E content Material

1. 1
MATHEMATICS
E Content On
Algebra & Trignomentry
Vinay M. Raut
Shri Shivaji Science College, Amravati

2. 2
Demoivre’s Theorem and its application
Content :
1) Introduction to complex number & compress plane.
2) Demoivre’s theorem (DMT)
3) Result on Demoivre’s theorem
4) Application of DMT Roots of complex number.
5) Relation between circular and hyperbolic method.
6) Logrithimic of complex number.
Question : State and prove Demoivre’s theorem. (DMT)
Statement : Whatever may be the value of n (positive or negative,
intergral or factional) the value or one of the values of
(cosθ + sinθ)n
is cos nθ + sin inθ where Rθ ∈
Proof : We consider there different cases
Case I : Let n is positive integer we prove this case by mathematical
induction method (Which is as follows)
1) Denote the given result by P(n)
2) The result is true for n = 1 i.e. P(1)
3) Assume the result is true for n = k
i.e. P(k) ⇒ the result is true for n = k + 1 i.e. P (k+1)
Then by mathematical Induction method the given result is true for all values of
n
Let demote P (n) :- ((cos sin ) cos sin )n
i n i nθ θ θ θ+ = + …….. (1)
Put n = 1 in eqn
(1)
∴ P (1) : (cos sin ) cos sini iθ θ θ θ+ = +
Hence P (1) is true for n = 1
Next Step
Assume P(n) for n = k i.e. P(k) is true
ie. (cos sin ) cos sinK
i k i kθ θ θ θ+ = + ……………………………………..(2)
To show that P(k+1) is true multiply eqn
(2) by cos sini nθ θ+ on both sides

3. 3
1
(cos sin ) (cos sin )(cos sin )k
i k i k iθ θ θ θ θ θ+
+ = + +
(cos cos sin sin ) (sin .cos cos .sin )k k i k kθ θ θ θ θ θ θ θ= − + +
cos( 1) sin( 1)k i kθ θ= + + +
Thus P(k+1) is true
ie. P(k) is true ⇒ P (k+1) is true
∴ By mathematical induction method given result is true for all value of n∈N
This proves the DMT ∀ n∈N
Case II : Let n1
be negative integer demote n =−m where m = +ve integer
Now (cos sin ) (cos sin )n m
i iθ θ θ θ −
+ = +
1 1
(cos sin ) cos sinm
m mθ θ θ θ
= =
+ +
(by case I)
Multiply to Nr
& Dr
by cos sinm i mθ θ− (conjugate of Dr
)
1 cos sin
(cos sin ) .
cos sin cos sin
n m i m
i
m i m m i m
θ θ
θ θ
θ θ θ θ
−
+ =
+ −
2
2 2
cos sin
( 1)
cos sin
m i m
i
m m
θ θ
θ θ
−
= = −
+
cos sinm i mθ θ= −
cos( ) sin( )m i mθ θ= − + −
cos sinn i nθ θ= +
cos( ) cos
sin( ) sin
θ θ
θ θ
− =
− =
{ )as n m= −
⇒ Thus DMT is true for n = negative integer
Case III
Let n = fraction = P/q = +ve integer and P is integer which may be positive
and negative
(cos / sin / ) cos . sin .q Q Q
q q q i q
q q
θ θ
   
+ = +   
   
cos siniθ θ= + (by DMT)
⇒
1
cos sin (cos sin ) q
i i
q q
θ θ
θ θ+ = +
⇒ cos sini
q q
θθ + is one of the roof of cos sinθ θ+
Note :
1
( q
If w V w= ⇒ is one of the th
q roof of v)

4. 4
i.e. one of the value of
1
(cos sin ) q
iθ θ+ is cos sin
Q Q
i
q q
+ Raising both sides
to the power P on both sides
(cos sin )
p
q
iθ θ+ is cos sin
P
Q Q
i
q q
 
