±(1 – 4i) (0, 1) EXERCISE 11.2 1. Express each of the following complex numbers in the form of a + ib a) (2 + 5i) + (1 – i) b) (2 + 5i) – (4 – i) c) (2 + 3i) (3 – 2i) d) 3 + 4i e) i f) 1–i 4 – 3i 2+i (1 + i)2 g) 2 – –25 h) 1+i 1 – –16 1–i 2. If z = 2 + 3i and w = 3 – 2i, find —z 2 + —w 2. 3. Compute the absolute values of the following a) 1 + 2i b) 1 + 3 i c) (1 + 2i) (2 + i) d) (3 + 4i) (3 – 4i) e) (1 + i)–1 f) 1+i 1–i 4. If z = 1 + 2i and w = 2 – i, verify that —— — ——  z  z b)  w  = — a) zw = z w w c) | zw | = | z | | w | d) | z + w | ≤ | z | + | w | — z 5. Prove that | z |2 is the multiplicative inverse of z. 6. a) If (3 – 4i) (x + iy) = 3 5 , show that 5x2 + 5y2 = 9. b) If x + iy = a – ib , show that x2 + y2 = 1. a + ib c) If 1 – ix = a – ib, prove that a2 + b2 = 1. 1 + ix d) If x – iy = 1 – i , prove that x2 + y2 = 1. 1 + i 7. If z and w are two complex numbers, prove that | z + w |2 = | z |2 + | w |2 + 2Re(z—w ) 8. Find the multiplicative inverse of the following complex numbers

  Complex Numbers  285  a) (3 + i)2 b) 2 – 5i 6+i 9. If z and w are two complex numbers, prove that | z – w | ≥ | z | – | w | 10. Determine the square roots of the following complex numbers a) 5 + 12i (T.U. 2051) H b) –5 + 12i c) 8 + 6i d) –8 + 6i e) 7 – 24i f) –7 + 24i g) i h) 12 – 5i i) 2 – 36i 2 + 3i j) (5, 12) (3, –4) Answers 1. a) 3 + 4i b) –2 + 6i c) 12 + 5i d) 0 + i e) 1 + 2 i f) – 1 – 1 i g) 22 + 3 i h) 1 + 1 2. 0 5 5 2 2 17 17 2 2i f) 1 3. a) 5 b) 2 c) 5 d) 25 e) 1 2 8. a) 2 – 3 i b) 7 + 32 i 25 50 29 29 10. a) ±(3 + 2i) b) ±(2 + 3i) c) ±(3 + i) d) ±(1 + 3i) e) ±(4 – 3i) f) ±(3 + 4i) g) ± 1 (1 + i) h) ± 1 (5 – i) i) ±(1 – 3i) j) ± 1 (4 + 7i) 2 2 5 11.7 The Cube Roots of Unity An important type of complex number arises when we consider the cube roots of unity. We shall find a number z such that its cube is unity, i.e. to find z such that z3 = 1. We have z3 = 1 or, z3 – 1 = 0 or, (z – 1) (z2 + z + 1) = 0 Either z – 1 = 0 or, z2 + z + 1 = 0  z=1 or –1 ± 1 – 4 2 i.e. z = 1 or –1 ± –3 2 i.e. z = 1, or –1 ± √3i 2

286  BASIC MATHEMATICS : GRADE XI  So the three cube roots of unity are 1, –1 + √3i and –1 – √3i . 2 2 The first one is a real number and the other two are imaginary or complex numbers, and these are often known as the imaginary cube roots of unity, any one of which is denoted by the greek letter  (omega). Properties of the Cube Roots of Unity i) Each imaginary cube root of unity is the square of the other. For, –1 + √3i 2 = 1 – 2√3i + 3i2  2  4 = 1 – 2√3i – 3 = –1 – √3i 4 2 and –1 – √3i 2 = 1 + 2√3i + 3i2  2 4 = –2 + 2√3i = –1 + √3i 4 2 Thus if we write  for any one of the imaginary cube roots, the other will be 2. Hence the three cube roots of unity are 1, , 2. ii) The product of the two imaginary cube roots of unity is equal to 1. For ·2 = –1 + √3i  –1 –√3i = 1 – 3i2 = 4 = 1 2 2 4 4 As a direct consequence, we have  3 = ·2 = 1, 3n = 1 for any integral value of n. Also 2 = 1 and  = 1  2 Thus, one imaginary cube root of unity is the reciprocal of the other. iii) The sum of the three cube roots of unity is zero. For 1 +  + 2 = 1 + –1 + √3i + –1 – √3i 2 2 = 2 – 1 + √3i – 1 – √3i = 0 2 Thus, we have two important relations 1 +  + 2 = 0 and 3 = 1 It may be noted here that any integral power of  will reduce to 1,  or 2. For examples, 4 = 3· = , 5 = 3·2 = 2    6 = 1, 20 = 18·2 = (3)62 = 2    –1 = 1 = 3 = 2 –10 = 1 = 1 = 1 = 2, etc.   10 9· 

