dr. nirav vyas complexnumbers1.pdf

8/20/2019 Dr. Nirav Vyas complexnumbers1.pdf

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Complex Numbers

N. B. Vyas

Department of Mathematics,

Atmiya Institute of Tech. and Science,Rajkot (Guj.)

N. B. Vyas Complex Numbers


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Definition of Complex Number

Complex Numbers

A number of the form x + iy, where x and y are

real numbers and i = √ −1 is called a complexnumber and is denoted by z . It is also denotedby an ordered pair(x, y).



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Complex Numbers

A number of the form x + iy, where x and y are

real numbers and i = √ −1 is called a complexnumber and is denoted by z . It is also denotedby an ordered pair(x, y).

Thus z = x + iy or z = (x, y)



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The set of complex numbers is denoted by .



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If z = x + iy is a complex number, then



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If z = x + iy is a complex number, thenx is called the real part of z and denoted byRe(z ).


C N


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If z = x + iy is a complex number, thenx is called the real part of z and denoted byRe(z ).y is called the imaginary part of z and is

denoted by Im(z ).


D C N


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denoted by Im(z ).

If x = 0 and y = 0 then z = 0 + iy = iyis called a purely imaginary number.


D C N


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denoted by Im(z ).If x = 0 and y = 0 then z = 0 + iy = iyis called a purely imaginary number.

If x = 0 and y = 0 then z = x + i0 = xis called a real number.


D C N


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denoted by Im(z ).If x = 0 and y = 0 then z = 0 + iy = iyis called a purely imaginary number.

If x = 0 and y = 0 then z = x + i0 = xis called a real number.

If x = 0 and y = 0 then z = 0 + i.0 = 0is the zero complex number.


Definition of Conjugate Complex Number


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Conjugate Complex Numbers

If two complex numbers differ only in the sign of theimaginary part then they are called conjugatecomplex numbers.




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Thus x + iy and x − iy are conjugate complexnumbers.




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The conjugate of a complex number z is denoted

by z̄ .




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The conjugate of a complex number z is denoted

by z̄ .The conjugate of real number is the real numberitself.


Properties of Complex Numbers


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1 If x + iy = 0 then x = 0, y = 0




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1 If x + iy = 0 then x = 0, y = 0

2 If x1 + iy1 = x2 + iy2 then x1 = x2 and y1 = y2




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1 If x + iy = 0 then x = 0, y = 0

2 If x1 + iy1 = x2 + iy2 then x1 = x2 and y1 = y23 If x1 + iy1 = x2 + iy2 then x1 − iy1 = x2 − iy2




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1 If x + iy = 0 then x = 0, y = 0

2 If x1 + iy1 = x2 + iy2 then x1 = x2 and y1 = y23 If x1 + iy1 = x2 + iy2 then x1 − iy1 = x2 − iy24 Sum, difference and quotient(division) of any two

complex numbers is a complex number.




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If z 1

= x1 + iy

1 and z

2 = x

2 + iy

2 then their




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If z 1

= x1 + iy

1 and z

2 = x

2 + iy

2 then their

Sum: z 1 + z 2 = (x1 + iy1) + (x2 + iy2)




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If z 1 = x1 + iy1 and z 2 = x2 + iy2 then their

Sum: z 1 + z 2 = (x1 + iy1) + (x2 + iy2)= (x1 + x2) + i(y1 + y2)




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Difference: z 1 − z 2 = (x1 + iy1) − (x2 + iy2)




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Difference: z 1 − z 2 = (x1 + iy1) − (x2 + iy2)= (x1 − x2) + i(y1 − y2)




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Product: z 1.z 2 = (x1 + iy1).(x2 + iy2)




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Product: z 1.z 2 = (x1 + iy1).(x2 + iy2)= (x1x2 − y1y2) + i(x1y2 + x2y1)


Polar form of a Complex Number


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Let P (x, y) be the point which representsz = (x, y) = x + iy




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Let P (x, y) be the point which representsz = (x, y) = x + iyLet OP = r and ∠P OM = θ.




