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Summary of the results below: you have,
z
arg zz arg z arg z; arg
z
arg z arg z ;
(1.1)
Section 7, problem 1: find the principal argument Arg z,
arg z arg z
3 i
6
i
arg i arg 2 2i
2 2i
arg z 6a 6 arg( 3 i) 6 tan 1
2 1 3
3
4
6
6
5 4
Arg z
3
4
(1.2)
Arg z 2
(1.3)
Section 7, problem 5: Use de Moivre's formula (Sec. 7) to derive de rive the following trigonometric identities: 3 2 2 3 cos 3 cos 3 cos sin and sin 3 3 cos sin sin .
Getting ambitious: Let’s do this for cos cos n and sin n , using the binomial expansion,
ein cos n i sin n (cos i sin ) n Analytic work: Consider n
even
n m 1
e
n
n/ 2
in
( )
even
n !cos 2 n
1 n n 2
odd
sin
n
n! n 0 n!( n n )!
(cos ) n ( isin ) n n
(1.4)
as separate cases, so that we can compute Re[ ] and Im[ ]
n 2n
(2n) !( n 2n) !
n 0
n/ 2
i ()
1 n n 2
n 1
n !cos 2 n 1 sin n 2 n 1 (2n 1) !( n 2n 1) 1) !
S1 iS 2 f e (n, ); (1.5)
e n e ( m1) e m e (S1 iS 2 ) cos i sin S1 cos S 2 sin i (S 1 sin S 2 co s ); (1.6) i
odd
and n
i
i
i
Specializing to n = 3: Considering Con sidering n 3 2 1 m 1 , we write, cos 3 Re ei3 cos S1 (2) S 2 (2) sin 1
cos ()
1 n
2!cos2 n
n 0
cos 3 co cos sin 2
sin 2 2 n
(2n) !(2 2n) ! 2 2!0!
1
sin ( ) n1
1 n
2!cos 2 n 1 sin 2 2 n 1
(2n 1) !(3 2n) !
2! cos sin cos3 3 cos sin 2 cos 3 1!1!
cos2 sin
(1.7)
sin 3 Im ei 3 S1 (2) sin S 2 (2) cos sin 2 cos2 sin 2 cos sin cos 3 cos2 sin sin3 (1.8)
