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COMPLEX FUNCTIONS AND TRIGONOMETRIC IDENTITIES
Revision E
By Tom Irvine
Email: [email protected]
September 14, 2006
Trigonometric Functions of Angle 
r
y

x
sin   
y
r
cos  
x
r
tan   
y
x
cot   
x
y
sec  
r
x
csc  
r
y
(1)
Trigonometric Expansion
sin x  x 
x3 x5


3! 5!
(2)
cos x  1 
x2 x4


2! 4!
(3)
1
x3 x5
sinh x  x 


3! 5!
(4)
x2 x4
cosh x  1 


2! 4!
(5)
Exponential Expansion
exp x   1  x 
x 2 x3


2! 3!
(6)
Trigonometric Identities
cos 2   
1 1
 cos2 
2 2
(7)
sin 2   
1 1
 cos2 
2 2
(8)
cos cos 
1
1
cos    cos  
2
2
(9)
1
1
sin  sin    cos    cos  
2
2
(10)
cos sin  
1
1
sin     sin   
2
2
sin  cos 
(11)
1
1
sin     sin   
2
2
(12)
sin 2    cos2    1
(13)
sec 2    tan 2    1
(14)
csc 2    cot 2    1
(15)
sin 2  2 sin cos
(16)
2
cos2   cos2    sin 2  
(17)
sin     sin cos  cossin 
(18)
cos    coscos  sin sin 
(19)
tan    tan 
1  tan   tan 
(20)
tan    
A sin t   B cost   A 2  B2  sin t   
where
 B
  arctan  
A
(21)
Euler's Equation
exp  j  cos  j sin 
(22)
sin   
exp  j   exp  j 
2j
(23)
cos  
exp  j   exp  j 
2
(24)
Hyperbolic Functions
sinh   
exp    exp   
2
(25a)
cosh   
exp    exp   
2
(25b)
3
cosh 2   sinh 2   1
(26)
Derivatives
d
sin u   cos u du
dx
dx
(27)
d
cos u    sin u du
dx
dx
(28)
d
tan u   sec 2 u du
dx
dx
(29)
d
cot u    csc u du
dx
dx
(30)
d
sec u   tan u sec u du
dx
dx
(31)
d
csc u    csc u cot u du
dx
dx
(32)
d
sinh u   cosh u du
dx
dx
(33)
d
cosh u   sinh u du
dx
dx
(34)
d
tanh u   1 2 du
dx
cosh u dx
(35)
Natural Logarithm of a Complex Number
b


ln a  j b   ln  a 2  b 2 exp  j ,   arctan  


a
4
(36)


ln a  j b   ln  a 2  b 2   ln  exp  j




ln a  j b   ln  a 2  b 2   j


(37)
(38)
b


ln a  j b   ln  a 2  b 2   j arctan  


a
(39)
APPENDIX A
The Square Root of a Complex Number.
Consider
x2  a  jb
(A-1)
Thus
x   a  jb
(A-2)
where a and b are real coefficients.
Solve for x.
Let
x1  c  jd
(A-3a)
x 2  c  jd
(A-3b)
where c and d are real coefficients.
Substitute equation (A-3a) into (A-1).
c  jd 2  a  j b
(A-4)
5
c  jdc  jd  a  j b
(A-5)
c2  d 2  j 2cd   a  j b
(A-6)
Equation (A-6) implies two equations. The first is
c2  d 2  a
(A-7)
The second implied equation is
2cd  b
(A-8)
Solve for d using equation (A-8).
d
b
2c
(A-9)
Substitute equation (A-9) into (A-8).
2
 b
2
c    a
(A-10)
2
 b
2
c a    0
(A-11)
 2c 
 2c 
Multiply through by 4c 2 .
4c4  4ac2  b 2  0
(A-12)
Apply the quadratic formula.
c2 
4a  16a 2  16b2
8
(A-13)
c2 
4a  4 a 2  b2
8
(A-14)
6
c2 
c
a  a 2  b2
2
(A-15)
a  a 2  b2
2
(A-16)
Require c to be real.
c
a  a 2  b2
2
(A-17)
Substitute equation (A-17) into (A-9).




b
d
  a  a 2  b2
2 
2
 
 








 
(A-18)




d







b




2
2
4 a  a  b 



2

(A-19)
7



d




b




2
2
2 a  a  b 

 
(A-20)
Substitute equations (A-20) and (A-17) into (A-3a).

2
2
 a a b
x1  
2










j




b




2
2
2 a  a  b 

 
(A-21a)
Substitute equations (A-20) and (A-17) into (A-3b).

2
2
 a a b
x 2  
2



 
 
  j
 
 



b




2
2
2 a  a  b 

 
(A-21b)
Note that equations (A-21a) and (A-21b) cannot be used for the special case:
a < 0 and b = 0.
For this special case, the roots are
x j a
(A-21c)
Example
x2  2  j 7
(A-22)
x   2 j7
(A-23)
8
Solve for x. Use equation (A-21a).
a2
(A-24)
b7
(A-25)
x 1  2.154  j 1.625
(A-26)
x 2  2.154  j 1.625
(A-27)
APPENDIX B
Arbitrary Root of a Complex Number
Let
x n  a  j b
(B-1a)
x  a  j b1 / n
(B-1b)
The coefficients a and b are real numbers. The denominator of the exponent n is also
real.
Take the natural logarithm.

