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Complex Numbers
N. B. Vyas
Department of Mathematics,
Atmiya Institute of Tech. and Science,Rajkot (Guj.)
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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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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).
Thus z = x + iy or z = (x, y)
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Definition of Complex Number
The set of complex numbers is denoted by .
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Definition of Complex Number
The set of complex numbers is denoted by .
If z = x + iy is a complex number, then
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Definition of Complex Number
The set of complex numbers is denoted by .
If z = x + iy is a complex number, thenx is called the real part of z and denoted byRe(z ).
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Definition of Complex Number
The set of complex numbers is denoted by .
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 ).
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Definition of Complex Number
The set of complex numbers is denoted by .
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 ).
If x = 0 and y = 0 then z = 0 + iy = iyis called a purely imaginary number.
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Definition of Complex Number
The set of complex numbers is denoted by .
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 ).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.
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Definition of Complex Number
The set of complex numbers is denoted by .
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 ).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.
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Definition of Conjugate Complex Number
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Definition of Conjugate Complex Number
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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Definition of Conjugate Complex Number
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Definition of Conjugate Complex Number
Conjugate Complex Numbers
If two complex numbers differ only in the sign of theimaginary part then they are called conjugatecomplex numbers.
Thus x + iy and x − iy are conjugate complexnumbers.
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Definition of Conjugate Complex Number
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Definition of Conjugate Complex Number
Conjugate Complex Numbers
If two complex numbers differ only in the sign of theimaginary part then they are called conjugatecomplex numbers.
Thus x + iy and x − iy are conjugate complexnumbers.
The conjugate of a complex number z is denoted
by z̄ .
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Definition of Conjugate Complex Number
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Definition of Conjugate Complex Number
Conjugate Complex Numbers
If two complex numbers differ only in the sign of theimaginary part then they are called conjugatecomplex numbers.
Thus x + iy and x − iy are conjugate complexnumbers.
The conjugate of a complex number z is denoted
by z̄ .The conjugate of real number is the real numberitself.
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Properties of Complex Numbers
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Properties of Complex Numbers
1 If x + iy = 0 then x = 0, y = 0
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Properties of Complex Numbers
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Properties of Complex Numbers
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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Properties of Complex Numbers
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Properties of Complex Numbers
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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Properties of Complex Numbers
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Properties of Complex Numbers
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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Properties of Complex Numbers
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Properties of Complex Numbers
If z 1
= x1 + iy
1 and z
2 = x
2 + iy
2 then their
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Properties of Complex Numbers
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Properties of Complex Numbers
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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Properties of Complex Numbers
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Properties of Complex Numbers
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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Properties of Complex Numbers
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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)
Difference: z 1 − z 2 = (x1 + iy1) − (x2 + iy2)
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Properties of Complex Numbers
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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)
Difference: z 1 − z 2 = (x1 + iy1) − (x2 + iy2)= (x1 − x2) + i(y1 − y2)
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Properties of Complex Numbers
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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)
Difference: z 1 − z 2 = (x1 + iy1) − (x2 + iy2)= (x1 − x2) + i(y1 − y2)
Product: z 1.z 2 = (x1 + iy1).(x2 + iy2)
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Properties of Complex Numbers
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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)
Difference: z 1 − z 2 = (x1 + iy1) − (x2 + iy2)= (x1 − x2) + i(y1 − y2)
Product: z 1.z 2 = (x1 + iy1).(x2 + iy2)= (x1x2 − y1y2) + i(x1y2 + x2y1)
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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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Polar form of a Complex Number
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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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Polar form of a Complex Number
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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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Polar form of a Complex Number
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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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Polar form of a Complex Number
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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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Polar form of a Complex Number
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r is called the absolute value or the modulus of z and is denoted by
|z |.
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Polar form of a Complex Number
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r is called the absolute value or the modulus of z and is denoted by
|z |.
