Complex Numbers

i. i⁴⁵⁷ = i ^{(456 + 1)}

= i⁴⁽¹¹⁴⁾ × i

= (1)¹¹⁴ × i = i since i⁴ = 1

ii. i⁵²⁸ = i⁴⁽¹³²⁾

= (1)¹³² =1 since i⁴ = 1

iii.

since i⁴ = 1

= – 1 since i² = – 1

iv.

[since i⁴ = 1]

v.

= (i – i) = 0

[since ]

vi. (i⁷⁷ + i⁷⁰ + i⁸⁷ + i⁴¹⁴ )³ = (i^{(76 + 1)} + i^{(68 + 2)} + i^{(84 + 3)} + i^{(412 + 2)} ) ³

(i⁷⁷ + i⁷⁰ + i⁸⁷ + i⁴¹⁴ )³ = (i + i² + i³ + i² )³

[since i³ = – i, i² = – 1]

= (i + (– 1) + (– i) + (– 1))³ = (– 2)³

(i⁷⁷ + i⁷⁰ + i⁸⁷ + i⁴¹⁴ )³ = –8

vii. i³⁰ + i⁴⁰ + i⁶⁰ = i^{(28 + 2) +} i⁴⁰ + i⁶⁰

= (i⁴)⁷ i² + (i⁴)¹⁰ + (i⁴)¹⁵

= i² + 1¹⁰ + 1¹⁵ = – 1 + 1 + 1 = 1

viii. i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁸ = i^{(48 + 1)} + i⁶⁸ + i^{(88 + 1)} + i^{(116 + 2)}

= (i⁴)¹²×i + (i⁴)¹⁷ + (i⁴)¹¹×i + (i⁴)²⁹×i²

= i + 1 + i – 1 = 2i

1 + i¹⁰ + i²⁰ + i³⁰ = 1 + i^{(8 + 2) +} i²⁰ + i^{(28 + 2)}

= 1 + (i⁴)² × i² + (i⁴)⁵ + (i⁴)⁷ × i²

= 1 – 1 + 1 – 1 = 0

[ since i⁴ = 1, i² = – 1]

Hence , 1 + i¹⁰ + i²⁰ + i³⁰ is a real number.

i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰ = i^{(48 + 1)} + i⁶⁸ + i^{(88 + 1)} + i^{(108 + 2)}

= (i⁴)¹² × i + (i⁴)¹⁷ + (i⁴)¹¹ × i + (i⁴)²⁷ × i²

= i + 1 + i – 1 = 2i

[since i⁴ = 1, i² = – 1]

i⁴⁹ + i⁶⁸ + i⁸⁹ + i¹¹⁰ = 2i

i³⁰ + i⁸⁰ + i¹²⁰ = i^{(28 + 2)} + i⁸⁰ + i¹²⁰

= (i⁴)⁷ × i² + (i⁴)²⁰ + (i⁴)³⁰

= – 1 + 1 + 1 = 1

[since i⁴ = 1, i² = – 1]

i³⁰ + i⁸⁰ + i¹²⁰ = 1

i + i² + i³ + i⁴ = i + i² + i²×i + i⁴

= i – 1 + (– 1)×i + 1

since i⁴ = 1, i² = – 1

= i – 1 – i + 1 = 0

i⁵ + i¹⁰ + i¹⁵ = i^{(4 + 1)} + i^{(8 + 2)} + i^{(12 + 3)}

= (i⁴)¹×i + (i⁴)²×i² + (i⁴)³×i³

= (i⁴)¹×i + (i⁴)²×i² + (i⁴)³×i²×i

= 1×i + 1×(– 1) + 1×(– 1)×i

= i – 1 – i = – 1

= i¹⁰

= i⁸ i²

= (i⁴)² i²

Since i⁴ = 1, i² = -1

= (1)² (-1)

= -1

1 + i² + i⁴ + i⁶ + i⁸ + ... + i²⁰ = 1 + (– 1) + 1 + (– 1) + 1 + ... + 1

= 1

(1 + i)⁶ + (1 – i) ³ = {(1 + i)² }³ + (1 – i)² (1 – i)

= {1 + i² + 2i}³ + (1 + i^{2 –} 2i)(1 – i)

= {1 – 1 + 2i}³ + (1 – 1 – 2i)(1 – i)

= (2i)³ + (– 2i)(1 – i)

= 8i³ + (– 2i) + 2i²

[since i³ = – i, i² = – 1]

= – 8i – 2i – 2

= – 10 i – 2

= – 2(1 + 5i)

Given:

⇒ a+ib = (1+i)(1+2i)

⇒ a+ib = 1(1+2i)+i(1+2i)

⇒ a+ib = 1+2i+i+2i²

We know that i²=-1

⇒ a+ib = 1+3i+2(-1)

⇒ a+ib = 1+3i-2

⇒ a+ib=-1+3i

∴ The values of a, b are -1, 3.

Given:

⇒

Multiplying and dividing with -2-i

⇒

⇒

We know that i²=-1

⇒

⇒

⇒

∴ The values of a, b are .

Given:

⇒

⇒

We know that i²=-1

⇒

⇒

Multiplying and diving with 3-4i

⇒

⇒

⇒

⇒

⇒

∴ The values of a, b is .

Given:

⇒

Multiplying and dividing by 1-i

⇒

⇒

We know that i²=-1

⇒

⇒

⇒

⇒

∴ The values of a, b is 0, -1.

