Ekeeda - Chemical Engineering- Applied Mathematics - Complex Number

Page 1

Ekeeda – Chemical Engineering Propof [[[[[ HISTORY The solution in radicals (without trigonometric functions) of a general cubic equation contains the square roots of negative numbers when all three roots are real numbers, a situation that cannot be rectified by factoring aided by the rational

root

irreducibilis).

test if

This

the

cubic

conundrum

led

is

irreducible Italian

(the

so-called casus

mathematician “Gerolamo

Cardano”to conceive of complex numbers in around 1545. Work on the problem of general polynomials ultimately led to the fundamental theorem of algebra, which shows that with complex numbers, a solution exists to every polynomial equation of degree one or higher. Complex numbers thus form an algebraically closed field, where any polynomial equation has a root. Many mathematicians contributed to the full development of complex numbers. The rules for addition, subtraction, multiplication, and division of complex numbers were developed by the Italian mathematician “Rafael Bombelli”. Complex numbers have practical applications in many fields, including Physics, Chemistry, Electrical Engineering, and Statistics. In the 18th century complex numbers gained wider use, as it was noticed that formal manipulation of complex expressions could be used to simplify calculations involving trigonometric functions. For instance, in 1730 Abraham de Moivre noted that the complicated identities relating trigonometric functions of an integer multiple of an angle to powers of trigonometric functions of that angle could be simply re-expressed by the following wellknown formula which bears his name, “DeMoivre's formula”

Definition


Ekeeda – Chemical Engineering A number in the form,

where x and y are real numbers and i is the

imaginary unit defined as

is called Complex Number and it is denoted

by Where, x is called real part & y is called imaginary part of complex number z, The complex number is purely real if imaginary part is zero and purely imaginary if real part zero. i.e. If y=0 then

is purely real.

If x=0 then

is purely imaginary.

ALGEBRA OF COMPLEX NUMBER A] Equality of Complex Number Two Complex numbers

and

are said to be equal if their real and

imaginary parts are respectively equal E.g. For i.e. For

and Or

,

If

and

have no meaning. and

B] Addition and Subtraction

To add or subtract two complex numbers, we add or subtract their real parts separately and imaginary part separately. C] Multiplication


Ekeeda – Chemical Engineering D] Division As such division by an imaginary quantity has no meaning. Therefore to make a meaningful quantity we multiply numerator and denominator by conjugate of denominator. i.e.

which is a complex number. MODULUS OR MAGNITUDE OF COMPLEX NUMBER Modulus of z is We write,

AMPLITUDE OR ARGUMENT OF COMPLEX NUMBER Amplitude or Argument of z is denoted by amp (z) or arg (z) =

.

To find argument we have the following four cases depending upon the position of a point corresponding to given complex number in a particular quadrant. 1. Given: ,if If corresponding points (x, y) lie in first quadrant Let ‘ ’ be the angle

For first quadrant; argument of 2. Given: ,if If the corresponding points (x, y) lie in second quadrant Let ‘ ’ be the angle

For second quadrant; argument of


Ekeeda – Chemical Engineering

3. Given:

,if

If the corresponding points (x, y) lie in third quadrant Let ‘

’ be the angle

For third quadrant; argument of

4. Given:

,if

If the corresponding points (x, y) lie in fourth quadrant Let ‘

’ be the angle

For fourth quadrant; argument of POLAR FORM OF COMPLEX NUMBER Polar form of complex number is EXPONENTIAL FORM OF COMPLEX NUMBER

DEMOIVRE’S THEOREM


Ekeeda – Chemical Engineering

Circular Functions of a Complex Number

Hyperbolic functions

1. 2.

3.

4.

5.

6. 7. 8. 9.


Ekeeda – Chemical Engineering 10. 11. 12. 13.

14.

15. 16. 17. 18.

19.

20.

21. 22. 23. 24. 25. 26.


Ekeeda – Chemical Engineering

27. 28. 29. Relationship between Hyperbolic & Circular Functions 1. 2. 3. 4. 5. 6. X

0 0 1 0

-1

1


Ekeeda – Chemical Engineering

Formulae 1. If

, then

2. If

then and

3. Expansion of

in powers of

4. Expansion of

in Terms of sines or cosines of Multiples of :-

5. Roots of a Complex Number:-

6. 7. 8. 9.

:-


Ekeeda – Chemical Engineering 10. 11. 12. 13. 14. 15. Inverse Hyperbolic Functions:a) b) c) � Proof a) Let

This is a quadratic in

Conventionally we take positive sign.

b) We leave this as an exercise. c) Let

By componendo and dividendo,


Ekeeda – Chemical Engineering

16. 17. 18. The general value of is given by

i.e. 19. 20. 21.