+ 
 
⇒ (cos sin )
p
q
iθ θ+ is cos sin
pp i
q q
θ θ+ (by case II)
(Put P/q = k)
⇒ (cos sin )n
iθ θ+ is cos sinn iθ θ+
⇒ Thus DMT is proved for n is fraction
Different form of the DMT
1) (cos sin ) cos( ) sin( ) cos sinn
i n i n n i nθ θ θ θ θ θ+ = − + − = −
2) (cos sin ) cos sinn
i i nθ θ θ θ− = −
3) (cos sin ) cos sinn
i n i nθ θ θ θ−
− = +
Example 1 : Shown that
1 1 2 2(cos sin )(cos sin ).............(cos sin )n nx i x x i x x i x+ + +
1 2 1 2cos( ____ ) sin( ____ )n nx x x i x x x= + + + + + + +
Solution : LHS = 1 2
. .........ix ix ixn
e e e (by euler formula cos sin )i
e iθ
θ θ= +
1 2( ____ )i
ne x x x= + + +
1 2 1 2cos( ___ ) sin( ___ )n nx x x i x x x= + + +
= RHS
Example 2 : Simplify
5 3
4 2
(cos3 sin3 ) (cos2 sin2 )
(cos5 sin5 ) (cos sin )
x i x x i x
x i x x i x
+ −
− −
Solution : Let
5 3
4 2
(cos3 sin3 ) (cos2 sin2 )
( )
(cos5 sin5 ) (cos sin )
x i x x i x
A x
x i x x i x
+ −
=
− −
3 5 2 3 15 6
5 4 2 20 2
( ) ( ) .
( ) .( ) .
ix ix ix ix
ix ix ix ix
e e e e
e e e e
− −
− − − −
= =
31
(15 6 20 2) cos 31 sin 31ix ix
e e x i x= − + + = = +

5. 5
Example 3: Express (1+7i) (2−i)-2
in the form (cos sin )r iθ θ+ prove that the
fourth power is real negative number.
Solution : Let 2
(1 7 )(2 1)Z i −
= + −
Then 2
1 7 1 7 1 7
(2 ) 4 4 1 3 4
i i i
Z
i i i
+ + +
= = =
− − − −
1 7 3 4
4 4 3 4
i i
i i
+ +
= ×
− +
(multiply and divided by 3+4i)
25 25
25
i− +
= 2 2
[ ( )( )]a b a b a b− = − +
1 i= − +
=
1 1 3 3
2 2 cos sin
4 42 2
i i
π π   
− + = +  
  
RHS is of the form (cos sin )r iθ θ+ where
3
2,
4
r
π
θ= =
⇒
4
4 3 3
2 cos sin 4(cos3 sin 3 )
4 4
z i i
π π
π π
  
= + = +  
  
= −4 [ cos 3 1, sin 3 0]π π= − =∵
= negative real number
Example 4 : Prove that
1 sin cos
sin cos
1 sin cos
i
i
i
θ θ
θ θ
θ θ
+ +
= +
+ −
Proof: Multiply and divided to LHS by (1 sin ) cosθ θ+ + we get
2 2
2 2
(1 sin ) 2 (1 sin )cos cos
(1 sin ) cos
iθ θ θ θ
θ θ
+ + + −
=
+ +
2sin (1 sin ) 2 (1 sin )cos
2(1 sin )
iθ θ θ θ
θ
+ + +
=
+
= RHS
Example 5: If α and β are the roots of equation 2
2 4 0x x− + = prove that
1
2 cos
3
n n n nπ
α β +
+ =
Solution : Solving the equation 3x i= ± we get

6. 6
Let 1 3iα = + and 1 3iβ = −
⇒ /3
2 i
e π
α = and / 3
2 . i
e π
β −
=
/31 3
1 3 2 2 cos sin 2
32 2 3
i
i i i eπππ
    
+ = + = + =         
∴ /3 /3
2n n n ni ni
e eπ π
α β −
 + = + 
2 2cos
3
n nπ
= × 2cosi i
e eθ θ
θ−
 + = ∵
1
2 .cos
3
n nπ+
= Where cos sini
e iθ
θ θ= +
= RHS
Example 6 : If
1
2 cos y x
x
= + then show that
1
2cos n
n
ny x
x
= + and
1
2 sin n
n
i ny x
x
= −
Solution : Let
1
2 cos y x
x
= +
then 1iy iy
e e x
x
−
+ = + [as 2 cosiy iy
e e y−
+ = ]
is
1 1iy
iy
e x
e x−
+ = +
⇒ iy
x e= and iy
x e−
=
For iy
x e= ,
1
( ) ( )n iy n iy n
n
x e e
x
−
+ = +
And
1
2 sinn iny iny
n
x e e i ny
x
−
+ = − =
Similarly we can show the result for iy
x e−
=
Example 7 : If cos sin
2 2
r r r
x i n
π π
= + them show that 1 2 3 .......... cos 1x x x π= = −
Example 8 : If sin sin sin cos cos cos 0p q r p q r+ + = + + = then show that
cos 3 cos 3 cos 3 3cos( )p q r p q r+ + = + +
and sin 3 sin 3 sin 3 3sin( )p q r p q r+ + = + +
Example 10 : Show that