∴ r = |z | = x2 + y2 = √ z.z̄
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Polar form of a Complex Number
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r is called the absolute value or the modulus of z and is denoted by
|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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Polar form of a Complex Number
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r is called the absolute value or the modulus of z and is denoted by
|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.It is denoted by argz or Ampz
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Polar form of a Complex Number
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r is called the absolute value or the modulus of z and is denoted by
|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.It is denoted by argz or Ampz
∴ θ = argz = tan−1 y
x
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Polar form of a Complex Number
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r is called the absolute value or the modulus of z and is denoted by
|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.It is denoted by argz or Ampz
∴ θ = argz = tan−1 y
xθ is the directed angle from the positive X-axis toOP.
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Polar form of a Complex Number
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r is called the absolute value or the modulus of z and is denoted by
|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.It is denoted by argz or Ampz
∴ 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.
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Multiplication and Division of Complex
Numbers in Polar Form
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Numbers in Polar Form
Let z 1 = r1(cosθ1 + isinθ1) and z 2 = r2(cosθ2 + isinθ2)be two complex numbers in polar form. Then
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Multiplication and Division of Complex
Numbers in Polar Form
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Numbers in Polar Form
Let z 1 = r1(cosθ1 + isinθ1) and z 2 = r2(cosθ2 + isinθ2)be two complex numbers in polar form. Then
Product: z 1z 2 = r1(cosθ1 + isinθ1)r2(cosθ2 + isinθ2)
N. B. Vyas Complex Numbers
Multiplication and Division of Complex
Numbers in Polar Form
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Numbers in Polar Form
Let z 1 = r1(cosθ1 + isinθ1) and z 2 = r2(cosθ2 + isinθ2)be two complex numbers in polar form. Then
Product: z 1z 2 = r1(cosθ1 + isinθ1)r2(cosθ2 + isinθ2)= r1r2[(cosθ1cosθ2
−sinθ1sinθ2) + i(cosθ1sinθ2 +
sinθ1cosθ2)]
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Multiplication and Division of Complex
Numbers in Polar Form
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Numbers in Polar Form
Let z 1 = r1(cosθ1 + isinθ1) and z 2 = r2(cosθ2 + isinθ2)be two complex numbers in polar form. Then
Product: z 1z 2 = r1(cosθ1 + isinθ1)r2(cosθ2 + isinθ2)= r1r2[(cosθ1cosθ2
−sinθ1sinθ2) + i(cosθ1sinθ2 +
sinθ1cosθ2)]= r1r2[cos(θ1 + θ2) + isin(θ1 + θ2)]
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Multiplication and Division of Complex
Numbers in Polar Form
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Numbers in Polar Form
Let z 1 = r1(cosθ1 + isinθ1) and z 2 = r2(cosθ2 + isinθ2)be two complex numbers in polar form. Then
Product: z 1z 2 = r1(cosθ1 + isinθ1)r2(cosθ2 + isinθ2)= r1r2[(cosθ1cosθ2
−sinθ1sinθ2) + i(cosθ1sinθ2 +
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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Multiplication and Division of Complex
Numbers in Polar Form
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Numbers in Polar Form
Division: z 1
z 2=
r1(cosθ1 + isinθ1)
r2(cosθ2 + isinθ2)
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Multiplication and Division of Complex
Numbers in Polar Form
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Numbers in Polar Form
Division: z 1
z 2=
r1(cosθ1 + isinθ1)
r2(cosθ2 + isinθ2)=
r1(cosθ1 + isinθ1)(cosθ2 − isinθ2)r2(cosθ2 + isinθ2)(cosθ2 − isinθ2)