  Complex Numbers  287  Worked Out Examples Example 1 Show that (1 –  + 2)4 + (1 +  – 2)4 = –16 Solution : Since 1 +  + 2 = 0, we have 1 +  = – 2 and 1 + 2 = –  L.H.S. = (1 –  + 2)4 + (1 +  – 2)4 = (–2)4 + (–22)4 = 164 + 168 = 16(4 + 8) = 16( + 2) = 16(–1) = –16 Example 2 a + b + c2 =  Prove that : b + c + a2 Solution : a + b + c2 b + c + a2 = a3 + b + c2 (˙.˙ 3 = 1) b + c + a2 = (a2 + b + c) b + c + a2 = EXERCISE 11.3 1. If  be a complex cube root of unity, show that a) (1 +  – 2)3 – (1 –  + 2)3 = 0 b) (2 +  + 2)3 + (1 + – 2)8 – (1 – 3 + 2)4 = 1 c) (1 –  + 2)4 ( 1 +  – 2)4 = 256 d) (1 – ) (1 – 2) (1 – 4) (1 – 8) = 9. e) a+ b + c2 + a+ b + c2 = –1 a + b2 +c a2 + b + c

288  BASIC MATHEMATICS : GRADE XI  2. a) If  = 1 ( – 1 + –3 ) and  = 1 ( –1 – –3) , show that 4 + 22 + 4 = 0 2 2 b) If  and  are the complex cube roots of unity, prove that 4 + 4 + –1–1 = 0 3. Prove that –1 + –3 9 – 1 – –3 6 –1 + –3 4 –1 – –3 4 2   2  2   2  a) + = 2 b) + = –1 4. If x = a + b, y = a + b2, z = a2 + b show that i) x + y + z = 0 ii) xyz = a3 + b3 iii) x3 + y3 + z3 = 3(a3 + b3) 11.8 Geometrical Representation of Complex Numbers A complex number, denoted by z, is defined as an ordered pair (x, y) of real numbers. Since every ordered pair of real numbers can be represented by a point in the Cartesian plane, the complex number z = (x, y) can be identified with a point P in the Cartesian plane with coordinates (x, y); and conversely every point in the plane represents a complex number. a) The Complex Plane When complex numbers are represented by points in the Cartesian plane, the plane is called the complex plane; and the set of such points form an Argand diagram Figure (a). The idea of expressing complex numbers geometrically was formulated by Argand (French) and Gauss (German). The name complex number is due to Gauss. In figure (b), we have z = (x, y) = x + iy and | z | = OP = x2 + y2 is the modulus of the complex number z. In other words, the modulus of a complex number is the distance of the point representing the complex number from the origin in the complex plane. 4i 4i 3i 3i P z = x + yi Complex 2i |y| 2i Plane |z| i i –4 –3 –2 –1 0 1234 –1 0 12 3 4 –i |x| –i –2i | z |2 = x2 + y2 –3i –2i –4i Pythagorean Theorem –3i (b) –4i (a)