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Let P (x, y) be the point which representsz = (x, y) = x + iyLet OP = r and ∠P OM = θ.Then from ∆OP M .




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Let P (x, y) be the point which representsz = (x, y) = x + iyLet OP = r and ∠P OM = θ.Then from ∆OP M .x = OM = rcosθ. y = P M = rsinθ.




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Let P (x, y) be the point which representsz = (x, y) = x + iyLet OP = r and ∠P OM = θ.Then from ∆OP M .x = OM = rcosθ. y = P M = rsinθ.∴ z = x + iy = r(cosθ + isinθ)




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r is called the absolute value or the modulus of z and is denoted by

|z |.




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|z |.

∴ r = |z | = x2 + y2 = √ z.z̄




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|z

|.

∴ r = |z | = x2 + y2 = √ z.z̄ Geometrically, |z | is the distance of point z fromthe origin.

θ is called the argument of z or amplitude of z.




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|z

|.


θ is called the argument of z or amplitude of z.It is denoted by argz or Ampz




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|z

|.



∴ θ = argz = tan−1 y

x




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|z

|.



∴ θ = argz = tan−1 y

xθ is the directed angle from the positive X-axis toOP.




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|z

|.



∴ θ = argz = tan−1

y

xθ is the directed angle from the positive X-axis toOP.

The value of θ lies in the interval −π < θ ≤ π iscalled principal value.


Multiplication and Division of Complex

Numbers in Polar Form


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Let z 1 = r1(cosθ1 + isinθ1) and z 2 = r2(cosθ2 + isinθ2)be two complex numbers in polar form. Then





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Product: z 1z 2 = r1(cosθ1 + isinθ1)r2(cosθ2 + isinθ2)





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Product: z 1z 2 = r1(cosθ1 + isinθ1)r2(cosθ2 + isinθ2)= r1r2[(cosθ1cosθ2

−sinθ1sinθ2) + i(cosθ1sinθ2 +

sinθ1cosθ2)]





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sinθ1cosθ2)]= r1r2[cos(θ1 + θ2) + isin(θ1 + θ2)]





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sinθ1cosθ2)]= r1r2[cos(θ1 + θ2) + isin(θ1 + θ2)]which shows modulus and amplitude of the product of

two complex numbers z 1 and z 2 is product of theirmoduli ( i.e. r1r2 ) and sum of their arguments ( i.eθ1 + θ2 ) respectively.





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Division: z 1

z 2=

r1(cosθ1 + isinθ1)






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Division: z 1

z 2=


r2(cosθ2 + isinθ2)=

r1(cosθ1 + isinθ1)(cosθ2 − isinθ2)r2(cosθ2 + isinθ2)(cosθ2 − isinθ2)





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Division: z 1

z 2=




= r1

r2[(cosθ1cosθ2 + sinθ1sinθ2) + i(sinθ1cosθ2

−cosθ1sinθ2)]





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Division: z 1

z 2=




= r1


−cosθ1sinθ2)]

= r1

r2[cos(θ1 − θ2) + isin(θ1 − θ2)]





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Division: z 1

z 2=




= r1


−cosθ1sinθ2)]

= r1

r2[cos(θ1 − θ2) + isin(θ1 − θ2)]

which shows that the modulus and amplitude of thequotient of two complex numbers z 1 and z 2 is the

quotient of their moduli ( i.e. r1

r2) and difference of

their arguments( i.e. θ1

−θ2 ) respectively.


Exponential form of Complex Numbers


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We know that

sinθ = θ− θ3

3! +

θ5

5! − θ7

7! . . .




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We know that

sinθ = θ− θ3

3! +

θ5

5! − θ7

7! . . .

cosθ = 1 − θ2

2! +

θ4

4! − θ

6

6! . . . and




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We know that

sinθ = θ− θ3

3! +

θ5

5! − θ7

7! . . .

cosθ = 1 − θ2

2! +

θ4

4! − θ

6

6! . . . and

eθ = 1 + θ + θ2

2! +

θ3

3! +

θ4

4! . . .