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Multiplication and Division of Complex
Numbers in Polar Form
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Numbers in Polar Form
Division: z 1
z 2=
r1(cosθ1 + isinθ1)
r2(cosθ2 + isinθ2)=
r1(cosθ1 + isinθ1)(cosθ2 − isinθ2)r2(cosθ2 + isinθ2)(cosθ2 − isinθ2)
= r1
r2[(cosθ1cosθ2 + sinθ1sinθ2) + i(sinθ1cosθ2
−cosθ1sinθ2)]
N. B. Vyas Complex Numbers
Multiplication and Division of Complex
Numbers in Polar Form
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Numbers in Polar Form
Division: z 1
z 2=
r1(cosθ1 + isinθ1)
r2(cosθ2 + isinθ2)=
r1(cosθ1 + isinθ1)(cosθ2 − isinθ2)r2(cosθ2 + isinθ2)(cosθ2 − isinθ2)
= r1
r2[(cosθ1cosθ2 + sinθ1sinθ2) + i(sinθ1cosθ2
−cosθ1sinθ2)]
= r1
r2[cos(θ1 − θ2) + isin(θ1 − θ2)]
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Multiplication and Division of Complex
Numbers in Polar Form
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Numbers in Polar Form
Division: z 1
z 2=
r1(cosθ1 + isinθ1)
r2(cosθ2 + isinθ2)=
r1(cosθ1 + isinθ1)(cosθ2 − isinθ2)r2(cosθ2 + isinθ2)(cosθ2 − isinθ2)
= r1
r2[(cosθ1cosθ2 + sinθ1sinθ2) + i(sinθ1cosθ2
−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.
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Exponential form of Complex Numbers
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We know that
sinθ = θ− θ3
3! +
θ5
5! − θ7
7! . . .
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Exponential form of Complex Numbers
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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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Exponential form of Complex Numbers
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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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Exponential form of Complex Numbers
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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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Exponential form of Complex Numbers
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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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Exponential form of Complex Numbers
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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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Exponential form of Complex Numbers
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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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Exponential form of Complex Numbers
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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)
N. B. Vyas Complex Numbers
Laws of Complex Numbers
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If z 1 and z 2 are two complex numbers, then
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Laws of Complex Numbers
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If z 1 and z 2 are two complex numbers, then
1 Triangle Inequality: |z 1 + z 2| ≤ |z 1| + |z 2|
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Laws of Complex Numbers
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If z 1 and z 2 are two complex numbers, then
1 Triangle Inequality: |z 1 + z 2| ≤ |z 1| + |z 2|2 |z 1 − z 2| ≥ | |z 1| − |z 2||
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Laws of Complex Numbers
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If z 1 and z 2 are two complex numbers, then
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)
N. B. Vyas Complex Numbers
Laws of Complex Numbers
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If z 1 and z 2 are two complex numbers, then
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)
4 |z 1z 2| = |z 1||z 2|
N. B. Vyas Complex Numbers
Laws of Complex Numbers
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If z 1 and z 2 are two complex numbers, then
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)
4 |z 1z 2| = |z 1||z 2|5 z 1z 2 = |
z 1
||z 2|
N. B. Vyas Complex Numbers