  Complex Numbers  289  b) The Unit Circle 1.5i Unit Circle 1i |z|=1 Of particular importance is the unit circle. This is the set of all points in the complex numbers with 0.5 absolute value 1. Of course, 1 is the absolute value of both 1 and –1, but it’s also the absolute value of both i –1.5 –1 –0.5 0 0.5 1 1.5 and –i since they’re both one unit away from 0 on the imaginary axis. The unit circle is the circle of radius 1 –0.5i centered at 0. It includes all complex numbers of absolute value 1, so it has the equation | z | = 1. –1i –1.5i c) The Triangle Inequality An important relation relating complex numbers 3i z+w with addition of absolute values is the triangle • |w| inequality. If z and w are any two complex numbers, • 2i then |z+w| |z+w|≤|z|+|w| i • |w| One can see this from the parallelogram rule for –2 –1 |z| addition. Consider the triangle whose vertices are 0, z and z + w. One side of the triangle, the one from 0 to z 0• 1 2 3 4 Triangle Inequality –i | z + w | – | z | + | w | + w has length | z + w |. A second side of the triangle, the one from 0 to z, has length | z |. And the third side of the triangle, the one from z to z + w, is parallel and equal to the line from 0 to w, and therefore has length | w |. Now, in any triangle, any one side is less than or equal to the sum of the other two sides, and, therefore, we have the triangle inequality displayed aside. 11.9 Polar Form of a Complex Number Let z = (x, y) = x + iy be a complex number. It can be Y P (x, y) represented by a point P in the complex plane with cartesian (0, y) coordinates (x, y). Let  be the angle in the standard position with OP as its terminal arm, and r the length of the line segment OP. So, we have  X x = r cos , O (x, 0) y = r sin  Thus, the complex number, z = x + iy, may be written in the following trigonometric form (polar form) z = r(cos  + i sin ) where r = x2 + y2 and tan  = y , x ≠ 0. x

290  BASIC MATHEMATICS : GRADE XI  Actually r = | z | is the modulus of z and the angle  is called the amplitude or the argument of z and is written as amp (z) or arg (z). Since sin  and cos  are both periodic with a period 2π or 360°, the complex number z = x + iy = r (cos  + i sin ) may be written in the general form as z = r[cos ( + 2nπ) + i sin ( + 2nπ], if  is in radians, = r[cos ( + n 360°) + i sin ( + n 360°)], if  is in degrees, n is an integer 11.10 Products and Quotients in Polar Form One of the important uses of the polar (or trigonometric) form of complex numbers is in the computation of products of complex numbers. It provides us a quick and efficient method for calculating the product and hence that of the quotient. Other important uses include the computation of powers and roots of complex numbers. Theorem The product and quotient of two complex numbers z1 = r1 (cos 1 + i sin 1) and z2 = r2 (cos 2 + i sin 2) are given by z1z2 = r1r2 [cos (1 + 2) + i sin (1 + 2)] and z1 = r1 [cos (1 – 2) + i sin (1 – 2)] z2 r2 Proof: z1z2 = r1 (cos 1 + i sin 1). r2 (cos 2 + i sin 2) = r1r2 {(cos 1 cos 2 – sin 1 sin 2) + i (sin 1 cos 2 + cos 1 sin 2)} = r1r2 {cos (1 + 2) + i sin (1 + 2)], thus proving the first part. To prove the second part, we have z1 = r1 (cos 1 + i sin 1) z2 r2 (cos 2 + i sin 2) = r1 (cos 1 + i sin 1) (cos 2 – i sin 2) r2 (cos 2 + i sin 2) (cos 2 – i sin 2) = r1 (cos 1 + i sin 1) [cos (–2) + i sin (–2)] r2 (cos2 2 + sin2 2) = r1 [ cos (1 – 2) + i sin (1 – 2)]. r2 If we now introduce a third factor in the product considered, we have r1 (cos 1 + i sin 1) . r2 (cos 2 + i sin 2) . r3 (cos 3 + i sin 3)

  Complex Numbers  291  = r1r2r3 [cos (1 + 2) + i sin (1 + 2) ]. (cos 3 + i sin 3) = r1r2r3 [cos (1 + 2 + 3) + i sin (1 + 2 + 3)], and we obtain, in a similar manner, the product of four or more complex numbers. In case of n such numbers, we obtain the formula: r1 (cos1 + i sin 1). r2 (cos 2 + i sin 2)..... rn(cos n + i sin n) = r1r2r3.... rn [cos (1 + 2 +....+ n) + i sin (1 + 2 + ... + n)]. This formula for the product of n complex numbers gives us the value of the nth power of a complex number which can be stated as a theorem known as De Moivre's Theorem. From the above results, we have the following relations. z1 = r1(cos 1 + i sin 1) and z2 = r2(cos 2 + i sin 2) give | z1 | = r1 , | z2 | = r2 and amp(z1) = 1, amp(z2) = 2 Now, z1z2 = r1r2 {cos (1 + 2) + i sin (1 + 2)} gives | z1z2 | = r1r2 = | z1 | | z2 | and amp(z1z2) = 1 + 2 = amp(z1) + amp(z2) Again, z1 = r1 {cos (1 – 2) + i sin (1 – 2)} z2 r2 gives,  z1  = r1 = | z1 | z2 r2 | z2 | and ampzz12 = 1 – 2 = amp(z1) – amp(z2) The product and the quotient of two complex numbers can be represented as follows: If the complex numbers z1 = r1(cos 1 + i sin 1) and z2 = r2 (cos 2 + i sin 2) with magnitudes r1, r2 and the amplitudes 1, 2 be represented in the complex plane by the points P(r1, 1) and Q(r2, 2) then the complex number z1z2 with magnitude r1r2 and the amplitude 1 + 2 will be represented by the point R such that XOR = 1 + 2 and OR = r1r2. YY RQ r1r2 Q r1 P r2 P 1 r1 R 2 r1/r2 2 r1 X X 1 O 1 – 2 O Fig. (ii) Fig. (i)