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We know that

sinθ = θ− θ3

3! +

θ5

5! − θ7

7! . . .

cosθ = 1 − θ2

2! +

θ4

4! − θ

6

6! . . . and

eθ = 1 + θ + θ2

2! +

θ3

3! +

θ4

4! . . .

Now θ = iθ




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We know that

sinθ = θ− θ3

3! +

θ5

5! − θ7

7! . . .

cosθ = 1 − θ2

2! +

θ4

4! − θ

6

6! . . . and

eθ = 1 + θ + θ2

2! +

θ3

3! +

θ4

4! . . .

Now θ = iθ

eiθ = 1 + iθ + (iθ)2

2! +

(iθ)3

3! +

(iθ)4

4! . . .




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We know that

sinθ = θ− θ3

3! +

θ5

5! − θ7

7! . . .

cosθ = 1 − θ2

2! +

θ4

4! − θ

6

6! . . . and

eθ = 1 + θ + θ2

2! +

θ3

3! +

θ4

4! . . .

Now θ = iθ

eiθ = 1 + iθ + (iθ)2

2! +

(iθ)3

3! +

(iθ)4

4! . . .

eiθ = 1 + iθ

−

θ2

2! −iθ3

3!

+ θ4

4!

. . .

{∵ i2 =

−1, i3 =

−i, i4 = 1




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We know that

sinθ = θ− θ3

3! +

θ5

5! − θ7

7! . . .

cosθ = 1 − θ2

2! +

θ4

4! − θ

6

6! . . . and

eθ = 1 + θ + θ2

2! +

θ3

3! +

θ4

4! . . .

Now θ = iθ

eiθ = 1 + iθ + (iθ)2

2! +

(iθ)3

3! +

(iθ)4

4! . . .

eiθ = 1 + iθ

−

θ2

2! −iθ3

3!

+ θ4

4!

. . .

{∵ i2 =

−1, i3 =

−i, i4 = 1

eiθ =

1 − θ

2

2! +

θ4

4! − . . .

+ i

θ − θ

3

3! +

θ5

5! − . . .




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We know that

sinθ = θ− θ3

3! +

θ5

5! − θ7

7! . . .

cosθ = 1 − θ2

2! +

θ4

4! − θ

6

6! . . . and

eθ = 1 + θ + θ2

2! +

θ3

3! +

θ4

4! . . .

Now θ = iθ

eiθ = 1 + iθ + (iθ)2

2! +

(iθ)3

3! +

(iθ)4

4! . . .

eiθ = 1 + iθ

−

θ2

2! −iθ3

3!

+ θ4

4!

. . .

{∵ i2 =

−1, i3 =

−i, i4 = 1

eiθ =

1 − θ

2

2! +

θ4

4! − . . .

+ i

θ − θ

3

3! +

θ5

5! − . . .

eiθ = (cosθ + isinθ)


Laws of Complex Numbers


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If z 1 and z 2 are two complex numbers, then




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1 Triangle Inequality: |z 1 + z 2| ≤ |z 1| + |z 2|




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1 Triangle Inequality: |z 1 + z 2| ≤ |z 1| + |z 2|2 |z 1 − z 2| ≥ | |z 1| − |z 2||




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1 Triangle Inequality: |z 1 + z 2| ≤ |z 1| + |z 2|2 |z 1 − z 2| ≥ | |z 1| − |z 2||3 Parellelogram equality:|z 1 + z 2|2 + |z 1 − z 2|2 = 2(|z 1|2 + |z 2|2)




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4 |z 1z 2| = |z 1||z 2|




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4 |z 1z 2| = |z 1||z 2|5 z 1z 2 = |

z 1

||z 2|


Ex


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1 Find complex conjugate of 3 + 2i

1 − i


Theorem


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DeMoivre’s TheoremIf n is a rational number than the value or one of thevalues of (cosθ + isinθ)n is cosnθ + isinnθ


Theorem


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∴ z = x + iy = r(cosθ + isinθ)


Theorem


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∴ z = x + iy = r(cosθ + isinθ)z n = rn(cosθ + isinθ)n


Theorem


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= rn(cosnθ + isinnθ)