22. 23. 24. 25. 26.

is denoted with

capital by

and


Ekeeda – Chemical Engineering

CLASSWORK PROBLEMS Part I : Basics of Complex Number 1. If

prove that

(i)

,

(ii)

.

2. Find the complex conjugate of 3. Find

.

.

4. Find the modulus and argument of Argument of

[Modulus of z = 1,

]

5. If 6. If

.

, show that and

.

are any two complex numbers, prove that .

7. If

and

are two complex numbers such that

difference of their amplitudes is

.(or prove that arg

, prove the .)


Ekeeda – Chemical Engineering

Part II : DeMoivre’s Theorem 1. If n is a positive integer, prove that

2. Prove that where

is equal to -1, if

and 2, if

is an integer.

3. Show that (4n)th power of

is equal to

, Where n is a positive

integer. 4. If

, prove that

(i) (ii) (iii) (iv) 5. If

are the roots of the equation

6. If

are the roots of

7. If

are the roots of the equation

, prove that , find the equation whose roots are

, prove that


Ekeeda – Chemical Engineering 8. If

then show that

the general value of 9. If

is then show that

and

.

10. By using De Moivre’s Theorem show that

.

11. Evaluate 12. If n is a positive integer and

prove that

(i) (ii) (iii) 13. Use De Moivre’s Theorem to show that Hence deduce that 14. Show that

. .

15. Using De Moivre’s Theorem prove that

16. Expand

where

in a series of cosines of multiples of

17. If

.

, prove that .

18. Using De Moivre’s Theorem prove that,


Ekeeda – Chemical Engineering 19. Find the cube roots of unity. If

is a complex cube root of unity, prove

that 20. If is a complex fourth root of unity, prove that . 21. Prove that the n nth roots of unity are in geometric progression. 22. Show that the sum of the n nth roots of unity is zero. 23. Prove that the product of the n nth roots of unity is

.

24. Solve 25. Solve completely the equation

26. If

.

is a root of the equation

, find all the other

roots of the equation.

27. Find all the values of

28. Show that the roots of 29. Solve the equation

and show that their continued product is

are given by and show that the real part of all roots is -

1/2. 30. If

, are the roots of

, find them and show that

31. Separate into real and imaginary parts

.


Ekeeda – Chemical Engineering

32. If

, prove that

.

Part III : Exponential form of Complex Number 1. Prove that

,

2. If 3. Express

4. If

.

, find tan hx. in terms of hyperbolic sines of multiples of x.

,

, show that

(i) (ii)

5. Prove that 6. If

. , prove that (i) (iii)

(ii) (iv)


Ekeeda – Chemical Engineering

7. If

, prove that

8. If

and

.

prove that (i)

,

(ii) (iii)

.

9. If

, prove that .

10. If

, prove that

11. If

.

, prove that

12. If

, prove that

.

13. Separate into real and imaginary parts

14. If

, show that

15. If

, or if

and

. express x and y in terms of

. Hence show that

16. If

are the roots of the equation

, prove that

,

and n is an integer, prove that 17. Prove that

.

.

. Further, if .


Ekeeda – Chemical Engineering

18. Prove that

.

19. If

, where a, b are real, prove that

20. If

.

, prove that .

21. Show that 22. If (i) r = 1

. , show that

(ii)

(iii)

Part IV : Logarithmic Form Of Complex Number 1. Show that

.

2. Prove that 3. If

, prove that

4. Find the principal value of

. and show that its real part is

.

5. If 6. If 7. Prove that

, find

,

.

, prove that , where

, where

. .


Ekeeda – Chemical Engineering

8. If

,

prove

that

,

when

. 9. Considering only the principal value, if value is

, is real, prove that its

.

10. If

, prove that the general value of x is given by where

and

.

11. If

, prove that

12. Prove that

where n is any positive integer. .

13. Prove that

.

HOMEWORK PROBLEMS Part I : Basics of Complex Number 1. Express the following in the form x + iy

(i)

(ii) 2. Find the modulus and the principal value of the argument of

(i)

(ii) 3. Find the square root of

(i)

(ii) 4. If

, prove that

.


Ekeeda – Chemical Engineering

5. If

and

, find z.

6. Prove that

.

7. If

, find

(i)

(ii)

(iv)

(v)

8. If z = x + iy, prove that

9. If

, prove that

Part II : DeMoivre’s Theorem 1. Show that (i) (ii) (iii) 2.

[Ans : z = 2]

(iii) (vi)

.

.


Ekeeda – Chemical Engineering

3. Hint: 4. Prove that

5. If

, Prove that

6. If

then show that

Hint:

Also 7. If

and where

in

the

Argand’s

diagram

if

then prove that

Hint: 8. If

are three complex numbers with modulus .