292  BASIC MATHEMATICS : GRADE XI  In the same way, the complex number z1 with magnitude r1 and amplitude 1 – 2 will be z2 r2 represented by the point R such that XOR = 1 – 2 and OR = r1 . r2 11.11 Integral Powers and Roots of Complex Numbers The product of a complex number z by itself, i.e., z. z is denoted by z2 and is called the square or second power of z. The cube or the third power of z, denoted by z3, is defined by z3 = z2. z. In general, for any positive integer n, the nth power of a complex number z is defined by zn = zn –1. z and z° = 1. Obviously, the nth power of a complex number is also a complex number. Let us denote it by w, so that zn = w. Conversely, if w ≠ 0, and if n is a positive integer, then any complex number z whose nth power is w, is known as the nth root of w. In other words, any complex number z such that zn = w, w ≠ 0, n = 1, 2, 3,..... is known as the nth root of w. The nth root of w is usually denoted by w1/n or nw. The computation of the nth power and the nth root of a complex number may be carried out with comparative case with the help of a theorem known as De Moivre's theorem, which uses the polar (or trigonometric) form of a complex number. The method of finding the nth roots of a complex number will be given after De-Moivre’s Theorem. De Moivre's Theorem If n is any positive integer, [r (cos  + i sin )]n = rn (cos n + i sin n). Proof. Obviously, when n = 1, [r (cos + i sin )]1 = r (cos  + i sin ). For n = 2, we have [r (cos  + i sin )]2 = r2 (cos2  + 2i cos  sin  + i2 sin2 ) = r2 (cos2  – sin2  + i 2 sin  cos ) = r2 (cos 2 + i sin 2). Thus the theorem is true for n = 1 and n = 2. We prove the theorem by induction.

  Complex Numbers  293  Let us assume that the theorem is true for some positive integer k. By assumption, [r (cos  + i sin )]k = rk (cos k + i sin k). Multiplying both sides by r (cos  + i sin ), we get [r (cos  + i sin )]k + 1 = rk + 1 (cos k + i sin k) (cos  + i sin) = rk + 1 [cos (k + ) + i sin ( k + )] = rk + 1 [cos (k + 1)  + sin (k + 1) ], which shows that the theorem is true for n = k + 1 whenever it is true for n = k. But we know that it is true for n = 1 and n = 2. When it is true for n = 2, the above proof shows that it is true for n = 3. Continuing this way, we come to the conclusion that the theorem is true for every positive integer n. This completes the proof. In fact, the theorem is true not only for a positive integer but also for any real number n. We shall assume this without proof. Let us now apply this theorem to compute the integral powers of complex numbers. nth root of a given complex number Let z = r (cos  + i sin ) be the given complex number whose nth root is required. Let w = R(cos ø + i sin ø) be the nth root of the complex number z. Then, wn = z  {R(cos ø + i sin ø)}n = r (cos  + i sin )  Rn(cos nø + i sin nø) = r (cos  + isin ) Since the two complex numbers are equal, so Rn = r  R = r1/n = n r and cos nø + i sin nø = cos  + i sin   cos nø = cos  and sin nø = sin   nø = 2k.π +  or, k.360° +   ø = k.360° +  n  w = R(cos ø + i sin ø) = n r cos k.360° +  + i sin k.360° +  n n   For k = 0, 1, 2, 3, ……, n – 1, we get n different roots of z. For other integral value of k will give the root obtained earlier i.e. roots will be repeated. Thus the kth root of z denoted by zk is given by zk = k r cos k.360° +  + i sin k.360° +  k = 0, 1, 1, 2, 3, ……, n – 1 n n  