Given:

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

Multiplying and dividing with 2-3i

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

∴ The values of a, b are .

Given:

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

Multiplying and dividing with 1+i

⇒

⇒

⇒

⇒

⇒

⇒

∴ The values of a, b are ,1.

Given:

⇒

Multiplying and dividing with 4-5i

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

∴ The values of a, b are .

Given:

⇒

⇒

We know that i²=-1

⇒

⇒

⇒

Multiplying and diving with 1-i

⇒

⇒

We know that i²=-1

⇒

⇒

⇒

⇒ a+ib=-3-i

∴ The values of a, b are -3, -1.

Given:

⇒ a+ib=(1+2i)^-3

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

⇒

⇒

Multiplying and dividing with 3-2i

⇒

⇒

⇒

⇒

⇒

∴ the values of a, b are .

Given:

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

Multiplying and dividing with 6-2i

⇒

⇒

⇒

⇒

⇒

⇒

∴ The values of a, b are .

Given:

⇒

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

⇒

⇒

⇒

⇒

Multiplying and dividing with 28+10i

⇒

⇒

⇒

⇒

⇒

∴ The values of a, b is

Given:

⇒

Multiplying and dividing with 1+i

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

⇒

∴ The values of a, b are 1, 2.

Given:

⇒ (x+iy)(2-3i)=4+i

⇒ x(2-3i)+iy(2-3i)=4+i

⇒ 2x-3xi+2yi-3yi²=4+i

We know that i²=-1

⇒ 2x+(-3x+2y)i-3y(-1)=4+i

⇒ (2x+3y)+(-3x+2y)=4+i

Equating Real and Imaginary parts on both sides, we get

⇒ 2x+3y=4 and -3x+2y=1

On solving we get,

⇒

∴ The real values of x and y are .

Given:

⇒ (3x-2iy)(2+i)²=10(1+i)

⇒ (3x-2yi)(2²+i²+2(2)(i))=10+10i

We know that i²=-1

⇒ (3x-2yi)(4+(-1)+4i)=10+10i

⇒ (3x-2yi)(3+4i)=10+10i

Dividing with 3+4i on both sides

⇒

Multiplying and dividing with 3-4i

⇒

⇒

⇒

⇒

⇒

Equating Real and Imaginary parts on both sides we get

⇒

⇒

∴ The values of x and y are .

Given:

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

⇒ (4+2i)x-3i-3+(9-7i)y=10i

⇒ (4x+9y-3)+i(2x-7y-3)=10i

Equating Real and Imaginary parts on both sides we get

⇒ 4x+9y-3=0 and 2x-7y-3=10

⇒ 4x+9y=3 and 2x-7y=13

On solving these equations we get

⇒ x=3 and y=-1

∴ Thee real values of x and y are 3 and -1

Given:

⇒ (1+i)(x+iy)=2-5i

Dividing with 1+i on both sides we get

⇒

Multiplying and dividing with 1-i

⇒

⇒

We know that i²=-1

⇒

⇒

⇒

Equating Real and Imaginary parts on both sides we get

⇒

∴ The real values of x and y are .

Given complex number is 4-5i

We know that conjugate of a complex number a+ib is a-ib

∴ The conjugate of 4-5i is 4+5i.

Given

complex number is

Let us convert this to standard form a+ib,

Multiplying and dividing with 3-5i

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

We know that complex conjugate of a complex number a+ib is a-ib.

⇒

∴ The conjugate of is .

Given complex number is

Let us convert this to the standard form a+ib

Multiplying and dividing with 1-i

⇒

⇒

We know that i²=-1

⇒

⇒

We know that complex conjugate of a complex number a+ib is a-ib.

⇒

∴ The conjugate of is

Given complex number is

Let us convert this to the standard form a+ib

⇒

⇒

We know that i²=-1

⇒

⇒

Multiplying and dividing with 2-i

⇒

⇒

We know that i²=-1

⇒

⇒

⇒

⇒ a+ib=2-4i

We know that the complex conjugate of a complex number a+ib is a-ib

⇒ a-ib=2+4i

∴ the conjugate of is 2+4i.

Given complex number is

Let us convert this to the standard form a+ib

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

Multiplying and dividing with 3-i

⇒

⇒

⇒

⇒

⇒

We know that complex conjugate of a complex number a+ib is a-ib

⇒

∴ The conjugate of is .

Given complex number is

Let us convert this into the standard form a+ib

⇒

⇒

We know that i²=-1

⇒

⇒

Multiplying and dividing with 4-3i

⇒

⇒

⇒

⇒

⇒

We know that the complex conjugate of a complex number a+ib is a-ib

⇒

∴ The conjugate of complex number is .

Given complex number is Z=1-i

We know that the multiplicative inverse of a complex number Z is .

⇒

Multiplying and dividing with 1+i

⇒

⇒

We know that i²=-1

⇒

⇒

∴ The Multiplicative inverse of 1-i is

Given complex number is Z=(1+i)²

⇒ Z=1²+(i)²+2(1)(i)

⇒ Z=1+3i²+2i

We know that i²=-1

⇒ Z=1+3(-1)+2i

⇒ Z=-2+2i

We know that the multiplicative inverse of a complex number Z is .

⇒

Multiplying and dividing with -2-2i

⇒

⇒

⇒

⇒

⇒

∴ The Multiplicative inverse of (1+i)² is .