Prove that (i)

(ii)

9. If Prove that (i) (ii) 10. Using De Moivre’s Theorem, prove the following. (i)

each and


Ekeeda – Chemical Engineering (ii) (iii) (iv) 11. If Find the values of

12. Prove that 13. Prove that

. Hence deduce that .

14. Prove that

.

Hint: Now Now,

Find

Similarly,

Find

15. (i) Prove that (ii) Expand

as a series of cosines of multiples of đ?œƒ .

(iii) Expand

as a series of sines of multiples of đ?œƒ .


Ekeeda – Chemical Engineering

16. (i) Express

in terms of

.

(ii) Show that 17. Show that 18. If

Prove that

19. If

.

. Show that

20. If x+1/x = 2 cos Îą, y+1/y = 2 cos β, z+1/z = 2 cos đ?›ž Show that xyz + √đ?‘Ľđ?‘Śđ?‘§ +

1 √đ?‘Ľđ?‘Śđ?‘§

= 2 cos (

) đ?‘š

21. If x -1/x = 2i sin đ?œƒ , y -1/y = 2i sin đ?›&#x; show that

22. If

√đ?‘Ľ

đ?‘›

√đ?‘Ś

+

đ?‘›

√đ?‘Ś = √đ?‘Ľ

đ?‘š

show that

23. If then by using De Moivre’s theorem simplify 24. If n is the + ve integer, show that

25. If Îą, β are the roots of quadratic equation x2- 2x+ 4 = 0, then đ?‘›đ?œ‹

(i)

Prove that ιn+ βn = 2n+1 cos ( 3 )

(ii)

Find the value of ι15+ β15Ans : -216

đ?œƒ

đ?œ™

2cos (đ?‘š − đ?‘› )


Ekeeda – Chemical Engineering

26. Find all the values of where

, k = 0,1,2,3]

27. Solve :

(i) (ii) 28. (i) x7+ x4+ x3 + 1 = 0 (ii) x10+ 11x5+ 10 = 0 (iii) x9- x5+ x4 - 1 = 0. 3đ?œ‹ 5

[Ans: -1, 1/2 Âą đ?œ‹

1√3 2

,

1 √2

đ?œ‹

Âą

1 √2

−1

,

√2

3đ?œ‹

1

Âą

]

√2

3đ?œ‹

[Ans: (-10)1/5, -1, cos 5 Âąi sin 5 , cos ( 5 ) Âąi sin ( 5 )] [Ans:Âą -1, Âąi, cos

đ?œ‹ 5

đ?œ‹

Âąi sin 5 , cos

3đ?œ‹ 5

Âąi sin

] đ?œ‹

(iv) x14+ 127x7- 128 = 0

[Ans: 2 [ cos (2k+ 1) 7 + i sin (2k+ 1)

đ?œ‹

]

7

k = o to 6] (v) x7+ x4+ i (x3+1) = 0 [Ans: -1, 1/2 Âą i

√3 , 2

Âą( cos

đ?œ‹ 8

- i sin

đ?œ‹ 8

), Âą (cos

3đ?œ‹ 8

3đ?œ‹

+ i sin

8

),] 29. Solve (i) x4- x2+ 1 = 0 [M96]

[Ans:Âą

√3 2

,

đ?‘–

Âą .] 2

(ii) x4- x3+ x2- x+ 1 = 0.

[Ans: cos

đ?œ‹

đ?œ‹

Âąi sin 5 ,cos 5

3đ?œ‹ 5

Âąi sin

3đ?œ‹ 5

]

30. Find the continued product of all the values of (i) [1+ i]2/3 (ii) [1+ i]1/5 (iii) (1+ i√3 )1/4 31. Show that the nth roots of unity are given by

[Ans: 2i ] [Ans: 1+i] [Ans: - ( 1+ i√3 )] where đ?œ† =

cos 2đ?œ‹/đ?‘› + i sin 2đ?œ‹/đ?‘›. Show that continued products of the all these nth roots is (-1)n+1


Ekeeda – Chemical Engineering 32. Prove that nth roots of unity are in geometric progression. Also find sum of nth root of unity. 33. Find the roots of

and show that the real part of all the roots is -

1/2 34. Solve

Hint : [Ans :

where

& k = 0, 1, 2]

35. Obtain the solution of the equation

Hint:

[Ans:

]

.State true of false.

[Ans: True]

where Solve

where k = 0,1, 2, 3, 4.

36. If

37. If arg (z+ 1) =

, then

and arg (z- 1) =

38. Find z if amp (z+ 2i) =

find z.

, amp (z- 2i) =

[Ans: z = 2+ i0 ]

39. If represents a point on the line 3x+ y = 0 in Argand’s diagram, find a. [Ans: a= 1 or ¾] 40. Find two complex numbers whose sum is 4 and product is 8. [Ans : z1 = 2+ 2i, z2 = 2- 2i]


Ekeeda – Chemical Engineering

41. If

where

.