Ex
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1 Find complex conjugate of 3 + 2i
1 − i
N. B. Vyas Complex Numbers
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
N. B. Vyas Complex Numbers
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
∴ z = x + iy = r(cosθ + isinθ)
N. B. Vyas Complex Numbers
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
∴ z = x + iy = r(cosθ + isinθ)z n = rn(cosθ + isinθ)n
N. B. Vyas Complex Numbers
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
∴ z = x + iy = r(cosθ + isinθ)z n = rn(cosθ + isinθ)n
= rn(cosnθ + isinnθ)
N. B. Vyas Complex Numbers
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
∴ z = x + iy = r(cosθ + isinθ)z n = rn(cosθ + isinθ)n
= rn(cosnθ + isinnθ)E.g.:
N. B. Vyas Complex Numbers
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
∴ z = x + iy = r(cosθ + isinθ)z n = rn(cosθ + isinθ)n
= rn(cosnθ + isinnθ)E.g.:
(cosθ + isinθ)2 = cos2θ + isin2θ
N. B. Vyas Complex Numbers
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
∴ z = x + iy = r(cosθ + isinθ)z n = rn(cosθ + isinθ)n
= rn(cosnθ + isinnθ)E.g.:
(cosθ + isinθ)2 = cos2θ + isin2θ
(cosθ + isinθ)−1 = cos(−θ) + isin(−θ) = cosθ − isinθ
N. B. Vyas Complex Numbers
Examples
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Ex. Simplify
(cos2θ + isin2θ)23 (cosθ
−isinθ)2
(cos3θ − isin3θ)2(cos5θ − isin5θ)13
N. B. Vyas Complex Numbers
Examples
Ex. Evaluate (1 + i√
3)90 + (1 − i√ 3)90
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Ex. Evaluate (1 + i√
3) + (1 i√
3)
N. B. Vyas Complex Numbers
Examples
Ex. Evaluate (1 + i√
3)90 + (1 − i√ 3)90
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( +√
) + (√
)
Sol. Let x = 1 and y =√
3
N. B. Vyas Complex Numbers
Examples
Ex. Evaluate (1 + i√
3)90 + (1 − i√ 3)90
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(√
) (√
)
Sol. Let x = 1 and y =√
3
r =
x2 + y2 = √ 1 + 3 = √ 4 = 2
N. B. Vyas Complex Numbers
Examples
Ex. Evaluate (1 + i√
3)90 + (1 − i√ 3)90
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( ) ( )
Sol. Let x = 1 and y =√
3
r =
x2 + y2 = √ 1 + 3 = √ 4 = 2θ = tan−1
yx
= tan−1
√ 3
1
=
π
3
N. B. Vyas Complex Numbers
Examples
Ex. Evaluate (1 + i√
3)90 + (1 − i√ 3)90
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( ) ( )
Sol. Let x = 1 and y =√
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 )
N. B. Vyas Complex Numbers
Examples
Ex. Evaluate (1 + i√
3)90 + (1 − i√ 3)90
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Sol. Let x = 1 and y =√
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 )
N. B. Vyas Complex Numbers
Examples
Ex. Evaluate (1 + i√
3)90 + (1 − i√ 3)90
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Sol. Let x = 1 and y =√
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
N. B. Vyas Complex Numbers
Examples
Ex. Evaluate (1 + i√
3)90 + (1 − i√ 3)90
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Sol. Let x = 1 and y =√
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
N. B. Vyas Complex Numbers
Examples
Ex. Evaluate (1 + i√
3)90 + (1 − i√ 3)90
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Sol. Let x = 1 and y =√
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
= 290(cos30π + isin30π) + 290(cos30π
−isin30π)
N. B. Vyas Complex Numbers
Examples
Ex. Evaluate (1 + i√
3)90 + (1 − i√ 3)90
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Sol. Let x = 1 and y =√
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
= 290(cos30π + isin30π) + 290(cos30π − isin30π)= 290(2cos30π)
N. B. Vyas Complex Numbers
Examples
Ex. Evaluate (1 + i√
3)90 + (1 − i√ 3)90
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Sol. Let x = 1 and y =√
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
= 290(cos30π + isin30π) + 290(cos30π − isin30π)= 290(2cos30π)
= 291cos30π
N. B. Vyas Complex Numbers
Examples
Ex. Evaluate (1 + i√
3)90 + (1 − i√ 3)90√
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Sol. Let x = 1 and y =√
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
= 290(cos30π + isin30π) + 290(cos30π − isin30π)= 290(2cos30π)
= 291cos30π
= 291(1) = 291
N. B. Vyas Complex Numbers