ln x  ln a  j b1 / n

(B-2)
ln x 
1
ln a  j b
n
(B-3)
ln x 
1  2
b 

ln  a  b2 exp  j arctan 
n 
a 

(B-4)
ln x 
1  2
b 
 1  
ln  a  b2   ln exp  j arctan 
n 
a 
 n  
9
(B-5)
1 

1
b
2
2

2
ln x  ln a  b n   j arctan


n
a




(B-6)
1 
 

1
b
  2

2
exp ln x  exp ln a  b 2n   j arctan 


n
a
 





(B-7)
1 
 
b
  2

 1
2
x  exp ln a  b 2n   exp  j arctan 

a
 n
 
 




(B-8)
1
b
 1
2
2
x  a  b 2n exp  j arctan 



 n
(B-9)
a
1
 1
b
b 
1
2
2
x  a  b 2n cos arctan   j sin  arctan 

n
a
n
a 
(B-10)
Note that equation (B-10) could be used for the special case of a square root.
APPENDIX C
Cube Root of a Complex Number
Consider
x3  a  j b
(C-1)
x  a  j b1 / 3
(C-2)
Equation (C-1) has three roots. The method in Appendix B yields the following formula
for one of the cube roots.
10
1
 1
b
b 
1
x 1 a 2  b 2 6 cos arctan   j sin  arctan 



3
a
3
a 
(C-3)
Rearrange equation (C-1).
x3  a  j b  0
(C-4)
Devise an equation for finding the other two roots.
x 3  a  j b  x  x1 x  x 2 x  x 3 
(C-5)
Expand the right-hand-side.


x 3  a  j b  x 2  x1  x 2 x  x1x 2 x  x 3 

 

x 3  a  j b  x 3  x1  x 2 x 2  x1x 2 x  x 3x 2  x 3 x1  x 2 x  x1x 2 x 3 
x 3  a  j b  x 3  x1  x 2 x 2  x1x 2 x   x 3x 2  x 3 x1  x 2 x  x1x 2 x 3 
x 3  a  j b  x 3  x1  x 2  x 3 x 2  x1x 2  x1x 3  x 2 x 3 x  x1x 2 x 3 
x 3  a  j b  x 2  x1  x 2 x  x1x 2 x   x 2  x1  x 2 x  x1x 2  x 3 
(C-6)
(C-7)
(C-8)
(C-9)
(C-10)
Equation (C-10) implies three separate equations.
x1  x 2  x3   0
(C-11)
x1x 2  x1x3  x 2x3   0
(C-12)
 x1x 2x3  a  j b
(C-13)
Continue with equation (C-11).
x 2   x1  x3
(C-14)
11
Substitute equation (C-14) into (C-12).
x1 x1  x3   x1x3   x1  x3 x3   0
(C-15)
 x12  x1x 3  x1x 3  x1x 3  x 32  0
(C-16)
 x12  x1x 3  x 32  0
(C-17)
x 32  x1x 3  x12  0
(C-18)
Use the quadratic formula.
x3 
 x1  x12  4x12
(C-19)
2
x3 
 x1   3x12
x3 
(C-20)
2
 x1  x1  3
2
(C-21)
1 j 3 
x 3  x1 

2


(C-22)
1 j 3 
x 3  x1 

2


(C-23)
Choose
Recall equation (C-14).
x 2   x1  x3
(C-24)
1 j 3 
x 2   x1  x1 

2


(C-25)
12

  1  j 3 


x 2  x11  

2





(C-26)

1  j 3  


x 2  x11  


 2 


(C-27)
 2 1 j 3 
x 2  x1 

2 
 2
(C-28)
1 j 3 
x 2  x1 

2


(C-29)
The roots x2 and x3 thus form a complex conjugate pair.
Summarize the roots.
1
 1
b
b 
1
2
2
x 1 a  b 6 cos arctan   j sin  arctan 



3
a
3
a 
(C-30)
1 j 3 
x 2  x1 

2


(C-31)
1 j 3 
x 3  x1 

2


(C-32)
Example
Solve for x.
x3  2  j7
(C-33)
x  2  j 71 / 3
(C-34)
a2
(C-35)
13
b7
(C-36)
n=3
(C-37)
There are three roots. The first root is
1
 1
b
b 
1
2
2
x 1 a  b 6 cos arctan   j sin  arctan 



3
a
3
a 
(C-38)
x1  1.938 0.909  j 418
(C-39)
x1  1.761  j 0.809
(C-40)
The second root is a coordinate transformation of the first root.
1 j 3 
x 2  x1 

2


(C-41)
1 j 3 
x 2  1.762  j 0.809 

2


(C-42)
x 2   1.581  j1.120
(C-43)
1 j 3 
x 3  x1 

2


(C-44)
1 j 3 
x 3  1.761  j 0.809

2


(C-45)
x3  0.180  j1.930
(C-46)
14
In summary, the cube roots of (2 + j 7) are
x1  1.761  j 0.809
(C-47)
x 2   1.581  j1.120
(C-48)
x3  0.180  j1.930
(C-49)
APPENDIX D
Derivation of the Quadratic Formula
ax 2  bx  c  0
(D-1)
x 2  b / a x  c / a   0
(D-2)
b 2 b2 c

x

 0

 
2
2a 
a

4a
(D-3)
b 2
b2 c

x





2
2a 
a

4a
(D-4)
b  2 b 2  4ac

x   
2a 

4a 2
(D-5)
b 
b 2  4ac

x    
2a 

4a 2
(D-6)
15
b   b 2  4ac

x   
2a 
2a

(D-7)
b 2  4ac
2a
(D-8)
 b  b 2  4ac
2a
(D-9)
b
x 

2a
x
16