Find

polar

form

of

. Hint: Divide N & D by z1 42. (a) Express

where in the form

. Find value of a & b in terms of

x and y.

(b) If 43. If

, prove that , Prove that

44. Prove that Hint: Let

45. If 46. If

where

and

, prove that , prove that

47. Prove that 48. If number.

. Prove that z lies on imaginary axis where z is a complex


Ekeeda – Chemical Engineering

Part III : Exponential form of Complex Number 2

1. If z = x+ iy and � � = a+ ib. Find the a and b. 2

2

Hint : a + ib = � � = � (�+��) = � � Ans: a = � �

2 −đ?‘Ś 2

cos 2xy, b = đ?‘’ đ?‘Ľ

2. If

2 −đ?‘Ś 2 +12đ?‘Ľđ?‘Ś

2 −đ?‘Ś 2

sin 2xy

find R and đ?›&#x;.

Ans: 3. If p = a + ib, q = a - ib where a and b are real then prove that pep + qeq is real. 4. Prove that (1 - đ?‘’ đ?‘–đ?œƒ )-1/2 + (1 - đ?‘’ đ?‘–đ?œƒ )-1/2 = ( 1 + cosec đ?œƒ/2)1/2.

5. Prove that ( 1 - sec đ?œƒ/2 )1/2 = ( 1 + đ?‘’ đ?‘–đ?œƒ )-1/2 - (1 + đ?‘’ đ?‘–đ?œƒ )-1/2 6. Show that

đ?‘ đ?‘–đ?‘›đ?œƒ 2

+

đ?‘ đ?‘–đ?‘›2đ?œƒ 22

+

đ?‘ đ?‘–đ?‘›3đ?œƒ 23

+ ‌‌‌‌..=

2đ?‘ đ?‘–đ?‘›đ?œƒ 5−4đ?‘?đ?‘œđ?‘ đ?œƒ

7. Solve the equation 7 cosh x + 8 sinh x = 1 for real values of x. 8. If tanh x = 1/2, find sinh 2x and cosh 2x

[Ans: -log 3]

[Ans: 4/3, 5/3]

9. If x = tanh-1 (0.5). show that sinh 2x = 4/3 Hint: sinh 2x = 2 tanh x/ 1tanh2 x 10. Prove that tanh (log√3 ) = 1/2. Hint: use definition of tanhx. 11. Prove that 16 sinh5 x = sinh 5 x – 5 sinh 3x + 10 sinh x. 12. Prove that 32 cosh6x- 10 = cosh 6x+ 6 cosh 4x+ 15 cosh 3x. 13. If cosh6x= a cosh 6x + b cosh 4x + c cosh 2x + d, prove that 5a+ 5b+ 3c- 4d = 0

14. Prove that 1+ ������ ℎ�

15. Prove that (i) [1−đ?‘Ąđ?‘Žđ?‘›â„Žđ?‘Ąđ?‘Žđ?‘›â„Ž đ?‘Ľ ]n = cosh2nx + sinh2nx (ii) (cos hx – sin hx)n = cosh nx – sinh nx đ?‘?đ?‘œđ?‘ â„Žđ?‘?đ?‘œđ?‘ â„Ž đ?‘Ľ+đ?‘ đ?‘–đ?‘›â„Žđ?‘ đ?‘–đ?‘›â„Ž đ?‘Ľ

16. Prove that [ đ?‘?đ?‘œđ?‘ â„Žđ?‘?đ?‘œđ?‘ â„Ž đ?‘Ľâˆ’đ?‘ đ?‘–đ?‘›â„Žđ?‘ đ?‘–đ?‘›â„Ž đ?‘Ľ ]n = cosh 2nx + sinh 2nx 17. If log ( tan x) = y, prove that (i) sinh ny = 1/2 (tannx – cotnx)


Ekeeda – Chemical Engineering (ii) 2 cosh ny cosec 2x = cosh (n+ 1) y + cosh (n- 1) y 18. If sin (đ?œƒ + iđ?›&#x;) = đ?‘’ đ?‘–đ?›ź , prove that sin đ?›ź = Âą cos2đ?œƒ = Âąsinh2đ?›&#x; 19. If cosh (đ?œƒ + iđ?›&#x;) = đ?‘’ đ?‘–đ?›ź , prove that sin2đ?›ź = sin4đ?›&#x; = sinh4đ?œƒ 20. If

prove that

21. If cos (x+iy) = eiĎ€/6, Prove that (i) 3sin2x-cos2x = 4sin2x.cos2x (ii)3sinh2y + cosh2y = 4sinh2y.cosh2y 22. If log [cos(x-iy)] = Îą + iβ, prove that Îą =

log

and find β.