Theorem


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= rn(cosnθ + isinnθ)E.g.:


Theorem


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(cosθ + isinθ)2 = cos2θ + isin2θ


Theorem


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(cosθ + isinθ)2 = cos2θ + isin2θ

(cosθ + isinθ)−1 = cos(−θ) + isin(−θ) = cosθ − isinθ


Examples


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Ex. Simplify

(cos2θ + isin2θ)23 (cosθ

−isinθ)2

(cos3θ − isin3θ)2(cos5θ − isin5θ)13


Examples

Ex. Evaluate (1 + i√

3)90 + (1 − i√ 3)90


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3) + (1 i√

3)


Examples


3)90 + (1 − i√ 3)90


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( +√

) + (√

)

Sol. Let x = 1 and y =√

3


Examples


3)90 + (1 − i√ 3)90


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(√

) (√

)


3

r =

x2 + y2 = √ 1 + 3 = √ 4 = 2


Examples


3)90 + (1 − i√ 3)90


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( ) ( )


3

r =

x2 + y2 = √ 1 + 3 = √ 4 = 2θ = tan−1

yx

= tan−1

√ 3

1

=

π

3


Examples


3)90 + (1 − i√ 3)90


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( ) ( )


3

r =

x2 + y2 = √ 1 + 3 = √ 4 = 2θ = tan−1

yx

= tan−1

√ 3

1

=

π

3

∴ 1 + i√ 3 = 2(cosπ

3 + isinπ

3 )


Examples


3)90 + (1 − i√ 3)90


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3

r =

x2 + y2 = √ 1 + 3 = √ 4 = 2θ = tan−1

yx

= tan−1

√ 3

1

=

π

3

∴ 1 + i√ 3 = 2(cosπ

3 + isinπ

3 ) and 1 − i√ 3 = 2(cosπ

3 − isinπ

3 )


Examples


3)90 + (1 − i√ 3)90


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3

r =

x2 + y2 = √ 1 + 3 = √ 4 = 2θ = tan−1

yx

= tan−1

√ 3

1

=

π

3

∴ 1 + i√ 3 = 2(cosπ

3 + isinπ

3 ) and 1 − i√ 3 = 2(cosπ

3 − isinπ

3 )∴ (1 + i

√ 3)90 + (1 − i√ 3)90


Examples


3)90 + (1 − i√ 3)90


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3

r =

x2 + y2 = √ 1 + 3 = √ 4 = 2θ = tan−1

yx

= tan−1

√ 3

1

=

π

3

∴ 1 + i√ 3 = 2(cosπ

3 + isinπ

3 ) and 1 − i√ 3 = 2(cosπ

3 − isinπ

3 )∴ (1 + i

√ 3)90 + (1 − i√ 3)90

=

2(cosπ3 + isinπ

3 )90

+

2(cosπ3 − isinπ3 )90


Examples


3)90 + (1 − i√ 3)90


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3

r =

x2 + y2 = √ 1 + 3 = √ 4 = 2θ = tan−1

yx

= tan−1

√ 3

1

=

π

3

∴ 1 + i√ 3 = 2(cosπ

3 + isinπ

3 ) and 1 − i√ 3 = 2(cosπ

3 − isinπ

3 )∴ (1 + i

√ 3)90 + (1 − i√ 3)90

=

2(cosπ3 + isinπ

3 )90

+


= 290(cos30π + isin30π) + 290(cos30π

−isin30π)


Examples


3)90 + (1 − i√ 3)90


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3

r =

x2 + y2 = √ 1 + 3 = √ 4 = 2θ = tan−1

yx

= tan−1

√ 3

1

=

π

3

∴ 1 + i√ 3 = 2(cosπ

3 + isinπ

3 ) and 1 − i√ 3 = 2(cosπ

3 − isinπ

3 )∴ (1 + i

√ 3)90 + (1 − i√ 3)90

=

2(cosπ3 + isinπ

3 )90

+


= 290(cos30π + isin30π) + 290(cos30π − isin30π)= 290(2cos30π)