7. 7
1 sin cos
(sin cos ) cos sin
1 sin cos 2 2
n
n x i x n n
x i x nx i nx
x i x
π π+ +     
+ = = − + −     
+ −     
Application of DMT (Demoivre’s Theorem)
Finding the nth
roots of complex number
Method 1 : Express the given z in a polar form (cos sin )Z r iθ θ= + when Z has
spetial forms it is convenient to write if by trial method.
For ex. 1)
1 1
, 2 2 cos sin
4 42 2
Z i i Z i i
π π   
= ± = + = ±  
  
2)
1 3
1 3 2 2 cos sin
2 2 3 3
Z i i i
π π   
= ± = ± = ±       
Put as cos cos(2 )kθ π θ= +
sin sin(2 )kθ π θ= +
(cos sin )Z r iθ θ∴ = +
1 1
{cos(2 ) sin(2 }n n
Z r k i kπ θ π θ= + + +
Then
1 1 1
{cos(2 ) sin(2 }n n n
Z r k i kπ θ π θ= + + +
1 2 2
cos sinn
k k
r i
n n
π θ π θ+ + 
= + 
 
(by DMT)
Denote
1 1 2 2
cos sinn n
k k
uk Z r i
n n
π θ π θ+ + 
= = + 
 
………….. (A)
2) Substitute K = o, 1, 2 …….. n-1 in equation (A) and n values u0, un-1 there will be
nth
roof of Z.
(Note : For K = n, n+1, ……..-1, -2 in equation (A) the values
un, un-1……. U-1, U-2..... are the repetitions of u0, un…….
un-1)
Note : 1) 1 cos0 sin0i= +
cos2 sin2k i kπ π= +
2) 1 cos siniπ π− = +
cos(2 1) sin(2 1)k i kπ π= + + +
3) cos sin cos 2 sin 2
2 2 2 2
i i k i k
π π π π
π π
   
± = ± = + ± +   
   

8. 8
Note : 1) sin( ) sinπ θ θ− = 2) cos( ) cosπ θ θ− =
3) cos( ) cosπ θ θ+ = − 4) sin( ) sinπ θ θ+ = −
Example 1: Find all the value or roots (-1)1/6
Let
1
6
( 1)Z = −
Solution : Now we know that 1 cos siniπ π− = +
cos(2 ) sin(2 )n i nπ π π π= + + +
1 1
6 6
( 1) [cos(2 ) sin(2 )]n i nπ π π π∴ − = + + + (by DMT
Let the RHS of above equation in Uk
cos(2 1) (2 1)
sin
6 6
k
n n
U i
π π+
= + ………….. (A)
Putting K = 0, 1, 2, 3, 4, 5 we obtain required six roots.
1) for k = 0 0 cos sin
6 6
u i
π π
= +
3 1
2 2
e= +
2) for k = 1 : .
0 cos sin
2 2
u i e
π π
= + =
3) for k = 2 :
.
.
2
5 5 3 1
cos sin
6 6 2 2
u i e
π π
= + = +
4) for k = 3 :
.
.
3
7 7 3 1
cos sin
6 6 2 2
u i e
π π −
= + = −
5) for k = 4 : 4
9 9
cos sin
6 6
u i i
π π
= + = −
6) for k = 5 :
.
.
5
11 11 3 1
cos sin
6 6 2 2
u i i
π π
= + = −
∴ roots are
. 3 3
, ,
2 2
i i
e
± − ±
∴±
Example 2: Find all the values of
1
6
( )i−
Let Z =
1
6
( )i
Solution : cos sin
2 2
i i
ππ− = −

