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Further complex numbers Multiplication and division in polar form If s and t are complex numbers with s p, arg s and t q, arg t , then st p (cos i sin ) q(cos i sin ) pq(cos cos i sin cos i cos sin i 2 sin sin ) pq cos cos sin sin i sin cos cos sin pq cos( ) i sin( ) . So st p q and it may be true that arg( pq) arg p arg q. Polar form It is readily seen the x r cos and y r sin . Hence, z x yi r cos r sin i r cos i sin . The non-zero complex number z can be written in modulus - argument form, or polar form, as z r cos i sin , where r z 0 is the modulus and arg z, with , is the argument. Examples Write in modulus-argument form z i, z 2, z 2 i and z 1 i. z i r 1, 2 z 2 r 2, z 1 cos i sin . 2 2 z 2 cos i sin . z 2 i r 5, 2.677 z 5 cos 2.677 i sin 2.677 . z 1 i r 2, 3 4 3 z 2 cos 4 3 i sin 4 . Multiplication and division in polar form If s and t are complex numbers with s p, arg s and t q, arg t , then st p (cos i sin ) q(cos i sin ) pq(cos cos i sin cos i cos sin i 2 sin sin ) pq cos cos sin sin i sin cos cos sin pq cos( ) i sin( ) . So st p q and it may be true that arg( pq) arg p arg q. Multiplication and division in polar form The rules for multiplication and division in modulus-argument from are st s t , arg( st ) arg( s) arg(t ) k (2 ), s arg arg s arg t k (2 ), t where in each case the number k ( 1, 0,1) is chosen to ensure that the argument lies in the interval . s s , t t De Moivre’s Theorem Note that cos i sin 2 cos i sin cos i sin cos 2 i sin 2 . De Moivre's Theorem: n , z r cos i sin z n r n cos n i sin n . When n is not an integer, then r n cos n i sin n is only one of the possible values. At A-Level, you are expected to be able to prove this statement for positive integers. Use induction! Examples (i) Find the value of 1 i . 10 (ii) Find the value of 1 4 4i 3 . (iii) Find an expression for cos 6 and sin 6 . Do Exercise 15A, p.334 Do Q1-Q15, pp.352-354 Some useful identities If z cos i sin then and so that z n cos n i sin n 1 cos n i sin n n z 1 z n 2 cos n z 1 z n n 2i sin n z n As a special case, when n 1 we have 1 z 2 cos z 1 z 2i sin z Example : Show cos 4 Show sin 6 1 cos 4 4 cos 2 3 . 8 1 10 15cos 2 6 cos 4 cos 6 . 32 Exponential form of a complex number ei i 1 i 2! 2 i 3 3! 3 5 i 3! 5! 2 4 1 2! 4! cos i sin . So rei r cos i sin . This is called the exponential form of a complex number. De Moivre's Theorem now becomes n , rei n r n e i n . Note that ei 1 0 What is i i ? Complex roots z n r cos i sin has the n distinct roots 2 k 2k z n r cos i sin , n n where k 0,1, 2, n 1. In exponential form: z re n i has the n distinct roots z n re i Example 1 : Solve z 3 1. 2 k n , where k 0,1, 2, n 1. Example 2 : Solve z 4 1. Solutions of complex equations Solve the following: Factorize the following: 16 z 4 ( z 1) 4 p( z ) z 4 1 z 4 3 4i 0 p( z ) z 5 1 ( z 1)5 z 5 2 Find the exact value of cos 5 3 Do Exercise 15B, pp.342-344 . Useful identities in exponential form If z cos i sin then and so that z n cos n i sin n 1 cos n i sin n n z 1 z n 2 cos n z 1 z n n 2i sin n z n As a special case, when n 1 we have 1 z 2 cos z 1 z 2i sin z ei n ei n 2cos n ei n ei n 2i sin n ei e i 2 cos ei e i 2i sin Summation of trigonometric series 2 sin (10 ) k 1 Show that, provided cos 0, (1) cos(2k 1) . cos k 1 10 Show that a1 a2 cos a3 cos 3 a4 cos 4 1 , cos k 40 32 cos k 0 2 where a1 , a2 , a3 and a4 are to be determined. 3 k