Examples


3)90 + (1 − i√ 3)90


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3

r =

x2 + y2 = √ 1 + 3 = √ 4 = 2θ = tan−1

yx

= tan−1

√ 3

1

=

π

3

∴ 1 + i√ 3 = 2(cosπ

3 + isinπ

3 ) and 1 − i√ 3 = 2(cosπ

3 − isinπ

3 )∴ (1 + i

√ 3)90 + (1 − i√ 3)90

=

2(cosπ3 + isinπ

3 )90

+



= 291cos30π


Examples


3)90 + (1 − i√ 3)90√


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3

r =

x2 + y2 = √ 1 + 3 = √ 4 = 2θ = tan−1

yx

= tan−1

√ 3

1

=

π

3

∴ 1 + i√ 3 = 2(cosπ

3 + isinπ

3 ) and 1 − i√ 3 = 2(cosπ

3 − isinπ

3 )∴ (1 + i

√ 3)90 + (1 − i√ 3)90

=

2(cosπ3 + isinπ

3 )90

+



= 291cos30π

= 291(1) = 291


Examples


3)90 + (1 − i√ 3)90√


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3

r =

x2 + y2 = √ 1 + 3 = √ 4 = 2θ = tan−1

yx

= tan−1

√ 3

1

=

π

3

∴ 1 + i√ 3 = 2(cosπ

3 + isinπ

3 ) and 1 − i√ 3 = 2(cosπ

3 − isinπ

3 )∴ (1 + i

√ 3)90 + (1 − i√ 3)90

=

2(cosπ3 + isinπ

3 )90

+



= 291cos30π

= 291(1) = 291


Example


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Ex. Evaluate 1 + sinθ + icosθ

1 + sinθ−

icosθn


Sol. We have sin2θ + cos2θ = 1


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Sol. We have sin2θ + cos2θ = 1∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ


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( i θ i θ)( i θ + i θ)


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= (sinθ − icosθ)(sinθ + icosθ)



(sinθ icosθ)(sinθ + icosθ)


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Now1 + sinθ + icosθ



= (sinθ icosθ)(sinθ + icosθ)


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Now1 + sinθ + icosθ= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)

N B Vyas Complex Numbers




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Now1 + sinθ + icosθ= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)= (sinθ + icosθ)(sinθ − icosθ + 1)





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Now1 + sinθ + icosθ= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)= (sinθ + icosθ)(sinθ − icosθ + 1)∴ 1 + sinθ + icosθ1 + sinθ − icosθ = sinθ + icosθ





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= cosπ

2 − θ

+ isinπ

2 − θ





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= cosπ

2 − θ

+ isinπ

2 − θ

∴ 1 + sinθ + icosθ1 + sinθ − icosθ

n

= cos π2 −

θ + isin π2 −

θn





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= cosπ

2 − θ

+ isinπ

2 − θ

∴ 1 + sinθ + icosθ1 + sinθ − icosθ

n

= cos π2 −

θ + isin π2 −

θn= cos

nπ

2 − θ

+ isinnπ

2 − θ


Example


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Ex. If z 1 = eiθ1 , z 2 = e

iθ2 and z 1 − z 2 = 0 prove that1

z 1 − 1

z 2= 0


Example

Sol. Hereθ


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z1 = eiθ1 = cosθ1 + isinθ1


Example

Sol. Hereiθ


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Example

Sol. Hereiθ


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z2 = eiθ2 = cosθ2 + isinθ2Given that z1 − z2 = 0


Example

Sol. Hereiθ θ i i θ


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z2 = eiθ2 = cosθ2 + isinθ2Given that z1 − z2 = 0 (cosθ1 + isinθ1) − (cosθ2 + isinθ2) = 0


Example

Sol. Hereiθ1 θ i i θ


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z2 = eiθ2 = cosθ2 + isinθ2Given that z1 − z2 = 0 (cosθ1 + isinθ1) − (cosθ2 + isinθ2) = 0(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0


Example

Sol. Hereiθ1 θ + i i θ


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z2 = eiθ2 = cosθ2 + isinθ2Given that z1 − z2 = 0 (cosθ1 + isinθ1) − (cosθ2 + isinθ2) = 0(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0Comparing real and imaginary parts on both the sides, we get