Examples
Ex. Evaluate (1 + i√
3)90 + (1 − i√ 3)90√
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Sol. Let x = 1 and y =√
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
= 290(cos30π + isin30π) + 290(cos30π − isin30π)= 290(2cos30π)
= 291cos30π
= 291(1) = 291
N. B. Vyas Complex Numbers
Example
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Ex. Evaluate 1 + sinθ + icosθ
1 + sinθ−
icosθn
N. B. Vyas Complex Numbers
Sol. We have sin2θ + cos2θ = 1
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N. B. Vyas Complex Numbers
Sol. We have sin2θ + cos2θ = 1∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ
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N. B. Vyas Complex Numbers
Sol. We have sin2θ + cos2θ = 1∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ
( i θ i θ)( i θ + i θ)
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= (sinθ − icosθ)(sinθ + icosθ)
N. B. Vyas Complex Numbers
Sol. We have sin2θ + cos2θ = 1∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ
(sinθ icosθ)(sinθ + icosθ)
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= (sinθ − icosθ)(sinθ + icosθ)
Now1 + sinθ + icosθ
N. B. Vyas Complex Numbers
Sol. We have sin2θ + cos2θ = 1∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ
= (sinθ icosθ)(sinθ + icosθ)
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= (sinθ − icosθ)(sinθ + icosθ)
Now1 + sinθ + icosθ= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)
N B Vyas Complex Numbers
Sol. We have sin2θ + cos2θ = 1∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ
= (sinθ − icosθ)(sinθ + icosθ)
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= (sinθ − icosθ)(sinθ + icosθ)
Now1 + sinθ + icosθ= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)= (sinθ + icosθ)(sinθ − icosθ + 1)
N B Vyas Complex Numbers
Sol. We have sin2θ + cos2θ = 1∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ
= (sinθ − icosθ)(sinθ + icosθ)
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= (sinθ icosθ)(sinθ + icosθ)
Now1 + sinθ + icosθ= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)= (sinθ + icosθ)(sinθ − icosθ + 1)∴ 1 + sinθ + icosθ1 + sinθ − icosθ = sinθ + icosθ
N B Vyas Complex Numbers
Sol. We have sin2θ + cos2θ = 1∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ
= (sinθ − icosθ)(sinθ + icosθ)
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= (sinθ icosθ)(sinθ + icosθ)
Now1 + sinθ + icosθ= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)= (sinθ + icosθ)(sinθ − icosθ + 1)∴ 1 + sinθ + icosθ1 + sinθ − icosθ = sinθ + icosθ
= cosπ
2 − θ
+ isinπ
2 − θ
N B Vyas Complex Numbers
Sol. We have sin2θ + cos2θ = 1∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ
= (sinθ − icosθ)(sinθ + icosθ)
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(sinθ icosθ)(sinθ + icosθ)
Now1 + sinθ + icosθ= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)= (sinθ + icosθ)(sinθ − icosθ + 1)∴ 1 + sinθ + icosθ1 + sinθ − icosθ = sinθ + icosθ
= cosπ
2 − θ
+ isinπ
2 − θ
∴ 1 + sinθ + icosθ1 + sinθ − icosθ
n
= cos π2 −
θ + isin π2 −
θn
N B Vyas Complex Numbers
Sol. We have sin2θ + cos2θ = 1∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ
= (sinθ − icosθ)(sinθ + icosθ)
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(sinθ icosθ)(sinθ + icosθ)
Now1 + sinθ + icosθ= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)= (sinθ + icosθ)(sinθ − icosθ + 1)∴ 1 + sinθ + icosθ1 + sinθ − icosθ = sinθ + icosθ
= 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
N B Vyas Complex Numbers
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
N B Vyas Complex Numbers
Example
Sol. Hereθ
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z1 = eiθ1 = cosθ1 + isinθ1
N B Vyas Complex Numbers
Example
Sol. Hereiθ
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z1 = eiθ1 = cosθ1 + isinθ1
z2 = eiθ2 = cosθ2 + isinθ2
N B Vyas Complex Numbers
Example
Sol. Hereiθ
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z1 = eiθ1 = cosθ1 + isinθ1
z2 = eiθ2 = cosθ2 + isinθ2Given that z1 − z2 = 0
N B Vyas Complex Numbers
Example
Sol. Hereiθ θ i i θ