23. If sin-1(ι+iβ) = Ν + iΟ. Prove that sin2 Ν and cosh2 Ο are the roots of the equations

x2 – (1+ Îą2 + β2)x + Îą2 = 0

24. Let P(z) where z = sin(ι+iβ). If ι is variable, show that the locus of the P(z) is

an ellipse

. Also show that x2 cosec2Îą – y2sec2 Îą = 1 if β is

variable. 25. If sinh (x+ iy) = eiĎ€/3,prove that (i) 3cos2y – sin2y = 4sin2y cos2y (ii)3sinh2 x + cosh2 x = 4sinh2x.cosh2 x 26. If u+ i v = cosh ( đ?›ź + i đ?œ‹/4 ).Find the value of u2 – v2

[Ans : ½]

27. If x+ iy = 2 cosh (đ?›ź+ i đ?œ‹/3), prove that 3x2- y2 = 3 28. If x = 2 sin đ?›ź cosh β, y = 2 cos đ?›źsinh β , Show that (i) cosec(đ?›ź − đ?‘– β ) + cosec (đ?›ź + đ?‘– β ) = (ii)cosec(đ?›ź − đ?‘– β ) - cosec (đ?›ź + đ?‘– β ) =

4đ?‘Ľ đ?‘Ľ 2 + đ?‘Ś2 4đ?‘–đ?‘Ś đ?‘Ľ2+ đ?‘Ś2

đ?œ‹

29. If tan( 6 + đ?‘–đ?›ź) = x+ iy, prove that x2+ y2 + 2x/√3 = 1. đ?œ‹

30. If cot ( 6 + đ?‘–đ?›ź) = x+ iy, prove that x2+ y2 - 2x/√3 = 1 31. Show that tan

=

32. If tan h (đ?›ź +i β ) = x+ iy, prove that x2+ y2- 2x cot 2 đ?›ź= 1, x2+ y2+ 2y coth


Ekeeda – Chemical Engineering 2 β + 1 = 0. đ?œ‹

33. If cot (đ?›ź +i β ) = i. Prove that β = 4 , đ?›ź = 0 34. If đ?›ź +i β = tan h ( x + i

đ?œ‹ 4

), prove that �2 + � 2 = 1

35. If tan h (a+ ib )= x+ iy, Prove that x2+ y2- 2x coth 2� + 1 = 0&x2+ y2+ 2y coth 2� - 1 = 0 36. If

, Show that

37. If

. Prove that

38. Separate into real and imaginary parts, (i) sec (x+ iy) (ii) tanh (x+ iy)

39. Show that (i) sinh-1 x = cosh-1 (√1 + đ?‘Ľ 2 ) (ii) tanh-1 (đ?›&#x;) = sinh-1 (

đ?œ™

√1−đ?œ™2

)

(iii) Prove that tanh-1 (sinđ?œƒ ) = cosh-1 (secđ?œƒ ). 40. Show that sech-1 (sinđ?œƒ ) = log (cot đ?œƒ/2) 41. Show that sinh-1 (tan x) = log [ tan ( 42. Prove that cosech-1 z = log (

1+√1+đ?‘§ 2 đ?‘§

đ?œ‹ 4

đ?‘Ľ

+ 2) ]

).Is defined for all values of z?

43. Show that cos-1 z = - i log ( zÂąâˆšđ?‘§2 − 1 ) 44. If cosh-1 a + cosh-1 b = cosh-1 x, then prove that a √đ?‘? 2 − 1 + b √đ?‘Ž2 − 1= √đ?‘Ľ 2 − 1. 45. If cosh-1 (x+ iy) + cosh-1 (x- iy) = cosh-1 a, prove that 2(a- 1) x2+ 2(a+ 1) y2 = a2- 1. 46. If A+ iB = C tan (x+ iy), prove that tan 2x =

2đ??śđ??´ đ??ś 2 −đ??´2 −đ??ľ2

47. Separate tan-1 (cosđ?œƒ + i sinđ?œƒ ) into real and imaginary parts 48. If tan (đ?œƒ + iđ?›&#x;) = cos đ?›ź + i sin đ?›ź, show that đ?œƒ =

đ?‘›đ?œ‹ 2

+

đ?œ‹ 4

1+đ?‘ đ?‘–đ?‘›đ?›ź

, đ?›&#x; = Âź log (1−đ?‘ đ?‘–đ?‘›đ?›ź)


Ekeeda – Chemical Engineering 49. If tan (đ?œƒ + i đ?›&#x; ) = đ?‘’ đ?‘–đ?›ź show that đ?œƒ = ( n+ 1/2) đ?œ‹/2 and đ?›&#x; = 1/2 log tan (đ?œ‹/4 + đ?›ź/2 ) 50. Separate into real and imaginary parts : tan-1(a+ iy)or Prove that tan-1 (a+ iy) = 1/2 tan-1 ( 2a/1- a2- y2 ) + i/4 log |