9. 9
0
2 2
2 2
i k i k
π π
π π
   
− = + − +   
   
0
cos 4 sin 4
2 2
i k i k
π π
π π
   
− = + − +   
   
1
61
6
4 1 4 1
( 1) cos sin
2 2
k k
iπ π
+ + 
∴ − = − 
 
4 1 4 1
cos sin
12 12
k k
iπ π
+ +
= − (by DMT)
Let the RHS of above equation in Uk
4 1 4 1
sin
12 12
k k
uk iπ π
+ +
= − ………….. (A)
Putting K = 0, 1, 2, 3, 4, 5 we obtain required six roots which are as follows
for k = 0, 0 cos sin
12 12
u i
π π
= +
for k = 1 :
.
1
5 5
cos sin
12 2
u i
π π
= −
for k = 2 : 2
9 9
cos sin
12 12
u i
π π 
= −  
 
for k = 3 : 3
13 13
cos sin
12 12
u i
π π
= −
cos sin
12 12
π π
π π
   
= + − +   
   
(by note)
0cos sin
12 12
u
π π
= + = −
for k = 4 : 4
17 17
cos sin
12 12
u i
π π
= +
5 5
cos sin
12 12
π π
π π
   
= + − +   
   
5 5
cos sin
12 12
π π
= + (by note)
1u= −
for k = 5 : 5
21 21
cos sin
12
u i
π π
π
= −
9 9
cos sin
12 12
π π
π π
   
= + − +   
   

10. 10
2
9 9
cos sin
12 12
u
π π
= − + = −
Hence the required roots are 0 2, ,u u u± ± ±

11. 11
Example 3 : Find all the values or roots of
1
5
(32)
Solution : 32 32 1 32(cos sin )iθ θ= = +
32(cos 2 sin 2 )k i kπ π= +
1 1 1
55 5 5
(32) (2 ) (cos 2 sin 2 )k i kπ π⇒ = +
( )52 2
2 cos sin 32 2
5 5
k k
i
π π 
= + = 
 
∵
The five value of
1
5
(32) are given by
2 2
2 cos sin
5 5
k
k k
U i
π π 
= + 
 
Put k = 0, 1, 2, 3, 4 to get five roots
for k = 0, 0 2(cos sin 0) 2u i= + =
for k = 1,
.
1
2 2
2 cos sin
5 5
u i
π π 
= + 
 
.
2 cos sin
5 5
i
π π 
= − + 
 
for k = 2,
.
2
4 4
2 cos sin
5 5
u i
π π 
= + 
 
.
2 cos sin
5 5
i
π π 
= − + 
 
Note
for k = 3,
.
3
6 6
2 cos sin
5 5
u i
π π 
= + 
 
2 cos sin
5 5
i
π π 
= − + 
 
for k = 4,
.
4
8 8
2 cos sin
5 5
u i
π π 
= + 
 
2 2
2 cos 2 sin 2
5 5
i
π π
π π
    
= + − +    
    
2 2
2 cos sin
5 5
i
π π 
= − 
 

12. 12
Example 4 : By using DMT find all the fourth root of 81
Solution : ut 81x =
Now 81 = 81.1 = 81 (cos 2 sin 2 )k i kπ π+
⇒
1 1
4 4
(81) [81(cos2 sin 2 )]k i kπ π= +
1 1
4 4
(81) (cos 2 sin 2 )k i kπ π= +
2 2
3 cos sin
4 4
k k
i
π π 
= + 
 
3 cos sin
2 2
k k
i
π π 
= + 
 
…………………. (1)
The four values k=0, 1, 2, 3 are as follows.
for k = 0, 0 3(cos sin 0) 3u = + =
for k = 1, 1 3 cos sin 3
2 2
u i i
π π 
= + = 
 
for k = 2, ( )2 3 cos sin 3u iπ π= + = −
for k = 3,
.
3
3 3
3 cos sin
2 2
u i
π π 
= + 
 
3 cos sin 3
2 2
i i
π π 
= − − = − 
 
Hence the roots are 3, 3i± ±
Example 5 : Show that the continued product of the four values of
( )
3
4
cos sin
3 3
iπ π+
Solution : Let ( )
3
4
cos sin
3 3
u iπ π= +
Then ( )
3
4
cos sin
3 3
u iπ π= +
4
cos sinu iπ π= +
⇒ ( )
1
4
cos sinu iπ π= +
1
4
{cos(2 ) sin(2 }k i kπ π π π= + + +