294  BASIC MATHEMATICS : GRADE XI  Worked Out Examples Example 1. Express 2 + 2 √3 i in the polar form. Represent the complex number in the complex plane. Solution. Let 2 + 2√3 i = r (cos  + i sin ). Then by the definition (4, /3) of the equality of two complex numbers 4 r cos  = 2 /3 and r sin  = 2√3. O This gives r2 = 4 + 12 = 16, or r = 4 (positive value only). Also, cos  = 1 and sin  = √3 , 2 2 giving  = 60° Hence 2 + 2√3i = 4 (cos 60° + i sin 60°). Example 2 Express 6 (cos 30° + i sin 30°) in the form x + iy. Solution Let 6 (cos 30° + i sin 30°) = x + iy. Equating real and imaginary parts, we have x = 6 cos 30° and y = 6 sin 30° = 6. √3 = 3√3 = 6. 1 = 3. 2 2 Hence 6 (cos 30° + i sin 30°) = 3√3 + 3i. Example 3 Find the value of 2(cos 70° + i sin 70°) cos 10° + i sin 10° Solution : 2 (cos 70° + i sin 70°) cos 10° + i sin 10° = 2 {cos (70° – 10°) + i sin (70° – 10)} = 2{cos 60° + i sin 60°} = 212 + i 3 2 =1+i 3

  Complex Numbers  295  Example 4 Prove that cos 8 + i sin 8 = cos 2 + i sin 2 (cos  + i sin )6 Solution : cos 8 + i sin 8 (cos  + i sin )6 = (cos  + i sin )8 (using De-Moivre’s theorem) (cos  + i sin )6 = (cos  + i sin )2 = cos 2 + i sin 2 Example 5 Using De-Moivre’s theorem, evaluate (1 – 3 i)6. Solution : Let 1 – 3 i = r (cos  + i sin ) Equating real and imaginary parts, r cos  (x) = 1, r sin  (y) = – 3 r = (1)2 + (– 3)2 = 2 tan  = –1 3 = – 3   = 300°  1 – 3 i = 2(cos 300° + i sin 300°) Now, (1 – 3 i)6 = { 2 cos (300° + i sin 300°)}6 = 26 {cos 1800° + isin 1800°} = 26 {cos 0° + isin 0°} = 26 = 64 Example 6 Show that z3 = 1, if z = – 1 + √3 i. 2 2 Solution. Let z = x + iy = r (cos  + i sin ) = – 1 + √3 i. 2 2 Here, x = – 1 and y= 3 . 2 2 Then, r = 1 + 3 = 1. 4 4

296  BASIC MATHEMATICS : GRADE XI  √3 ians txhiessneecgoantdivqeuaanddrant tan  = 2 = – √3, or  = 120° y is positive  –12 Thus, z = cos 120° + i sin 120°. Again, z3 = (cos 120° + i sin 120°)3 = cos 3 (120°) + i sin 3 (120°) = cos 360° + i sin 360° = 1. Another important use of De Moivre's theorem is in the computation of the roots of complex numbers. We shall illustrate the method with some examples. Example 7 Find the square roots of 2 + 2√3i Solution : Let z = 2 + 2√3i Here, x = 2 and y = 2 3 To write z in polar form we note that r = 22 + (2√3)2 = 4 and tan  = 2√3 = √3 or,  = 60° 2  in polar form z = 4(cos 60° + i sin 60°) = 4[cos (60° + n 360°) + i sin (60° + n 360°)] in general polar form  z = z1/2 = 4 [cos (60° + n 360°) + i sin (60° + n 360°)]1/2 = 2[ cos (30° + n 180°) + i sin (30° + n 180°)] where n = 0, 1 (Using De Moivre’s theorem) When n = 0, z1/2 = 2(cos 30° + i sin 30°) = √3 + i When n = 1, z1/2 = 2(cos 210° + i sin 210°) = –√3 – i For n = 2, 3, … the values obtained above will repeat, as the angles differ by multiples of 360°. Note: The square roots can be obtained directly using the formula given in Art. 11.11 Example 8 Find the cube roots of 1 using De Moivre’s theorem. Solution : Let z3 = 1 so that we need to find z. Writing in polar form