(1+đ?‘Ś)2 +đ?‘Ž2 (1−đ?‘Ś)2 +đ?‘Ž2

|

51. Prove that one value of tan-1 (x+ iy/x- iy) is đ?œ‹/4 + đ?‘–/2 log x+ y/x- y where x > y > 0. 52. If tan (x+ iy) = i, x, y ∈R. Show that x is indeterminate and y is infinite. Hint: tan(x- iy) – I, then tan 2x=tan[ (x+ iy)+(x- iy)] & tan 2iy = tan [(x+ iy)-(x-iy)] 53. If tan (u+ iv) = x+ iy then prove that curves u = constant and v = constant are families of circles.


Ekeeda – Chemical Engineering

Part IV : Logarithmic Form of Complex Number

1. Show that 2. Solve for z if 3. 4. Prove that 5. Prove that

.

6. Show that

7. Prove that 8. Show that 9. Show that 10. If

, prove that (i) (ii)

11. Separate into real and imaginary parts :

(i) (ii)


Ekeeda – Chemical Engineering

(iii) (iv)

12. Separate into real and imaginary parts

(consider principal

values only)

13. Prove that the real value of principal of 14. Prove that the general values of

is is Hence find the principal value.

15. If

, show that


Ekeeda – Chemical Engineering

EXTRA SOLVED PROBLEMS Q.1. If Z1 , Z2 are non-zero complex numbers of equal modulus and Z1 ≠Z2 then prove that

is purely imaginary.

Solution: Since Z1 and Z2 are two complex numbers with equal modulus (say r), Let &


Ekeeda – Chemical Engineering

Also

Dividing (i) by (ii) we get,

which is purely imaginary.

Q.2. If

prove that

(i) Solution: Now (i)

Then

(ii)


Ekeeda – Chemical Engineering

[For G.P., Sum= (ii) Also

Q.3. If Solution: (i) L.H.S.

prove that

and

]


Ekeeda – Chemical Engineering

(ii)

L.H.S.

[Multiplying Numerator& Denominator by i]

Q.4. If

Prove that (i) Solution: Now

(ii)


Ekeeda – Chemical Engineering

Comparing both sides, we get,

Squaring and adding (i) & (ii) we get,

Dividing (ii) by (i) we get,

Q.5. If

,

and

, prove that


Ekeeda – Chemical Engineering Solution: Now

,

,

Similarly

[As above]

Subtracting (ii) from (i) we get,


Ekeeda – Chemical Engineering

Q.6. Prove that Solution: Let

Then

[Multiplying N & D by -i]

From (i) & (ii) we get,


Ekeeda – Chemical Engineering

Q.7. If

find the value of a, b, c.

Hence show that Solution: Now

Comparing imaginary part on both sides, we get, Comparing above equation with the given equation we get,

Deduction:


Ekeeda – Chemical Engineering

Q.8. If

,prove that

Solution: Let Also

Then

,

Hence

Comparing this with the given equation we get,

,


Ekeeda – Chemical Engineering

Q.9. Show that the Solution:

Now

Hence

power of

is

where n is a positive integer.


Ekeeda – Chemical Engineering

Q.10. Find the roots common to

and

.

Solution: We have [General polar form]

Putting k=0,1,2,3,4,5, we get the roots as,

Also [General polar form]


Ekeeda – Chemical Engineering

Putting k=0,1,2,3, we get the roots as,

From (I) and (II) we get the common roots as

Q.11. If

are the roots of

, find their values and show

that Solution: Now [Multiplying both sides by (x-1)]

where k=0,1,2,3,4. When k=0,

Root

When k=1,

Root

(say)

When k=2,

Root

(say)

When k=3,

Root

(say)


Ekeeda – Chemical Engineering

When k=4,

Root

Since

are the roots of

Putting x=1,

we get,

Note: Hence

Q.12. Prove that Solution: Consider

When k= 0, k = 1,

k=2,

k=3,

(say) , we have


Ekeeda – Chemical Engineering

k= 4,

Also

From(I) and (II) we have

Q.13. If Solution: (i) Now

Then

then

prove

that

(i)

(ii)


Ekeeda – Chemical Engineering

(ii) Now

Comparing both sides we get,

Then

Also


Ekeeda – Chemical Engineering From (I) and (II) we have,

Q.14. If Solution:

then show that


Ekeeda – Chemical Engineering Now

,

Adding two equations we get,

……………………(i) Similarly subtracting we get,

[as above]

……………………(ii) From (i) & (ii) we get, Alternately,


Ekeeda – Chemical Engineering

Comparing both sides we get, ……………………(i) And ……………………(ii) From (i) and (ii) we get,

Q.15. If log (tan x) = y then prove that (i)


Ekeeda – Chemical Engineering (ii) Solution: Now log

(i) sinh ny

(ii)

Alternately,

But


Ekeeda – Chemical Engineering

Subs .in (I) we get,


Ekeeda – Chemical Engineering

Q.16. If Solution: Let Now

& Then

Also i tanh

Prove that


Ekeeda – Chemical Engineering

Subs. in (I) we get, z = Q.17. Find the sum of the series Solution: Let ……… .....