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z1 = eiθ1 = cosθ1 + isinθ1
z2 = eiθ2 = cosθ2 + isinθ2Given that z1 − z2 = 0 (cosθ1 + isinθ1) − (cosθ2 + isinθ2) = 0
N. B. Vyas Complex Numbers
Example
Sol. Hereiθ1 θ i i θ
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z1 = eiθ1 = cosθ1 + isinθ1
z2 = eiθ2 = cosθ2 + isinθ2Given that z1 − z2 = 0 (cosθ1 + isinθ1) − (cosθ2 + isinθ2) = 0(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0
N. B. Vyas Complex Numbers
Example
Sol. Hereiθ1 θ + i i θ
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z1 = eiθ1 = cosθ1 + isinθ1
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
N. B. Vyas Complex Numbers
Example
Sol. Hereiθ1 θ + i i θ
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z1 = eiθ1 = cosθ1 + isinθ1
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
N. B. Vyas Complex Numbers
Example
Sol. Herez eiθ1 cosθ + isinθ
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z1 = eiθ1 = cosθ1 + isinθ1
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
N. B. Vyas Complex Numbers
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
cosθ1 + isinθ1− 1
cosθ2 + isinθ2
= cosθ1 − isinθ1cos2θ1 + sin2θ1
− cosθ2 − isinθ2cos2θ2 + sin2θ2
N. B. Vyas Complex Numbers
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
cosθ1 + isinθ1− 1
cosθ2 + isinθ2
= cosθ1 − isinθ1cos2θ1 + sin2θ1
− cosθ2 − isinθ2cos2θ2 + sin2θ2
N. B. Vyas Complex Numbers
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Ex. (1 + i)n + (1 − i)n = 2n2 +1cosnπ4
N. B. Vyas Complex Numbers
Root of a Complex Number
De Moivre’s theorem is useful for finding the roots of a complexnumber.
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N. B. Vyas Complex Numbers
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
N. B. Vyas Complex Numbers
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 theoremcos
θ
n + isin
θ
n
n= cos
θ
nn
+ isin
θ
nn
= cosθ + isinθ
N. B. Vyas Complex Numbers
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 theoremcos
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.,
N. B. Vyas Complex Numbers
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 theoremcos
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.,
(cosθ + isinθ)1
n = cosθ
n + isin
θ
n
N. B. Vyas Complex Numbers
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 theoremcos
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.,
(cosθ + isinθ)1
n = cosθ
n + isin
θ
nThe remaining roots may be obtained by periodic nature of thetrigonometric functions, i.e.,
N. B. Vyas Complex Numbers
Root of a Complex Number
De Moivre’s theorem is useful for finding the roots of a complexnumber.
-
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113/126
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
Thus cosθ
n + isin
θ
n is one of the nth root of cosθ + isinθ, i.e.,
(cosθ + isinθ)1
n = cosθ
n + isin
θ
nThe remaining roots may be obtained by periodic nature of thetrigonometric functions, i.e.,
cos(2kπ + θ) = cosθ and sin(2kπ + θ) = sinθ
N. B. Vyas Complex Numbers
Root of a Complex Number
De Moivre’s theorem is useful for finding the roots of a complexnumber.
-
8/20/2019 Dr. Nirav Vyas complexnumbers1.pdf
114/126
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
Thus cosθ
n + isin
θ
n is one of the nth root of cosθ + isinθ, i.e.,
(cosθ + isinθ)1
n = cosθ
n + isin
θ
nThe remaining roots may be obtained by periodic nature of thetrigonometric functions, i.e.,
cos(2kπ + θ) = cosθ and sin(2kπ + θ) = sinθHence the general form of complex number cosθ + isinθ form restof all the roots.
N. B. Vyas Complex Numbers
Root of a Complex Number
De Moivre’s theorem is useful for finding the roots of a complexnumber.
-
8/20/2019 Dr. Nirav Vyas complexnumbers1.pdf
115/126
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
Thus cosθ
n + isin
θ
n is one of the nth root of cosθ + isinθ, i.e.,
(cosθ + isinθ)1
n = cosθ
n + isin
θ
nThe remaining roots may be obtained by periodic nature of thetrigonometric functions, i.e.,
cos(2kπ + θ) = cosθ and sin(2kπ + θ) = sinθHence the general form of complex number cosθ + isinθ form restof all the roots.