13. 13
2 1 2 1
cos sin
4 4
k k
iπ π
+ +
= + (by DMT)
Thus the four values of u are given by
2 1 2 1
cos sin
4 4
k
k k
u iπ π
+ +
= +
Put k = 0, 1, 2, 3
Then we get
4
0 cos / 4 sin
4
i
u i e
ππ
π= + =
3
4
1
3
cos 3 / 4 sin
4
i
u i e
ππ
π= + =
5
4
2
5 5
cos sin
4 4
i
u i e
ππ π −
= + =
7
4
3
7 7
cos sin
4 4
i
u i e
ππ π
= + =
3 5 7
4 4 4 4
0 1 2 3 . . .
i i i i
u u u u e e e
π π π π
∴ =
4
(1 3 5 7 ) / 4i i
e eπ π
= + + + =
cos 4 sin 4 1i unityπ π= + = =
Find the nth root of unity
If 1n
u = or
1
(1) n
u = then u is called nth
root of unity
Proof : Now cos2 sin 2l k i kπ π= +
Then
1 1
(1) (cos2 sin 2 )n n
u k i kπ π= = +
22 2
cos sin
k i
n
k k
i e
n n
ππ π
= + =
The number of the nth
roof are
2k i
n
kU e
π
= ……………….. (A)
Put k =0, 1, 2, _______n-1 to find the nth
root.
Now for k = 0 eqn
(1) 0 1u e⇒ = =
For k = 1, eqn
(1)
2 /
1
i n
u e sayπ
ω⇒ = = =
For k = 2, eqn
(1)
4 / 2
2
i n
u e π
ω⇒ = =
For k = 3, eqn
(1)
6 / 2
3
i n
u e π
ω⇒ = =

14. 14
and so on
For k = n−1, eqn
(1)
2( 1) / 1
1
n i n n
nu e π
ω− −
−⇒ = =
Thus the nth
roots are 1, ω, ω2
…….. ωn-1
they form a geometric progression with
common ratio
2k i
n
w e
π
=
Example 6 : If
2
0 1 2(1 ) ........n
x P Px Px+ = + + +
Show that 2 4
0 2 4 ............ 2 .cos
n n
P P P
π
+ + − =
Solution : Let
2
0 1 2 4(1 ) ......n n
x P Px Px Px+ = + + +
Take x i=
2 3 4
0 1 2 3 4(1 ) ......n
x P Pi Pi Pi Pi∴ + = + + + + +
( )0 2 4 6 1 3 5 7....... ( ......)P P P P i P P P P= − + − + + − + − +
Now (1 ) ( 2) {cos sin }
4 4
n n nn
i i
π π
+ = + ………………(1)
/2
2 {cos sin }
4 4
n n n
i
π π
= + ………………(2)
Equation (1) and (2) ⇒
( )0 2 4 6 1 3 5 7....... ( ......)P P P P i P P P P= − + − + + − + − +
/2
2 {cos sin }
4 4
n n n
i
π π
= +
Equating real part only
/2
0 2 4 6 ....... 2 cos
4
n n
P P P P
π
− + − + =
Relation between complex circular and hyperbolic function
We know that complex circular function are sin , cos
2 2
iz iz iz iz
e e e e
z Z
i
− −
− −
= =
tan
( )
iz iz
iz iz
e e
z
i e e
−
−
−
=
−
……….……so on

15. 15
Hyperbolic function :
1) The quantity
2
y y
e e−
− where y be real or complex is called the hyperbolic sine of y
and it is written as sin by similarly the quantity
2
y y
e e−
+ is called hyperbolic cosine of
y and is written as cos ny
sinh
2
y y
e e
y
−
−
∴ = and cos
2
y y
e e
y
−
+
=
By using this we can find tan hy, cos echy , cot hy etc.
Now relation between circular and hyperbolic function
Theorem 1 : Prove : sin( ) sinhyi y=
Proof : we have sin
2
yi y
e e i
y
i
−
−
= , cos
2
yi yi
e e
y
−
+
=
(1)
( ) ( )
1
sin
2 2 2
yi i yi i y y y y
i
e e e e e e
y
i i i
− − −
 − − −
= = = −  
 