Example

Sol. Hereiθ1 θ + i i θ


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z2 = eiθ2 = cosθ2 + isinθ2Given that z1 − z2 = 0 (cosθ1 + isinθ1) − (cosθ2 + isinθ2) = 0(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0Comparing real and imaginary parts on both the sides, we get

cosθ1 − cosθ2 = 0 and sinθ1 − sinθ2 = 0


Example

Sol. Herez eiθ1 cosθ + isinθ


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z2 = eiθ2 = cosθ2 + isinθ2Given that z1 − z2 = 0 (cosθ1 + isinθ1) − (cosθ2 + isinθ2) = 0(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0Comparing real and imaginary parts on both the sides, we getcosθ

1 −cosθ

2 = 0 and sinθ

1 −sinθ

2 = 0Now

1

z1− 1

z2=

1

cosθ1 + isinθ1− 1

cosθ2 + isinθ2


Example

Sol. Herez1 = e

iθ1 = cosθ1 + isinθ1


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z1 = e 1 = cosθ1 + isinθ1z2 = eiθ2 = cosθ2 + isinθ2

Given that z1 − z2 = 0 (cosθ1 + isinθ1) − (cosθ2 + isinθ2) = 0(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0Comparing real and imaginary parts on both the sides, we getcosθ

1 −cosθ

2 = 0 and sinθ

1 −sinθ

2 = 0Now

1

z1− 1

z2=

1


cosθ2 + isinθ2

= cosθ1 − isinθ1cos2θ1 + sin2θ1

− cosθ2 − isinθ2cos2θ2 + sin2θ2


Example

Sol. Herez1 = e

iθ1 = cosθ1 + isinθ1


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z1 = e 1 = cosθ1 + isinθ1z2 = eiθ2 = cosθ2 + isinθ2

Given that z1 − z2 = 0 (cosθ1 + isinθ1) − (cosθ2 + isinθ2) = 0(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0Comparing real and imaginary parts on both the sides, we getcosθ

1 −cosθ

2 = 0 and sinθ

1 −sinθ

2 = 0Now

1

z1− 1

z2=

1


cosθ2 + isinθ2

= cosθ1 − isinθ1cos2θ1 + sin2θ1

− cosθ2 − isinθ2cos2θ2 + sin2θ2



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Ex. (1 + i)n + (1 − i)n = 2n2 +1cosnπ4


Root of a Complex Number

De Moivre’s theorem is useful for finding the roots of a complexnumber.


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If n is any positive integer, then by De Moivre’s theorem





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If n is any positive integer, then by De Moivre’s theoremcos

θ

n + isin

θ

n

n= cos

θ

nn

+ isin

θ

nn

= cosθ + isinθ





110/126


θ

n + isin

θ

n

n= cos

θ

nn

+ isin

θ

nn

= cosθ + isinθ

Thus cosθ

n + isin

θ

n is one of the nth root of cosθ + isinθ, i.e.,





111/126


θ

n + isin

θ

n

n= cos

θ

nn

+ isin

θ

nn

= cosθ + isinθ

Thus cosθ

n + isin

θ


(cosθ + isinθ)1

n = cosθ

n + isin

θ

n





112/126


θ

n + isin

θ

n

n= cos

θ

nn

+ isin

θ

nn

= cosθ + isinθ

Thus cosθ

n + isin

θ


(cosθ + isinθ)1

n = cosθ

n + isin

θ

nThe remaining roots may be obtained by periodic nature of thetrigonometric functions, i.e.,





113/126


θ

n + isin

θ

n

n= cos

θ

nn

+ isin

θ

nn

= cosθ + isinθ

Thus cosθ

n + isin

θ


(cosθ + isinθ)1

n = cosθ

n + isin

θ


cos(2kπ + θ) = cosθ and sin(2kπ + θ) = sinθ





114/126


θ

n + isin

θ

n

n= cos

θ

nn

+ isin

θ

nn

= cosθ + isinθ

Thus cosθ

n + isin

θ


(cosθ + isinθ)1

n = cosθ

n + isin

θ


cos(2kπ + θ) = cosθ and sin(2kπ + θ) = sinθHence the general form of complex number cosθ + isinθ form restof all the roots.