Given complex number is Z=4-3i

We know that the multiplicative inverse of a complex number Z is .

⇒

Multiplying and dividing with 4+3i

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

∴ The Multiplicative inverse of 4-3i is .

Given complex number is Z=+3i

We know that the multiplicative inverse of a complex number Z is .

⇒

Multiplying and dividing with -3i

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

∴ The Multiplicative inverse of +3i is .

Given:

⇒ z₁=2-i and z₂=1+i

We have to find

We know that is

⇒

⇒

⇒

We know that |a+ib| is .

⇒

⇒

⇒

∴ The value of is .

Given:

⇒ z₁=2-i and z₂=-2+i

i. We need to find

⇒

⇒

⇒

∴ The Real part of is -2.

ii. We need to find

We know that

⇒

We know that for a complex number Z=a+ib it’s magnitude is given by

⇒

⇒

⇒

⇒

∴ The Imaginary part of the is .

Given complex number is

⇒

We know that i²=-1

⇒

⇒

⇒ Z=2i

We know that for a complex number Z=a+ib it’s magnitude is given by

⇒

⇒ |Z|=2

∴ The modulus of is 2

Given:

⇒

We know that for a complex number Z=a+ib it’s magnitude is given by

We know that is

Applying Modulus on both sides we get,

⇒

⇒

⇒

⇒

⇒

Squaring on both sides

⇒

⇒ x²+y²=1

∴ Thus Proved.

Let us assume the given complex number be

Multiplying and dividing with 1+i

⇒

⇒

We know that i²=-1

⇒

⇒

⇒

⇒

We know that i^2k is real for k>0.

So, the smallest positive integral ‘n’ that can make real is 2.

∴ The smallest positive integral value of ‘n’ is 2.

Let us assume the given complex number as

Multiplying and dividing with 1+2icos

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

For a complex number to be purely real, the imaginary part equals to zero.

⇒

⇒ 3cos=0 (∵ 1+4cos²θ≥1)

⇒ cos θ=0

⇒ , for nI

∴ The values of θ to get the complex number to be purely real is for nI.

Let us write the given complex number as

Multiplying and dividing with (1-i)²

⇒

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

⇒

⇒

⇒

We know that i^2k is real for k≥0.

∴ The least positive integral of n is 1.

Given:

⇒

Rationalising denominator

⇒

⇒

We know that i²=-1

⇒

⇒

⇒

⇒ i³–(-i)³=x+iy

⇒ 2i³=x+iy

⇒ 2i².i=x+iy

⇒ 2(-1)i=x+iy

⇒ -2i=x+iy

Equating Real and Imaginary parts on both sides we get

⇒ x=0 and y=-2

∴ The values of x and y are 0 and -2.

Given:

⇒

⇒

We know that i²=-1

⇒

⇒

Multiplying and dividing with 2+i

⇒

⇒

⇒

⇒

Equating Real and Imaginary parts on both sides we get

⇒

⇒

⇒

∴ The value of x+y is .

Given:

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

⇒

⇒ (-i)¹⁰⁰=a+ib

⇒ i¹⁰⁰=a+ib

⇒ (i²)⁵⁰=a+ib

⇒ (-1)⁵⁰=a+ib

⇒ 1=a+ib

Equating Real and Imaginary parts on both sides we get

⇒ a=1 and b=0

∴ The values of a and b are 1 and 0.

Given:

⇒ a=cosθ+isinθ

⇒

We know that 1+cos2θ=2cos²θ, 1-cos2θ=2sin²θ and sin2θ=2sinθcosθ

⇒

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

⇒

∴ The value of is

Given:

⇒

⇒ 2x³+2x²-7x+72

⇒

⇒

We know that i²=-1

⇒

⇒

⇒

⇒

⇒

⇒ -68+72

⇒ 4

∴ 2x³+2x²-7x+72=4

Given:

⇒ x=3+2i

⇒ x⁴-4x³+4x²+8x+44

⇒ (3+2i)⁴-4(3+2i)³+4(3+2i)²+8(3+2i)+44

⇒ (3⁴+4(3)³(2i)+6(3)²(2i)²+4(3)(2i)³+(2i)⁴)-4(3³+3(3)²(2i)+3(3)(2i)²+(2i)³)+4(3²+(2i)²+2(3)(2i))+24+16i+44

⇒ 81+216i+216i²+96i³+16i⁴-108-216i-144i²-32i³+36+16i²+48i+24+16i+44

We know that i²=-1

⇒ 77+64i+88i²+64i³+16i⁴

⇒ 77+64i+88(-1)+64(-1)(i)+16(-1)²

⇒ 5

∴ x⁴-4x³+4x²+8x+44=5

Given:

⇒ x=-1+i

⇒ x+1=i

⇒ (x+1)⁴=(i)⁴

⇒ x⁴+4x³+6x²+4x+1=2i⁴

We know that i²=-1

⇒ x⁴+4x³+6x²+4x+1=2(-1)²

⇒ x⁴+4x³+6x²+4x+1=2

⇒ x⁴+4x³+6x²+4x+1+8=2+8

⇒ x⁴+4x³+6x²+4x+9=10

∴ x⁴+4x³+6x²+4x+9=10

Given:

⇒

⇒ x⁶+x⁴+x²+1

⇒ x⁴(x²+1)+1(x²+1)

⇒ (x⁴+1)(x²+1)