  Complex Numbers  297  z3 = 1 = 1 + 0i = cos 0 + i sin 0 [r = 1,  = 0 ] = cos 2nπ + i sin 2nπ, in general polar form.  z = (cos 2nπ + i sin 2nπ)1/3 = cos 2nπ + i sin 2nπ , where n = 0, 1, 2. 3 3 When n = 0, z = cos 0 + i sin 0 = 1 When n = 1, z = cos 2π + i sin 2π = – 1 + √3 i 3 3 2 2 When n = 2, z = cos 4π + i sin 4π = – 1 – √3 i 3 3 2 2 Hence 1, –1 + √3i , –1 – √3i are the three cube roots of 1. 2 2 [Remember that we have computed these cube roots earlier without using De-Moivre’s Theorem] Note: The cube roots can be obtained directly using the formula given in Art. 11.11 Example 9 If z be a complex number, prove that  1  = | 1 | and amp1z = – amp(z)  z  z Solution : Let z = r (cos  + i sin ) Then, |z| =r and amp (z) =  1 = r(cos  1 i sin ) z + = r(cos  1 i sin )  cos  – isin  + (cos  – i sin ) = cos  – isin  r(cos2 + sin2) = 1 {cos (–) + i sin (–)} r Now, 1 = 1 = | 1 | z r z Again, amp 1 = – = – amp (z) z Example 10 If z = cos  + i sin , prove that zn + 1 = 2 cos n zn Solution :

298  BASIC MATHEMATICS : GRADE XI  z = cos  + i sin  Then, zn = (cos  + i sin )n = cos n + i sin n and 1 = z–n = cos n – i sin n Now, zn zn + 1 = cos n + i sin n + cos n – i sin n zn = 2 cos n EXERCISE 11.4 1. Express the following complex numbers in the polar form: (a) 2 + 2i (b) √3 + i (c) 2i (d) i – √3 (e) 1 – i (f) (1 + i) (g) i (h) 1–i (1 – i) (1 + i) 1+i 2. Express the following in the form x + iy: (b) 3 (cos 120° + i sin 120°) (a) 3 (cos 60° + i sin 60°) (d) 2 cos (–45°) + i 2 sin (–45°). (c) 2 (cos 150° + i sin 150°) 3. Simplify : a) (cos 32° + i sin 32°) (cos 13° + i sin 13°) b) (sin 40° + i cos 40°) (cos 40° + i sin 40°) c) cos 80° + i sin 80° cos 20° + i sin 20° d) (cos 3 + i sin 3) (cos  – i sin ) (cos  + i sin )2 4. Apply De Moivre’s Theorem to compute b) [ 3(cos 120° + i sin 120°)]3 a) [ 2(cos 15° + i sin 15°)]6 d) [cos 9° + i sin 9°]40 c) (cos 18° + i sin 18°)5 f) (–1 + i)14 e) (1 + i)20 g) 21 + i √23 7 i) i2 5. Using De–Moivre’s Theorem, find the square roots of a) 4 + 4√3i b) –1 + √3i c) –2 – 2√3i d) 2i e) –i 6. Determine the cube roots of –1. 7. Solve the following equations a) z4 = 1 b) z6 = 1 c) z4 + 1 = 0 d) z3 = 8i

  Complex Numbers  299  8. Find the fourth roots of – 1 + i √3 . 2 2 —— 9. If z be the conjugate of the complex number z, prove that Arg( z ) = 2π – Arg(z). 10. If z = cos  + i sin , prove that zn – 1 = 2 sin ni. zn Answers 1. (a) 2√2 (cos 45° + i sin 45°) (b) 2(cos 30° + i sin 30°) (c) 2(cos 90° + i sin 90°) (f) (cos 90° + i sin 90°) (d) 2(cos 150° + i sin 150°) (e) √2 (cos 315° + i sin 315°) (d) √2 – i√2 (g) 1 (cos 45° + i sin 45°) (h) cos 315° + i sin 315° √2 2. (a) 3 + i 3√3 (b) – 3 + i 3√3 (c) –√3 + i 2 2 2 2 3. a) 1 + 1 b) i c) 1 + 3 d) 1 2 2i 2 2i 4. a) 64i b) 27 c) i d) 1 e) –210 f) 27i g) 1 + i √3 h) –1 2 2 5. a) ± (√6 + i√2) b) ± 1 (1 + i√3) c) ±(–1 + i√3) d) ±(1 + i) e) ± 1 –i √2 2 6. –1, 1 (1 + i√3) , 1 (1 – i√3) 2 2 7. a) ±1, ±i b) ±1, 1 (1 ± i√3) , 1 (–1 ± i√3) c) ± 1 +i , ± 1 –i 2 2 2 2 d) 3 + i, – 3 + i, –2i 8. ± √23 + 1 i , ± 21 – √3 i 2 2 Additional Questions 1. Define a complex number. If z = 2 + i , find the real and the imaginary parts of 3 – i (i) 1+z (ii) z + 1 . 1–z z 2. If z1 = 3 – 2i and z2 = 4 + 3i, find z1–1, z2–1, z1 and z2 . z2 z1 3. Find the multiplicative inverse of : (i) (2 + i)2 (ii) 3 + 4i 4 – 5i 4. If a + ib = 1 + i , prove that a2 + b2 = 1. 1 – i 5. If x + iy = a + ib, prove that x – iy = a – ib.

