F08 - Further Complex Numbers - Lecture Notes (Teacher s Version)

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Teacher’s Version) Page 1 of 24 National Junior College 2016 – 2017 H2 Further Mathematics Topic F8: Further Complex Numbers Key Questions to Answer: Complex Numbers in Polar & Exponential Form 1. How do we interpret geometrically the effects of conjugating a complex number, and adding/subtracting/multiplying/dividing two complex numbers? - What is the geometrical effect of multiplication by i? 2. What is de Moivre’s Theorem? - How do we apply de Moivre’s Theorem to find the powers and nth roots of a complex number, and to derive trigonometric identities? Complex Loci How do we sketch the loci of simple equations and inequalities involving a complex variable in the Argand diagram such as   , and arg ? z c r z a z b z a         §1 Geometrical Effects of Operations on Complex Numbers 1.1 Addition and Subtraction of Complex Numbers Recall that c omplex numbers can alternatively be represented by position vectors, i.e., a complex number iz x y  can be represented as a position vector xOP y       in the x-y plane. Techniques and operations used in coordinate geometry and vectors can be applied to adding and subtracting complex numbers , corresponding to the parallelogram law of vector addition and subtraction respectively. Let 1z = a + ib be represented by 1P . 2z = c + id be represented by 2P . 1 2z z z  = (a + c) + i(b + d) be represented by P. In terms of vectors, 1 2OP OP OP       . Re O z2 z1 Im P1 P2 P www.KiasuExamPaper.com 771

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Teacher’s Version) Page 2 of 24 The subtraction 2 1z z is the same as the addition of 2z and 1z . Thus, let 2P P   represents 1.z Then OP   represents 2 1z z . 2 1z z z  = (c – a) + i (d – b) is represented by P. In terms of vectors, 1 2 2 1OP P P OP OP           . What are some possible limitations of considering complex numbers as vectors on the Argand Diagram? Every complex number can only be represented by position vectors and not displacement vectors. So for example when finding the difference between two complex numbers (e.g. 2 1z z ), the vector representing the complex number 2 1z z must be translated to start from the origin first before the real and imaginary parts of the complex number can be obtained. Finding differences of vectors on the other hand, has no such restriction. 1.2 Conjugation of Complex Numbers Example 1.2.1 Represent the following complex numbers and their corresponding conjugates in an Argand diagram: 1 4ia   , 2 3ib  , 3 ic  , 2 4id   What is the geometrical effect of conjugating a complex number? Solution: z2 z1 Re Im O P1 P2 P Let’s be intellectually careful. 1z Re Im a b d c d* a* b* c* From the Argand diagram, the conjugate is obtained by reflecting the corresponding complex number about the real axis. Can you give an explanation of the effect by considering the modulus and argument of the conjugate? www.KiasuExamPaper.com 772

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Teacher’s Version) Page 3 of 24 1.3 Multiplication and Division of Complex Numbers We first consider the special case of multiplication of a complex number by i. Let z = 3 + 2i. iz = i(3 + 2i) = −2 + 3i i2z = −(3 + 2i) = −3 − 2i = −z i3z = −i(3 + 2i) = 2 − 3i i4z = 3 + 2i = z Using the same scale on both axes on the Argand diagram, we observe that if a point P represents a complex number z, then the point representing i z is obtained by rotating OP 90o anti-clockwise about the origin. Similar geometrical effect applies to the points representing 2 3 4i ,i and iz z z . In general, let us consider multiplying a complex number 2i 2 2 ez r  to another complex number 1i 1 1 ez r  . The result is  1 2i 1 2 1 2 ez z rr     . Geometrically, the length of 1 2z z equals to the product of the lengths of 1z and 2z i.e. 1 2 1 2z z rr  , and the argument of 1 2z z equals to the sum of the arguments of 1z and 2z i.e.  1 2 1 2arg z z     . It is observed that 1z has been rotated by 2 radians anti- clockwise (assuming 2 is positive) and its length scaled by factor 2r to give 1 2z z . Now, let us consider dividing a complex number 1i 1 1 ez r  by another complex number 2i 2 2 ez r  . The result is  1 2i1 1 2 2 ez r z r    . How would you illustrate this division geometrically?  1z has been rotated by 2 radians clockwise (assuming 2 is positive) and its length scaled by factor 2 1 r to give 1 2 .z z Im Re O Im Re O z = 3 + 2i iz = −2 + 3i i2z = −3 − 2i i3z = 2 − 3i www.KiasuExamPaper.com 773

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Teacher’s Version) Page 4 of 24 §2 Powers and nth Roots of a Complex Number 2.1 de Moivre’s Theorem If z is a complex number with z r and  arg z  , then    cos i sin cos i sin nn nz r r n n         for all real values of n. Try using mathematical induction to prove the de Moivre’s Theorem for positive integer values of n. Let’s be broad and adventurous! www.KiasuExamPaper.com 774

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Teacher’s Version) Page 5 of 24 Example 2.1.1 (Power of a Complex Number) Find the exact value of   6 1 3i . Solution:      6 6 6 π π1 3i 2 cos i sin 3 3 2 cos 2 π i sin 2π by De Moivre's Theorem 64              What other methods can we apply to answer the above question? Example 2.1.2 Prove that  41 (cos isin ) 2cos 2 cos 2 isin 2 ,        and that, provided the denominator is not zero,     4 4 1 cos isin cos 4 isin 4 1 cos isin             . Solution:       4 2 1 (cos isin ) 1 cos 4 isin 4 1 2cos 2 1 i 2sin 2 cos 2 2cos 2 cos 2 isin 2 .                                  4 4 2 2 2 2 2 1 cos isin 2cos 2 cos 2 isin 2 2cos 2 cos 2 isin 21 cos isin cos 2 isin 2 cos 2 isin 2 cos 2 isin 2 cos 2 isin 2 cos 2 isin 2 cos 2 sin 2 cos 4 isin 4 since cos 2 sin 2 1 .                                         Let’s be broad and adventurous! www.KiasuExamPaper.com 775

National Junior College Mathematics Department 2017 2016 – 2017 / H2 FMaths / Further Complex Numbers (Teacher’s Version) Page 6 of 24 Example 2.1.3 Using de Moivre’s Theorem, prove that sin 20 cos18sin 4 sin8 sin12 ..... sin 36 ,cos 2           provided cos 2 0.  Solution:                                           i 4 i 8 i 12 i 36 9i 4 i 4 i 4 i 4 i 36 i 4 i 4 i 18 i 18 i 18 i 2 i 2 i 2 i 20 sin 4 sin8 sin12 ..... sin 36 Im e e e ..... e e e 1Im e 1 e e 1Im e 1 e e e eIm e e e e 2cos18Im 2cos 2                                                                  cos 20 isin 20 2cos18 sin 20 cos18Im .2cos 2 cos 2                   2.2 nth Roots of a Complex Number Learning Task I (i) Write down the exponential form of 32 . (ii) State the number of complex roots of the equation 5 32 0z   . (iii) Hence solve the equation 5 32 0z   . www.KiasuExamPaper.com 776

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