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

⇒ (-1+1)(i+1)

⇒ (0)(i+1)

⇒ 0

∴ x⁶+x⁴+x²+1=0

Given:

⇒ x=-2-i

⇒ 2x⁴+5x³+7x²-x+41

⇒ 2(-2-)⁴+5(-2-i)³+7(-2-i)²-(-2-i)+41

⇒ 2(2⁴+4(2)³(i)+6(2)²(i)²+4(2)(i)³+(i)⁴)-5(2³+3(2)²(i)+3(2)(i)²+(i)³)+7(2²+2(2)(i)+(i)²)+2+i+41

⇒ 16+64i+144i²+48i³+18i⁴-40-60i-90i²-15i³+28+28i+21i²+i+43

We know that i²=-1

⇒ 127+33i+75i²+33i³+18i⁴

⇒ 127+33i+75(-1)+33(-1)(i)+18(-1)²

⇒ 70

∴ 2x⁴+5x³+7x²-x+41=70

Given:

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

⇒

∴ The values of .

Given:

⇒ (1+i)z=(1-i)

Dividing with (1+i) on both sides we get,

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

⇒

⇒ z=-i

∴ Thus proved

Given:

⇒ Re(z²)=0 and |z|=2

Let us assume Z=x+iy

⇒ Re(z²)=0

⇒ Re((x+iy)²)=0

⇒ Re(x²+(iy)²+2(x)(iy))=0

⇒ Re(x²+i²y²+i(2xy))=0

We know that i²=-1

⇒ Re(x²-y²+i(2xy))=0

⇒ x²-y²=0----------------------(1)

⇒ |z|=2

⇒

⇒ (x²+y²)=2²

⇒ (x²+y²)=4-------------------(2)

Solving (1) and (2) we get

⇒ x= and y=.

∴ .

Given:

⇒ is purely imaginary

⇒ Let us assume , where K is any real number

Let us assume z=x+iy

⇒

Multiplying and dividing with (x+1)-iy

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

Equating Real and Imaginary parts on both sides we get

⇒

⇒ x²+y²-1=0

⇒ x²+y²=1

⇒

⇒ |z|=1

∴ |z|=1

Given:

⇒

Let us assume z₁=x+iy

⇒ |Z₁|=1

⇒

⇒ x²+y²=1-------------------(1)

⇒

⇒

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

⇒

⇒

⇒

∴ z₂ is an imaginary one.

Given:

⇒ |z+1|=z+2(1+i)

Let us assume z=x+iy

⇒ |x+iy+1|=x+iy+2+2i

⇒

Equating Real and Imaginary parts on both sides

⇒ y+2=0

⇒ y=-2----------------(1)

⇒

⇒ (x+1)²+y²=(x+2)²

⇒ x²+2x+1+(-2)²=x²+4x+4

⇒ 2x=1+4-4

⇒ 2x=1

⇒ .

∴ .

Given:

⇒ |z|=z+1+2i

Let us assume z=x+iy

⇒ |x+iy|=x+iy+1+2i

⇒

Equating Real and Imaginary parts on both sides we get

⇒ y+2=0

⇒ y=-2-----------------------(1)

⇒

⇒ x²+(-2)²=(x+1)²

⇒ x²+4=x²+2x+1

⇒ 2x=3

⇒

∴ .

Given:

⇒ (1+i)²ⁿ=(1-i)²ⁿ

⇒ ((1+i)²)ⁿ=((1-i)²)ⁿ

⇒ (1²+i²+2(1)(i))ⁿ=(1²+i²-2(1)(i))ⁿ

We know that i²=-1

⇒ (1-1+2i)ⁿ=(1-1-2i)ⁿ

⇒ (2i)ⁿ=(-2i)ⁿ

We can see that the Relation holds only when n is an even integer.

∴ The smallest positive integer n is 2.

Given:

⇒

⇒

We know that z=|z|²

⇒

⇒

We know that |z|=||

⇒

⇒

∴ |z₁+z₂+z₃|=1.

Given:

⇒ z²+|z|²=0

Let us assume z=x+iy

⇒

⇒ x²+(iy)²+2(x)(iy)+x²+y²=0

⇒ 2x²+y²+i²y²+i2xy=0

We know that i²=-1

⇒ 2x²+y²-y²+i2xy=0

⇒ 2x²+i2xy=0

Equating Real and Imaginary parts on both sides we get,

⇒ 2x²=0 and 2xy=0

⇒ x=0 and yR

∴ z=0+iy where yR. i.e, Infinite solutions.

Given:

⇒ x+iy=-5+12i

Here y>0

We know that for a complex number z=a+ib

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

∴ .

Given:

⇒ x+iy=-7+24i

Here y<0

We know that for a complex number z=a+ib

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

∴ .

Given:

⇒ x+iy=1-i

Here y<0

We know that for a complex number z=a+ib

⇒

⇒

⇒

⇒

∴ .

Given:

⇒ x+iy=-8-6i

Here y<0

We know that for a complex number z=a+ib

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

∴ .

Given:

⇒ x+iy=8-15i

Here y<0

We know that for a complex number z=a+ib

⇒

⇒

⇒

⇒

⇒

⇒

⇒

∴ .

Given:

⇒

⇒ x+iy=-11-60i

Here y<0

We know that for a complex number z=a+ib

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

∴ .