2
2
y y
i e e
i
−
 −
=  
 
(Multiply divided by i)
2
sinh ( 1)
2
y y
e e
i i y i
−
 −
= = = − 
 
Similarly we can prove (2) cos coshyi y=
(3) cos cosiy i echy= −
(4) sec seciy iy=
(5) tan tanhiy i y=
Theorem 2 : Prove 2 2
sin cos 1z z+ =
Proof : LHS 2 2
sin cosz z= +
2 2
2 2
iz iz iz iz
e e e e
i
− −
   − +
= +   
   
2 21
{ ( ) ( ) }
4
iz iz iz iz
e e e e− −
= − − + +
1
{4 . }
4
iz iz
e e−
= (as 2 2
( )( )a b a b a b− = − + )

16. 16
Theorem 3 : 1 2 1 2 2 2sin ( ) sin .cos co s .sinz z z z z z+ = +
1 2 1 2 1 2cos( ) cos .cos sin .sinz z z z z z+ = +
Proof : be have 1 2 1 2( )
.i z z iz iz
e e e+
=
1 2 1 2 1 1cos( ) sin( ) (cos sin )z z i z z z i z⇒ + + + = + 2 2(cos sin )z i z+ (by Euler)
1 2 1 2 1 2 1 2(cos cos sin , sin ) (sin , cos cos sin )z z z z i z z z z= − + +
Equate the real and imaginary part we get the required result
Theorem 4 : Prove sin( ) sinz z− = −
cos( ) cosz z− =
Separate the real and imaginary part
Example 1: sin( )x iy+
Solution: sin( ) sin .cos cos sinx iy x iy x iy+ = +
sin cosh cos ( sinh )x y x i y= +
sin cosh cos sinhx y i x y= +
( sin sinh . cos cosh )iy i y iy i y= =∵
Example 2: tan( )x iy+
Solution: sin( ) 2 sin( ) cos( )
tan( )
cos( ) 2 cos( ).cos( )
x iy x iy x iy
x iy
x iy x iy x iy
+ + −
+ = =
+ + −
[multiply & divided by cos x iy− ]
sin[( ) ( )] sin[( ) ( )]
cos[( ) ( )] cos[( ) ( )]
x iy x iy x iy x iy
x iy x iy x iy x iy
+ + − + + − +
=
+ + + + + − −
(use 2sin cos & 2cos cosA B A B formula)
sin 2 sin(2 )
cos 2 cos(2 )
x iy
x iy
+
=
+
sin 2 sinh 2
cos 2 cosh 2
x i y
x y
+
=
+
(by hyperbolic relation)
sin sinh 2
cos 2 cosh 2 cos 2 cosh 2
x y
i
x y x y
= +
+ +
(Real Part) (Imaginar part)
Example 3 : 1
sec( )
cos( )
x iy
x iy
+ =
+

17. 17
2 cos( )
2 cos( ).cos( )
x iy
x iy x iy
−
=
+ −
[Multiply & divided by 2 cos( )x iy−
2[cos cos sin sin ]
cos 2 cos 2
x iy x iy
x iy
+
=
+
2[cos cosh sin sinh ]
cos 2 cosh 2
x y i x y
x y
+
=
+
2 cos cosh 2 sin sinh
cos 2 cosh 2 cos 2 cosh 2
x y x y
i
x y x y
= +
+ +
(Real Part) (Imaginary part)
Example 4: cosh( )x iy+
Solution: cosh( ) cos ( )x iy i x iy+ = + as [cos cosh ]iy y=
cos( ).[cos cosh ]ix y iy y= − =
cos cos sin .sinix y ix y= +
cosh .cos sin sinx y ix y= + (by hyperbolic relation)
cosh .cos sinh .sinx y i x y= +
Example 5 : If ( ( )A B x iy+ = + prove that
Solution: 2 2
2
tan 2
1
x
A
x y
=
− −
and 2 2
2
tanh 2
1
y
B
x y
=
+ +
OR 2 2
2 cot 2 1x y x A+ + =
2 2
2 coth 2 1 0x y y B+ − + =
Solution : Let tan( )A iB x iy+ = +
Them tan( )A iB x iy− = −
tan 2 tan[( ) ( )]A A iB A iB∴ = + + −
2 2 2 2
( ) ( ) 2 2
1 ( )( ) 1 ( ) 1
x iy x iy x x
x iy x iy x y x y
+ + −
= = =
− + − − + − −
2 2
1 2 cot 2x y x A⇒ − − =
2 2
2 cot 1x y x A+ + =
Similarly tan(2 ) tan[( ) ( )]iB A iB A iB= + − −
tan( ) tan( )
1 tan( ) tan( )
A iB A iB
A iB A iB
+ − −
=
+ + −
2 2
( ) ( ) 2
tanh 2
1 ( )( ) 1 ( )
x iy x iy iy
i B
x iy x iy x y
+ − −
⇒ = =
+ + − + +