………

[

for a Geom. Series]


Ekeeda – Chemical Engineering

Equating the imaginary parts, we get, Equating the real parts, we get,

Q.18. If u + iv =

prove that

Solution:

Now

=

=

= =

= =

and


Ekeeda – Chemical Engineering

= = =

But

………(i)

= =

[Subs. from (i)]

= Comparing both sides we get,

Q.19. Find the value of log [ sin(x+ iy) ] Solution: sin (x+ iy) = sin x.cos iy + cos x. sin iy = sin x. cosh y + i cos x. sinh y [sin(x+ iy)] = log (sin x. cosh y + i cos x. sinh y) = =

= = =


Ekeeda – Chemical Engineering

Q.20. If Solution:

, find

and

Now

…… (I) But

Subs in (I) we get,

(say)

Hence

,

Q.21. Prove that Solution: Let a – b = x, a + b = y

Then

where


Ekeeda – Chemical Engineering

Q.22. Show that if

has real values then one of them is

Solution:

For the given expression to be real we must have ………(i) Then value of expression [ Subs. from(i) ]

Q.23. If

Then show that the general value of θ is Solution: Now

……[By comparing imaginary parts]


Ekeeda – Chemical Engineering

Q.24. If Z1 , Z2 and Z3, Z4 are two pairs of conjugate complex numbers

Then show that (i)

(ii)

Solution: Let

and and

Then &

……(i)

&

………(ii)

Also

Hence from (i) & (ii), &

Q.25. If

Show that

Solution: Let

But

…..(given)

=1


Ekeeda – Chemical Engineering

Q.26. If

Show that (i)


Ekeeda – Chemical Engineering

(Putting n = p – q) (ii) Solution: Let Where &

p=1,2,3…..n

Let Where & Now,

(given)

(i) Comparing amplitude we get,

i.e. (ii) Comparing modulii we get,

∴ ∴

Q.27. Prove that Solution:

(Squaring both thet sides)


Ekeeda – Chemical Engineering Let

Then &

Hence Q.28. If

then prove that

Solution: Let

,

Comparing both the sides

Then

Q.29. If Solution: Now

and

then prove that


Ekeeda – Chemical Engineering

Then

But

[By equal ratio theorem] From (i) & (ii) we get, Q.30. Find two complex numbers such that their difference is 10i and their product is 29 Solution: Since that difference between two complex numbers is imaginary and their product is real, The two numbers must be conjugates Let the numbers be

and

Now

Also

Hence the two numbers are Q.31. If

and

&

Or , find z

&


Ekeeda – Chemical Engineering Solution: Let

Now

Also

From (i) & (ii) we get, Then


Ekeeda – Chemical Engineering

Q.32. If

then Show that

(i) (ii) Solution: Now Putting

we get,

(i) Comparing real parts we get, (ii) comparing imaginary part we get,


Ekeeda – Chemical Engineering

Q.33. If

then prove that

Solution:

Now

, Hence Then

,


Ekeeda – Chemical Engineering

Q.34. If

and

show that

Solution: Now

Let But

………(i) ………(ii) Then

using (i) and (ii) Q.35. If Show that

and


Ekeeda – Chemical Engineering

Solution: Let

Comparing both that sides we get, Q.36. If

Prove that Solution:

Now

Hence

Then (i) + (ii) gives, Q.37. If Solution:

then show that


Ekeeda – Chemical Engineering

Q.38. Prove that Solution: L.H.S.


Ekeeda – Chemical Engineering

Q.39. If

prove that

Solution: Now

Using componendo - dividendo


Ekeeda – Chemical Engineering

Dividing N & D by

Q.40. Using De Moivre’s Theorem show that

where

Solution: Now, Expanding R.H.S. by Binomial Theorem and Comparing real parts we get,

Squaring both the sides


Ekeeda – Chemical Engineering

Q.41. Show that Solution: Now

Multiplying N & D by -i

Where

&

.