Given:

⇒

⇒

⇒

Here y>0

We know that for a complex number z=a+ib

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

∴

Given:

⇒ x+iy=4i

Here y>0

We know that for a complex number z=a+ib

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

∴ .

Given:

⇒ x+iy=-i

Here y<0

We know that for a complex number z=a+ib

⇒

⇒

⇒

⇒

⇒

⇒

⇒

∴ .

Given complex number is Z=1+i

We know that the polar form of a complex number Z=x+iy is given by Z=|Z|(cosθ+isinθ)

Where,

|Z|=modulus of complex number=

θ =arg(z)=argument of complex number=

Now for the given problem,

⇒

⇒

⇒

⇒

Since x>0,y>0 complex number lies in 1^st quadrant and the value of θ will be as follows 0⁰≤θ≤90⁰.

⇒

⇒ .

⇒

∴ The Polar form of Z=1+i is .

Given Complex number is Z=+i

We know that the polar form of a complex number Z=x+iy is given by Z=|Z|(cosθ+isinθ)

Where,

|Z|=modulus of complex number=

θ =arg(z)=argument of complex number=

Now for the given problem,

⇒

⇒

⇒

⇒

⇒

Since x>0,y>0 complex number lies in 1^st quadrant and the value of θ will be as follows 0⁰≤θ≤90⁰.

⇒ .

⇒

∴ The Polar form of Z=+i is .

Given complex number is z=1-i

We know that the polar form of a complex number Z=x+iy is given by Z=|Z|(cosθ+isinθ)

Where,

|Z|=modulus of complex number=

θ =arg(z)=argument of complex number=

Now for the given problem,

⇒

⇒

⇒

⇒

Since x>0,y<0 complex number lies in 4^th quadrant and the value of θ will be as follows -90⁰≤θ≤0⁰.

⇒

⇒ .

⇒

⇒

∴ The Polar form of Z=1+i is .

Given complex number is

⇒

⇒

We know that i²=-1

⇒

⇒

⇒

⇒ z=0-i

We know that the polar form of a complex number Z=x+iy is given by Z=|Z|(cosθ+isinθ)

Where,

|Z|=modulus of complex number=

θ =arg(z)=argument of complex number=

Now for the given problem,

⇒

⇒

⇒

⇒ |z|=1

⇒

Since x≥0,y<0 complex number lies in 4^th quadrant and the value of θ will be as follows -90⁰≤θ≤0⁰.

⇒

⇒ .

⇒

⇒

∴ The Polar form of is .

Given complex number is .

⇒

⇒

We know that i²=-1

⇒

⇒

We know that the polar form of a complex number Z=x+iy is given by Z=|Z|(cosθ+isinθ)

Where,

|Z|=modulus of complex number=

θ =arg(z)=argument of complex number=

Now for the given problem,

⇒

⇒

⇒

⇒

Since x>0,y<0 complex number lies in 4^th quadrant and the value of θ will be as follows -90⁰≤θ≤0⁰.

⇒

⇒ .

⇒

⇒

∴ The Polar form of is .

Given complex number is .

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

⇒

We know that the polar form of a complex number Z=x+iy is given by Z=|Z|(cosθ+isinθ)

Where,

|Z|=modulus of complex number=

θ =arg(z)=argument of complex number=

Now for the given problem,

⇒

⇒

⇒

⇒

⇒

Since x<0,y>0 complex number lies in 2^nd quadrant and the value of θ will be as follows 90⁰≤θ≤180⁰.

⇒

⇒ .

⇒

∴ The Polar form of is .

Given complex number is z=sin120⁰-icos120⁰

⇒

⇒

We know that the polar form of a complex number Z=x+iy is given by Z=|Z|(cosθ+isinθ)

Where,

|Z|=modulus of complex number=

θ =arg(z)=argument of complex number=

Now for the given problem,

⇒

⇒

⇒

⇒ |z|=1

⇒

Since x>0,y>0 complex number lies in 1^st quadrant and the value of θ will be as follows 0⁰≤θ≤90⁰.

⇒

⇒ .

⇒

∴ The Polar form of Z=sin120⁰-icos120⁰ is .

Given complex number is

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

⇒ z=-4+i4

We know that the polar form of a complex number Z=x+iy is given by Z=|Z|(cosθ+isinθ)

Where,

|Z|=modulus of complex number=

θ =arg(z)=argument of complex number=

Now for the given problem,

⇒

⇒

⇒

⇒ |z|=8

⇒

Since x<0,y>0 complex number lies in 2^nd quadrant and the value of θ will be as follows 90⁰≤θ≤180⁰.

⇒

⇒ .

⇒

∴ The Polar form of is .

Given Complex number is Z=(i²⁵)³

⇒ Z=i⁷⁵

⇒ Z=i⁷⁴.i

⇒ Z=(i²)³⁷.i

We know that i²=-1

⇒ Z=(-1)³⁷.i

⇒ Z=(-1).i

⇒ Z=-i

⇒ Z=0-i

We know that the polar form of a complex number Z=x+iy is given by Z=|Z|(cosθ+isinθ)

Where,

|Z|=modulus of complex number=

θ =arg(z)=argument of complex number=

Now for the given problem,

⇒

⇒

⇒

⇒ |z|=1

⇒

Since x>0,y<0 complex number lies in 4^th quadrant and the value of θ will be as follows -90⁰≤θ≤0⁰.

⇒

⇒ .