N. B. Vyas Complex Numbers
Root of a Complex Number
(cosθ + isinθ)1
n = [cos(2kπ + θ) + isin(2kπ + θ)]1
n
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N. B. Vyas Complex Numbers
Root of a Complex Number
(cosθ + isinθ)1
n = [cos(2kπ + θ) + isin(2kπ + θ)]1
n
= cos
2kπ + θ
n
+ isin
2kπ + θ
n
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N. B. Vyas Complex Numbers
Root of a Complex Number
(cosθ + isinθ)1
n = [cos(2kπ + θ) + isin(2kπ + θ)]1
n
= cos
2kπ + θ
n
+ isin
2kπ + θ
n
1
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which gives all the roots of (cosθ + isinθ)n
fork = 0, 1, 2, . . . (n− 1).
N. B. Vyas Complex Numbers
Root of a Complex Number
(cosθ + isinθ)1
n = [cos(2kπ + θ) + isin(2kπ + θ)]1
n
= cos
2kπ + θ
n
+ isin
2kπ + θ
n
1
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which gives all the roots of (cosθ + isinθ)n
fork = 0, 1, 2, . . . (n− 1).This roots are as follows:
N. B. Vyas Complex Numbers
Root of a Complex Number
(cosθ + isinθ)1
n = [cos(2kπ + θ) + isin(2kπ + θ)]1
n
= cos
2kπ + θ
n
+ isin
2kπ + θ
n
1
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which gives all the roots of (cosθ + isinθ)n
fork = 0, 1, 2, . . . (n− 1).This roots are as follows:
For k = 0,cosθ
n + isin
θ
n
N. B. Vyas Complex Numbers
Root of a Complex Number
(cosθ + isinθ)1
n = [cos(2kπ + θ) + isin(2kπ + θ)]1
n
= cos
2kπ + θ
n
+ isin
2kπ + θ
n
1
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which gives all the roots of (cosθ + isinθ)n
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
N. B. Vyas Complex Numbers
Root of a Complex Number
(cosθ + isinθ)1
n = [cos(2kπ + θ) + isin(2kπ + θ)]1
n
= cos
2kπ + θ
n
+ isin
2kπ + θ
n
1
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which gives all the roots of (cosθ + isinθ)n
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
N. B. Vyas Complex Numbers
Root of a Complex Number
(cosθ + isinθ)1
n = [cos(2kπ + θ) + isin(2kπ + θ)]1
n
= cos
2kπ + θ
n
+ isin
2kπ + θ
n
( θ θ)
1
n
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which gives all the roots of (cosθ + isinθ)n
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
. . . . . .
N. B. Vyas Complex Numbers
Root of a Complex Number
(cosθ + isinθ)1
n = [cos(2kπ + θ) + isin(2kπ + θ)]1
n
= cos
2kπ + θ
n
+ isin
2kπ + θ
n
h h ll h f ( θ i i θ)
1
n
f
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which gives all the roots of (cosθ + isinθ)n
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
. . . . . .
. . . . . .
N. B. Vyas Complex Numbers
Root of a Complex Number
(cosθ + isinθ)1
n = [cos(2kπ + θ) + isin(2kπ + θ)]1
n
= cos
2kπ + θ
n
+ isin
2kπ + θ
n
hi h i ll h f ( θ i i θ)
1
n
f
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which gives all the roots of (cosθ + isinθ)n
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
N. B. Vyas Complex Numbers
Root of a Complex Number
(cosθ + isinθ)1
n = [cos(2kπ + θ) + isin(2kπ + θ)]1
n
= cos
2kπ + θ
n
+ isin
2kπ + θ
n
hi h i ll th t f ( θ i i θ)
1
n
f
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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.
N. B. Vyas Complex Numbers