115/126


θ

n + isin

θ

n

n= cos

θ

nn

+ isin

θ

nn

= cosθ + isinθ

Thus cosθ

n + isin

θ


(cosθ + isinθ)1

n = cosθ

n + isin

θ


cos(2kπ + θ) = cosθ and sin(2kπ + θ) = sinθHence the general form of complex number cosθ + isinθ form restof all the roots.



(cosθ + isinθ)1

n = [cos(2kπ + θ) + isin(2kπ + θ)]1

n


116/126



(cosθ + isinθ)1


n

= cos

2kπ + θ

n

+ isin

2kπ + θ

n


117/126



(cosθ + isinθ)1


n

= cos

2kπ + θ

n

+ isin

2kπ + θ

n

1


118/126

which gives all the roots of (cosθ + isinθ)n

fork = 0, 1, 2, . . . (n− 1).



(cosθ + isinθ)1


n

= cos

2kπ + θ

n

+ isin

2kπ + θ

n

1


119/126


fork = 0, 1, 2, . . . (n− 1).This roots are as follows:



(cosθ + isinθ)1


n

= cos

2kπ + θ

n

+ isin

2kπ + θ

n

1


120/126



For k = 0,cosθ

n + isin

θ

n



(cosθ + isinθ)1


n

= cos

2kπ + θ

n

+ isin

2kπ + θ

n

1


121/126



For k = 0,cosθ

n + isin

θ

n

k = 1, cos

2π + θn

+ isin

2π + θ

n



(cosθ + isinθ)1


n

= cos

2kπ + θ

n

+ isin

2kπ + θ

n

1


122/126



For k = 0,cosθ

n + isin

θ

n

k = 1, cos

2π + θn

+ isin

2π + θ

n

k = 2, cos

4π + θ

n

+ isin

4π + θ

n



(cosθ + isinθ)1


n

= cos

2kπ + θ

n

+ isin

2kπ + θ

n

( θ θ)

1

n


123/126



For k = 0,cosθ

n + isin

θ

n

k = 1, cos

2π + θn

+ isin

2π + θ

n

k = 2, cos

4π + θ

n

+ isin

4π + θ

n

. . . . . .



(cosθ + isinθ)1


n

= cos

2kπ + θ

n

+ isin

2kπ + θ

n

h h ll h f ( θ i i θ)

1

n

f


124/126



For k = 0,cosθ

n + isin

θ

n

k = 1, cos

2π + θn

+ isin

2π + θ

n

k = 2, cos

4π + θ

n

+ isin

4π + θ

n

. . . . . .

. . . . . .



(cosθ + isinθ)1


n

= cos

2kπ + θ

n

+ isin

2kπ + θ

n

hi h i ll h f ( θ i i θ)

1

n

f


125/126



For k = 0,cosθ

n + isin

θ

n

k = 1, cos

2π + θn

+ isin

2π + θ

n

k = 2, cos

4π + θ

n

+ isin

4π + θ

n

. . . . . .

. . . . . .

k = n− 1, cos

2(n− 1)π + θn

+ isin

2(n− 1)π + θ

n



(cosθ + isinθ)1


n

= cos

2kπ + θ

n

+ isin

2kπ + θ

n

hi h i ll th t f ( θ i i θ)

1

n

f


126/126

which gives all the roots of (cosθ + isinθ) fork = 0, 1, 2, . . . (n− 1).This roots are as follows:

For k = 0,cosθ

n + isin

θ

n

k = 1, cos

2π + θn

+ isin

2π + θ

n

k = 2, cos

4π + θ

n

+ isin

4π + θ

n

. . . . . .

. . . . . .

k = n− 1, cos

2(n− 1)π + θn

+ isin

2(n− 1)π + θ

n

The further values of k will give the same roots as above in order.


dr. nirav vyas complexnumbers1.pdf

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