⇒

⇒

∴ The Polar form of Z=(i²⁵)³ is .

Given Complex number is Z=1+itanα

We know that the polar form of a complex number Z=x+iy is given by Z=|Z|(cosθ+isinθ)

Where,

|Z|=modulus of complex number=

θ =arg(z)=argument of complex number=

We know that tanα is a periodic function with period .

We have lying in the interval

Case1:

⇒

⇒

⇒

⇒

Since sec is positive in the interval

⇒

⇒

⇒

Since tan is positive in the interval

⇒ θ=

∴ The polar form is z=sec(cos+isin).

Case2:

⇒

⇒

⇒

⇒

Since sec is negative in the interval .

⇒

⇒

⇒

Since tan is negative in the interval .

⇒ .(∵ θ lies in 4^th quadrant)

⇒ z=-sec(cos()+isin())

⇒ z=-sec(-cos-isin)

⇒ z=sec(cos+isin)

∴ The polar form is z=sec(cos+isin)

Given Complex number is tan-i

We know that the polar form of a complex number Z=x+iy is given by Z=|Z|(cosθ+isinθ)

Where,

|Z|=modulus of complex number=

θ =arg(z)=argument of complex number=

We know that tanα is a periodic function with period .

We have lying in the interval

Case1:

⇒

⇒

⇒

⇒

Since sec is positive in the interval

⇒

⇒

⇒

Since cot is positive in the interval

⇒ (∵ θ lies in 4^th quadrant)

⇒

⇒ z=sec(sin-icos)

∴ The polar form is z=sec(sin-icos)

Case2:

⇒

⇒

⇒

⇒

Since sec is negative in the interval .

⇒

⇒

⇒

Since cot is negative in the interval .

⇒ . (∵ θ lies in3^rd quadrant)

⇒

⇒ z=-sec(-sin+icos)

⇒ z=sec(sin-icos)

∴ The polar form is z=sec(sin-icos).

Given Complex number is z=1-sin+icos

We know that sin²θ+cos²θ=1, sin2θ=2sinθcosθ, cos2θ=cos²θ-sin²θ.

⇒

⇒

e know that the polar form of a complex number Z=x+iy is given by Z=|Z|(cosθ+isinθ)

Where,

|Z|=modulus of complex number=

θ =arg(z)=argument of complex number=

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

We know that sine and cosine functions are periodic with period 2

Here We have 3 intervals as follows:

(i)

(ii)

(iii)

Case(i):

In the interval , and also

so,

⇒

⇒

⇒

⇒ .(∵ θ lies in 1^st quadrant)

∴ The polar form is .

Case(ii):

In the interval , and also

so,

⇒

⇒

⇒

⇒

⇒ . (∵ θ lies in 4^th quadrant)

⇒

∴ The polar form is .

Case(iii):

In the interval , and also

so,

⇒

⇒

⇒

⇒

⇒ .(since θ presents in first quadrant and tan’s period is )

⇒ .

∴ The polar form is .

Given complex number is

⇒

⇒

⇒

⇒

We know that i²=-1

⇒

⇒

⇒

We know that the polar form of a complex number Z=x + iy is given by Z=|Z|(cos θ+ i sin θ)

Where,

|Z|=modulus of complex number

θ = arg(z)=argument of complex number

Now for the given problem,

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

Since x<0,y<0 complex number lies in 3^rd quadrant and the value of θ will be as follows -180⁰≤θ≤-90⁰.

⇒

⇒ .

⇒

⇒

∴ The Polar form of is .

Given:

⇒ |z₁|=|z₂| and arg(z₁)+arg(z₂)=

Let us assume arg(z₁)=θ

⇒ arg(z₂)=-θ

We know that z=|z|(cosθ+isinθ)

⇒ z₁=|z₁|(cosθ+isinθ)-----------------(1)

⇒ z₂=|z₂|(cos(-θ)+isin(-θ))

⇒ z₂=|z₂|(-cosθ+isinθ)

⇒ z₂=-|z₂|(cosθ-isinθ)

Now we find the conjugate of z₂

⇒ =-|z₂|(cosθ+isinθ) (∵ )

Now,

⇒

⇒ (∵ |z₁|=|z₂|)

⇒ z₁=-

∴ Thus proved.

Given:

⇒

⇒

We know that

⇒

⇒

⇒

We know that arg(z)+arg()=0

⇒

⇒

∴ Thus proved.

Given Complex number is

We know that sin2θ=2sinθcosθ and 1-cos2θ=2sin²θ

⇒

⇒

∴ The Polar form of is .

Let ……………….1

Squaring both sides, we get

i² = (a²-b²) +2aib

By comparing real and imaginary term, we get

2ab = 1 and a²-b² = 0

By solving these we get

By putting value of a and b in 1, we get

Squaring both side

-i = (x + iy)²

= (x²-y²)+2ixy

x²-y² = 0

2xy = -1

As we know all that,

(x²+y²)² = (x²-y²)²+4x²y²

(x²+y²)² = 0+1

(x²+y²)² = 1

x²+y² = 1

X²-y² = 0 ……………………. (1)

x²+y² = 1 ……………….(2)

From (1)

x² = y²…………….(3)

2x² = 1 (because x² = y²)