18. 18
2 2
2
tanh 2
1
y
i B
x y
⇒ =
+ +
2 2
1 2 coth 2x y y B⇒ + + =
OR 2 2
2 coth 2 1 0x y y B+ − + =
Example 6 : If sin( )i x iyα β− + prove that
2 2
2 2
1
cosh sinh
x y
β β
+ = and
2 2
2 2
1
sin cos
x y
α α
− =
Solution : Let sin( )i x iyα β+ = +
Then sin .cos cos .sini i x iyα β α β+ = +
sin .cosh cos sinhi x iyα β α β⇒ + = + (by hyperbolic relations)
Equating real and imaginary part
sin cosh xα β = and cos .sinh yα β =
sin
cosh
x
α
β
⇒ = and cos
sinh
y
α
β
=
and cosh
sin
x
β
α
= and cosh
sin
y
β
α
=
squaring and adding the tems of eqn
(1)
2 2
2 2
2 2
sin cos 1
cosh sinh
x y
α α
β β
+ = + =
similarly squaring and substracting equation (2)
2 2
2 2
2 2
cosh sin 1
sin cos
x y
hβ β
α α
− = − =
Example 7 : prove that sin( ) cos sini iθ φ α α+ = +
also 2
cos sinθ α±
Example 8: Prove that
A) ( )1 2
sinh log 1x x x−
= + +
B) ( )1 2
cosh log 1x x x−
= + +

19. 19
C)
1 1 1
tanh log
2 1
x
x
x
− + 
=  
− 
D) Prove that 1 1
2
tanh sinh
1
x
x
x
− −  
=  
− 
Solution (A): Let 1
sinh x y−
= ……… (1)
sinx ny⇒ =
2 2
1 sinh 1 coshx y y∴ + = + =
2
1 sinh coshx x y y⇒ + + = +
2 2
y y y y
e e e e− −
− +
= +
2
1 y
x x e∴ + + =
( )2
log 1x x y⇒ + + = ………. (2)
For eqn
(1) & (2) ( )1 2
sinh log 1x x x−
⇒ = + +
Similarly we can prove B, C & D
Logarithmic of complex number
Example 1 : Show that log( 5) log(5) (2 1)i nπ− = + +
Solution : log( 5) log( 5 ¯ 0)i− = − +
2 1 0
log ( 5) 2 tan
5
i nπ − 
= − + + − 
tan 5 (2 )i nπ π= + + [ tan 0]π =∵
log 5 (2 1)i nπ= + +
2) 2 2
log(1 ) log 1 1i+ = +
1 1
2 tan
1
i nπ − 
= +  

20. 20
log 2 2
4
i n
π
π
 
= + +  
1
log 2 (8 1)
2 4
i n
π
= + +
3) log log( ) log( )
x iy
x iy x iy
x iy
+
= + − −
−
2 2 1 2 2 1
log tan log tany yx y i x y i
x x
− − = + + − + +
  
1 1
tan tani y x y x− −
 = + 
1
2 tani y x−
=
1 1
tan ( ) tanx x− −
 − = − ∵
Objective Question :
1) coshθ is equal to __________
2) The value i
e θ
is __________
3) The value of sinix is ___________
4) The value i
e π−
is __________
5) If i
e iθ
= − the value of θis ________
6) If 1 3Z i= − then z is _________
7) The period of coshz________ for all Z
8) DMT theorem is proved for n may be ________
9) The nth
root of unity are ________ when
2 i
n
e
π
ω=
10) (cos sin ) n
iθ θ −
− = ______________