Q.42. If

are the root of the equation

Hence, deduce that Solution: Now are its roots we have

Let

Hence

, prove that


Ekeeda – Chemical Engineering

Similarly Hence Putting Q.43. If

,

We get

are the roots of

Solution:

Now

Adding we get,

Q.44. Find the continued product of Solution: Now

When When When

. Prove that


Ekeeda – Chemical Engineering

When Continued product od all the values

Now 1+3+5+…….

is an A.P. with

Its

Hence, Required Product of Values

Q.45. Find the cube root of Solution: Let


Ekeeda – Chemical Engineering

When

When

When

Q.46. Solve Solution: Now

Multiply by (x+1) on both the side


Ekeeda – Chemical Engineering

When k=0,

Root

When k=1,

Root

When k=2,

Root

When k=3,

Root

When k=4, Discarded,

Root (as we have taken it in the equation)

Also & Hence Required Roots are

,


Ekeeda – Chemical Engineering Q.47. Given that

is one root of the equation

the other roots. Solution: Since

is one root of the equation. is the other root.

The equation with this root is

For finding the other factor we have to divide Then

Hence the required roots are

.

. Find


Ekeeda – Chemical Engineering

Q.48. Show that all the roots of

are given by

Solution:

[By Componendo-Dividendo]


Ekeeda – Chemical Engineering

[Multiplying N & D by - 1]

When

not defined (Hence discard)

Hence the solution are given by

where


Ekeeda – Chemical Engineering

Q.49. Show that the points representing the roots of the equation on Argand’s diagram are collinear. Solution:

Now


Ekeeda – Chemical Engineering

[Multiplying N & D by-i]

For k = 0, 1, 2 we get three values of z. All these values have the same real parts i.e. Hence the points represented by the 3 numbers are collinear.

Q.50. If and n is an integer, prove that multiple of 3. Solution: Now

Similarly Hence If n is not a multiple then, Let Where

where k is an integer

is not a


Ekeeda – Chemical Engineering

Value of expression

When Values of expression

Subs in (i) we get,

Q.51. Show that

Solution:


Ekeeda – Chemical Engineering

Q.52. Show that

Solution: L.H.S.

Q.53. If

Prove that

Solution: Now

(i) Now But (eq. of ellipse is

is constant)


Ekeeda – Chemical Engineering

(ii) Also But (eq. of hyperbola is Q.54. Show that Solution: Let

Comparing both sides, we get From (i), Also

Hence

Q.55. Prove that Solution: (i)Let

From (i) & (ii) we have, (ii) Now

&

is constant)


Ekeeda – Chemical Engineering

From (i) & (ii) we have, Q.56. Prove that Solution: Let

From (i) & (ii) we have, Q.57. Prove that

Solution: Let

hence deduce that


Ekeeda – Chemical Engineering

[By Componendo - Dividendo]

From (i) & (ii) we get, Putting

Q.58. If Solution: Now Let

Then R.H.S.

and

resp. in (i) and then adding we get,

then prove that


Ekeeda – Chemical Engineering

Q.59. If

a prove that

Solution: Let

&

Adding we get Subtracting we get Also

[given]

T.P.T. i.e.

Q.60. If (i) (ii)

(iii) Solution:

[Dividing by

Prove that

]


Ekeeda – Chemical Engineering (i) Now

(ii)

(iii) Now,

Q.61. If (ii) Solution: (i) Now

Prove that (i) (iii)


Ekeeda – Chemical Engineering

(ii)

[from (i)]

Where

Hence (iii) Now

[By componendo-Dividendo] Hence Q.62. Find the sum of the series ……… to n terms Solution:


Ekeeda – Chemical Engineering

Let

n terms terms

[By Binomial Expression of

]

[By De Moivre’s Theorem]

Equating imaginary parts we get,

Q.63. Prove that Solution: Let


Ekeeda – Chemical Engineering

Comparing real parts we get

Q.64. If Solution: Now

Comparing both the sides

& Dividing (ii) by (i)

Q.65. Prove that Solution:


Ekeeda – Chemical Engineering Now

Then Q.66. Prove that Solution: Now

Hence Q.67. If

prove that

Solution: Now

Taking log (general of both sides) we get,

Hence


Ekeeda – Chemical Engineering

Q.68. If

prove that

Solution: Now

But

Q.69. If

prove that

Solution: Now

Comparing imaginary parts of both the side we get,


Ekeeda – Chemical Engineering

Q.70. If

prove that

Solution: Now

(i) Comparing imaginary parts we get

(ii) Comparing imaginary parts we get

Q.71. If Solution: Now Then

But

show that


Ekeeda – Chemical Engineering

Hence Q.72. Find the principle value of

& show that it is purely real if

is multiple of Solution: Now

If

is entirely real then (i.e. multiple of

)


Ekeeda – Chemical Engineering

Q.73. If

, prove that the general value of x is given by

Where Solution: If

Taking log (general) of both sides we get,

Comparing both side we get,

Then

Also

gives,

gives,


Turn static files into dynamic content formats.

Create a flipbook
Issuu converts static files into: digital portfolios, online yearbooks, online catalogs, digital photo albums and more. Sign up and create your flipbook.