2xy = -1

it means xy<0

Either x<0 , y > 0

Or x>0, y<0

X and y have different sign

By squaring both sides, we get

On comparing real and imaginary parts, we get

We know that,

As we all know that,

z = 1 + cos θ + i sin θ

a = (1+cosθ) and b = sin θ

π< θ < 2π it means z lies in second quadrant

z = -θ

Explanation

As we know that i² = -1, i³ = -i , i⁴ = 1

= i

Explanation

= -2

_{Z = 1-i = a+ib}

_{So, a = 1 , b = -1}

tan α a>0 , b<1

∴z lies in forth quadrant

arg (z) = θ

Required polar form

=

_{So, a = 1 , b = -1}

tan α a<0 , b>1

∴z lies in second quadrant

Required polar form

=

_{So, a = 0 , b = -1}

a = 0 , b = -1

∴z lies in forth quadrant

arg(z) = θ

= iⁿ

As we know that i² = -1

And value of n is real number so,

n = 2

_{As we know that, z = a+ib}

= 1+i²+2i√3

= 1-3+2i√3

= -2+2i√3

a = -2 b = 2√3

= |√3|

α<0 , b>1

∴z lies in second quadrant

arg(z) = θ

r = |z| = 4 ,

z = r(cos θ + i sin θ)

z = -2√3+2i

z = a + ib

|a+ib-5i| = |a+ib+5i|

|a+ib-5i|² = |a+ib+5i|²

|a +i(b-5)|² = |a + i(b+5)|²

a²+(b-5)² = a²+(b+5)²

a²+b²+25-10b = a²+b²+25+10b

20b = 0

b = 0

b is a imaginary part of z

= y

= 0

Im (z) = 0

So, the locus point is real axis

_{Comparing real and imaginary part, we get}

_{So, X}²+Y²

= 5i×3i

= 15i²

= -15

0

Explanation

Here,

= i

n = 1000 terms

= 0

As we all know that,

z = r(cos θ + i sin θ ) , θ = arg(z)

= r(cos(-θ)+sin(-θ)

So,

= 0

6 and 0

Explanation

As we all know that,

|z₁ +z₂ |≤|z₁ |+|z₂ | and |z₁ +z₂ |≥|z₁ |-|z₂ |

Suppose,

Z₁ = z+4

Z₂ = -3

| z₁|-| z₂|≤| z₁+ z₂|≤ | z₁|+| z₂|

|z+4|-|-3|≤|z+4-3|≤ |z+4|+|-3|

|z+4|-3≤|z+1|≤ |z+4|+3

3-3≤|z+1|≤ 3+3 (Given-|z + 4| ≤ 3)

0≤|z+1|≤ 6

|az₁-bz₂|² + |az₂+bz₁|²

= | z₁|² (a²+b² )+| z₂|² (a²+b² )

= (a²+b² )(| z₁|²+| z₂|²)

As we know that,

i = √(-1 ), i² = -1, i³ = -i, i⁴ = 1

z = iⁿ + iⁿ⁺¹ + iⁿ⁺² + iⁿ⁺³

= iⁿ (1+i¹+i²+i³ )

= iⁿ (1+i-1-i)

= iⁿ(0)

= 0

a = 2

Explanation

Z is a purely real, it means Im (z) = 0

Z = 3i³-3ai²+ (1-a) i+5

= -3i+3a+ (1-a) i+5

= (3a+5)+i(-3+1-a)

= (3a+5)+i(-2-a)

Re(z) = 3a+5 , Im(z) = (-2-a)

Z is a real so, Im (z) = 0

-2-a = 0

a = -2

r = |z| = 2 ,

z = r(cos θ + i sin θ)

z = √2 (1+i)

As we know that,

arg (z₁z₂) = arg(z₁)+arg(z₂) so,

arg (z₁z₂z₃) = arg(z₁)+arg(z₂)+arg(z₃)

arg((1+√3 i)(1+i)(cos θ+ i sin θ)

= arg(1+√3 i)+arg(1+i)+arg(cos θ + i sin θ).......... (1)

∴arg(z₁ ) = θ

z₂ = arg(1+i)

z₃ = arg(cos θ + i sin θ)

⇒1(cos θ + i sin θ)

arg(z₃ ) = θ

∵ In r(cos θ + i sin θ) , arg(z) = θ

By putting the value of all arg in 1, we get

We know that

i= √(-1)

i²= i × i

= -1

(1+i)(1+i²)(1+i³)(1+i⁴) = (1+i)(1+(-1))(1+i³)(1+i⁴)

= (1+i)(0)(1+i³)(1+i⁴)

= 0

For real number, imaginary part should be 0

⇒ 8 sin θ = 0

⇒ θ = n π

As θ belongs to (0,2π) so θ = π

Given that (1 + i) (1 + 2i) (1 + 3i) …. (1 + n i) = a + i b …(1)

We can also say that

(1 - i) (1 - 2i) (1 - 3i) …. (1 - n i) = a - i b …(2)

Multiply and divide the eq no. 2 with eq no. 1

((1)² – (i)²)((1)² – (2i)²)……((1)² – (ni)²) = ((a)² – (ib)²)

2 × 5 × 10 × …… × (1 + n²) = a² + b²

Square both sides

a + ib = (x + iy)² = x² + i2xy -y²

So, we can say that a = x² – y² and b = 2xy

a – ib = (x² – y²) – i(2xy)

= (x)² + 2(x)(-iy) + (-iy)²

= (x + (-iy))²

= (x – iy)²

z = (i²⁵)³ = i⁷⁵ = i^4×18+3

We know that i⁴ =1 and i³ = -i

z = i^4×18.i³ = 0 – i

z = |z|(cos θ + i sin θ)

We know that

i⁴ⁿ⁺¹ = i

i⁴ⁿ⁺² = i² = -1

i⁴ⁿ⁺³ = i³ = -i

i⁴ⁿ⁺⁴ = i⁴ = 1

i⁴ⁿ⁺¹ + i⁴ⁿ⁺² + i⁴ⁿ⁺³ + i⁴ⁿ⁺⁴ = i + (-1) + (-i) + 1

= 0

S = i + i² + i³ + …… upto 1000 terms

We can make the pair of 4 terms because we know that value is repeat after every 4^th terms. So, there are total 250 pairs are made and each pair have value equal to 0.