300  BASIC MATHEMATICS : GRADE XI  6. If x + iy = u + iv , prove that x2 + y2 = 1. u – iv Solution: x + iy = u + iv u – iv or, x + iy = u + iv  u + iv u – iv u + iv or, x + iy = (u + iv)2 u2 + v2 x + iy = u2 – v2 + i 2uv u2 + v2 u2 + v2 Equating real and imaginary parts, x = u2 – v2 and y = 2uv u2 + v2 u2 + v2 Now, x2 + y2 = u2 – v2 2 +  2uv  2 u2 + v2 u2 + v2 = (u2 – v2)2 + 4u2v2 = (u2 + v2)2 = 1 (u2 + v2)2 (u2 + v2)2 7. State De' Moivre’s Theorem. Use it to find the cube roots of unity. (HSEB 2056, 2058) 8. State De-Moivre’s theorem and use it to find the cube roots of unity. Verify that the sum of the three cube roots of unity is zero. (T.U. 2052) 9. Simplify (Use De-Moivre’s theorem) i) (cos 40° + i sin 40°) (cos 50° + i sin 50°) Solution : (cos 40° + i sin 40°) (cos 50° + i sin 50°) = cos (40° + 50°) + i sin (40° + 50°) = cos 90° + i sin 90° = 0 + i.1 = i ii) (cos 75° + i sin 75°) (cos 15° – i sin 15°) iii) (cos  + i sin )4 (cos  – i sin )5 10. If  be one of the complex cube roots of unity, prove that (i) 3 = 1,(ii) 1 +  + 2 = 0 11. Prove that each complex cube root of unity is the reciprocal of the other. 12. Express a complex number in a polar form. State De-Moivre’s theorem. Using De- Moivre’s theorem find the cube roots of unity. If = –1+ 3i be a complex cube root of 2 unity, prove that 2 = –1 – 3i 2

  Complex Numbers  301  13. What do you mean by a complex number ? Express it in the polar form. Find the square root of 2 – 36i (T.U. 2053) 2 + 3i 14. Prove that : i) –1 + –3 6 + –1 – –3 9 = 2 ii) –1 + –3 5 + –1 – –3 5= –1 2   2  2   2  15. Show that a + ib will be real if ad = bc. c + id Solution: a + ib = a + ib  c – id c + id c + id c – id = (ac + bd) + i(bc – ad) c2 + d2 = ac + bd + i bc – ad c2 + d2 c2 + d2 The given complex number wil be real if bc – ad = 0 c2 + d2 i.e. ad = bc 16. If z and w are two complex numbers, prove that a) | z – w | ≤ | z | + | w | b) | z + w | ≥ | z | – | w | 17. Prove that –1 + –3 n –1 – –3 n 2   2  + = 2, if n is a multiple of 3 if n is any other integer = –1, Answers 1. i) 1, 2 ii) 3 , – 1 2. 3 + 2 i , 4 – 3 i, 6 – 17 i, 6 + 17 i 2 2 13 13 25 25 25 25 13 13 3. i) 3 – 4 i ii) – 8 – 31 I 9. ii) 1 + √3 i iii) cos 9 + i sin 9 25 25 25 25 2 2 13. i) 1 +i , –1 – i ii) –1 + i , 1 –i 13. ±(3i – 1) 2 2 2 2 •