S = 0

Given that (1 + i) (1 + 2i) (1 + 3i) …. (1 + n i) = a + i b …(1)

We can also say that

(1 - i) (1 - 2i) (1 - 3i) …. (1 - n i) = a - i b …(2)

Multiply and divide the eq no. 2 with eq no. 1

((1)² – (i)²)((1)² – (2i)²)……((1)² – (ni)²) = ((a)² – (ib)²)

2 × 5 × 10 × …… × (1 + n²) = a² + b²

We know that the principal value of amplitude is value of argument lie between (-π,π]

arg(z) = tan^-1(1)

So, is called the principal value of the amplitude of (1 + i) because it lies between (-π,π]

Let check the value of (1 + i)ⁿ for different value of n

at n =1 , 1+ i (no)

at n =2 , (1 + i)² = 1 + i² + 2i = 2i (no)

at n =3 , (1 + i)²(1 + i) = (1 + i)(2i) = 2i – 2 (no)

at n =4 , (1 + i)²(1 + i)² = (2i)² = -4 (no)

at n =5 , (1 + i)⁴(1 + i) = -4(1 + i) (no)

at n =6 , (1 + i)⁴(1 + i)² = -4(2i) (no)

at n =7 , (1 + i)⁶(1 + i) = -8i(1 + i) = -8i + 8 (no)

at n =8 , (1 + i)⁴(1 + i)⁴ = (-4)(-4) = 8 (yes)

So, we can say that n = 8 is the least positive integer for which is positive integer.

Let, z = re^iθ

=1

Solve option A

=|e^-iθ |

=1

a² = (1 + i)(1 + i)

= 1² + 2i + i²

= 1 – 1 + 2i

= 2i

(x + iy)^1/3 = a + ib

x + iy = (a + ib)³

= a³ + (ib)³ + 3a²(ib) + 3a(ib)²

= a³ – ib³ + i3a²b – 3ab²

= (a³ – 3ab²) + i(3a²b – b³)

x = a³ – 3ab² and y = 3a²b – b³

= a² – 3b² + 3a² – b²

= 4(a² – b²)

√-2= √2 i and √-3= √3 i

= i² √6

= -√6

=60⁰

But answer is going in 3^rd quadrant because tan θ is positive but sin θ and cos θ both are negative and it is possible only in 3^rd quadrant.

So, answer is π + 60⁰ = 180⁰

=i

z⁴= i⁴

=1

=1+i0

=0

Given that (1 + i) (1 + 2i) (1 + 3i) = x + i y …(1)

We can also say that

(1 - i) (1 - 2i) (1 - 3i) = x - i y …(2)

Multiply and divide the eq no. 2 with eq no. 1

((1)² – (i)²)((1)² – (2i)²)((1)² – (3i)²) = ((x)² – (iy)²)

x² + y² = 2 × 5 × 10 = 100

=1

= √2

=0+i(-1)

=0+i(-1)

We know that

i⁴ⁿ⁺¹ = i

i⁴ⁿ⁺² = i² = -1

i⁴ⁿ⁺³ = i³ = -i

i⁴ⁿ⁺⁴ = i⁴ = 1

i⁵ + i⁶ + i⁷ + i⁸ + i⁹ = i + (-1) + (-i) + 1 + i

= i

= -i

We know that

i⁴ⁿ⁺¹ = i

i⁴ⁿ⁺² = i²

= -1

i⁴ⁿ⁺³ = i³

= -i

i⁴ⁿ⁺⁴ = i⁴

= 1

i⁵⁹² = i^4(147)+4

= 1

i⁵⁸² = i^4(145)+2

= -1

i⁵⁹⁰ = i^4(147)+2

= -1

i⁵⁸⁰ = i^4(144)+4

= 1

i⁵⁸⁸ = i^4(146)+4

= 1

i⁵⁷⁸ = i^4(144)+2

= -1

i⁵⁸⁶ = i^4(146)+2

= -1

i⁵⁷⁶ = i^4(143)+4

= 1

i⁵⁸⁴ = i^4(145)+4

= 1

i⁵⁷⁴ = i^4(143)+2

= -1

= -2

(1 + i)⁴ + (1 - i)⁴ = ((1 + i)²)² + ((1 - i)²)²

= (2i)² + (-2i)²

= -4 + -4

= -8

If z = a + ib lies in third quadrant then a and b both are less than zero

a² – b² < 0 and ab > 0 because a² + b² is always greater than zero

(a – b)(a + b) < 0

Here a and b both are less than zero that means (a + b) is always less than zero

So, a – b > 0 ⇒ a > b

Then, final answer is b < a < 0