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Let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂

⇒ z₁ + z₂ = (x₁ + x₂) + i (y₁ + y₂) which is real

⇒ y₁ + y₂ = 0

⇒ y₁ = - y₂

Assuming x₁ = x₂

Since z₂ = x₁ – iy₁

∴ z₂ = z̅ ₁

√−16 + 3√−25 + √−36 − √−625

= i√16 + 3i√25 + i√36 −i√625 

= 4i+ 15i + 6i - 25i 

= 25i - 25i 

= 0

The given vertices are z, iz, z + iz ⇒ z, iz are ⊥r to each other. 

Area of triangle = (1/2) bh = 50 

⇒ (1/2) |z| |iz| = 50 

⇒ (1/2) |z| |z| = 50 

⇒ |z|² = 100 

⇒ |z| = 10

i√−16 + i√−25 + √49 − i√−49+14

=i² √16 + i²√25 + √49 − i²√49 + 14 

= 4i² + 5i² + 7 − 7i² + 14

 = -4 – 5 + 7 + 7 + 14 

= -9 + 28 

= 19

Given |z₁| = |z₂|

If |z₁| = |z₂| then z₁ and z₂ are at the same distance from origin.

But if arg (z₁) ≠ arg (z₂) then z₁ and z₂ are different.

So, if |z₁| = |z₂|, then it is not necessary that z₁ = z₂.

For example: z₁ = 3 + 4i and z₂ = 4 + 3i

Here |z₁| = |z₂| = 5 but z₁ ≠ z₂.

(a) (1/2) |z|

Area of triangle = (1/2) bh

= (1/2) |z| |iz|

= (1/2) |z|²

(i⁷⁷ + i⁷⁰+i⁸⁷ + i⁴¹⁴)³

= [(i⁴ )¹⁹i + (i⁴ )¹⁷ i² + (i⁴ )²¹i³ + (i⁴ )¹⁰³i² ]³

 = [ i+i² + i³ + i² ] ³    [∵i⁴ = 1] 

= [ i− 1 − i − 1] ³      [∵ i² = −1and i³ = −]

 = [-2]³ = -8

(i) (x + iy) (2 – 3i) = 4 + i

Given as

(x + iy) (2 – 3i) = 4 + i

Now let us simplify the expression we get,

x(2 – 3i) + iy(2 – 3i) = 4 + i

2x – 3xi + 2yi – 3yi² = 4 + i

2x + (-3x + 2y)i – 3y(-1) = 4 + i [since, i² = -1]

2x + (-3x + 2y)i + 3y = 4 + i [since, i² = -1]

(2x + 3y) + i(-3x + 2y) = 4 + i

By equating real and imaginary parts on both sides, we get

2x + 3y = 4… (i)

And -3x + 2y = 1… (ii)

Now multiply (i) by 3 and (ii) by 2 and add

By solving we get,

6x – 6x – 9y + 4y = 12 + 2

13y = 14

y = 14/13

Then substitute the value of y in (i) we get,

2x + 3y = 4

2x + 3(14/13) = 4

2x = 4 – (42/13)

= (52 - 42)/13

2x = 10/13

x = 5/13

x = 5/13, y = 14/13

Thus the real values of x and y are 5/13, 14/13

(ii) (3x – 2i y) (2 + i)² = 10(1 + i)

Given as

(3x – 2i y) (2 + i)² = 10(1 + i)

(3x – 2yi) (2² + i² + 2(2)(i)) = 10 + 10i

(3x – 2yi) (4 + (-1) + 4i) = 10 + 10i [since, i² = -1]

(3x – 2yi) (3 + 4i) = 10 + 10i

Now let us divide with 3 + 4i on both sides we get,

(3x – 2yi) = (10 + 10i)/(3 + 4i)

Then multiply and divide with (3 - 4i)

= [10(3 - 4i) + 10i(3 - 4i)]/(3² – (4i)²)

= [30 - 40i + 30i - 40i²]/(9 – 16i²)

= [30 - 10i - 40(-1)]/(9 - 16(-1))

= [70 - 10i]/25

Then, equating real and imaginary parts on both sides we get

3x = 70/25 and -2y = -10/25

x = 70/75 and y = 1/5

x = 14/15 and y = 1/5

Thus the real values of x and y are 14/15, 1/5

Given, (x + iy) (2 − 3i) = 4 + i 

⟹ (2x + 3y) + i(−3x + 2y) = 4 + i

⟹ 2x + 3y = 4 and − 3x + 2y = 1 

On solving, we get,

x = \(\frac{5}{13}\) and y = \(\frac{14}{13}\)

\(\frac{x+iy}{2+3i}=7-i\) 

x + iy = (7 – i)(2 + 3i)

x + iy = 14 + 21i – 2i – 3i²

x + iy = 14 + 19i – 3(-1)

x + iy = 17 + 19i

Equating real and imaginary parts, we get

∴ x = 17 and y = 19

Given:  \(\frac{(x+3i)}{(2+iy)}\) = (1 - i)

⇒ x + 3i = (1 – i)(2 + iy)

⇒ x + 3i = 1(2 + iy) – i(2 + iy)

⇒ x + 3i = 2 + iy – 2i – i 2y

⇒ x + 3i = 2 + i(y – 2) – (-1)y [i² = -1]

⇒ x + 3i = 2 + i(y – 2) + y

⇒ x + 3i = (2 + y) + i(y – 2)

Comparing the real parts, we get

x = 2 + y

⇒ x – y = 2 …(i)

Comparing the imaginary parts, we get

3 = y – 2

⇒ y = 3 + 2

⇒ y = 5

Putting the value of y = 5 in eq. (i), we get

x – 5 = 2

⇒ x = 2 + 5

⇒ x = 7

Hence, the value of x = 7 and y = 5

(b) 2

|z₁ + z₂ + z₃| = 2

(i) Let z = x + iy 

|z – 2 – i| = 3 

⇒ |x + iy – 2 – i| = 3 

⇒ |(x – 2) + i(y – 1)| = 3

= √((x - 2)² + (y -1)²) = 3

Squaring on both sides

(x – 2)² + (y – 1)² = 9 

⇒ x² – 4x + 4 + y² – 2y + 1 – 9 = 0 

⇒ x² + y² – 4x – 2y – 4 = 0 represents a circle 

2g = -4 ⇒ g = -2

2f = -2 ⇒ f = -1 

c = -4 

(a) Centre (-g, -f) = (2, 1) = 2 + i (b) 

(b) Radius = √(g² + f² - c) = √(4 + 1 + 4) = 3

Aliter: |z – (2 + i)| = 3 

Centre = 2 + i 

Radius = 3

(ii) |2(x + iy) + 2 – 4i| = 2 

⇒ |2x + i2y + 2 – 4i| = 2 

⇒ |(2x + 2) + i(2y – 4)| = 2 

⇒ |2(x + 1) + 2i(y – 2)| = 2 

⇒ |(x + 1) + i(y – 2)| = 1

= √((x + 1)² + (y - 2)²) = 1

Squaring on both sides, 

x² + 2x + 1 + y² + 4 – 4y – 1 = 0

⇒ x² + y² + 2x – 4y + 4 = 0 represents a circle 

2g = 2 ⇒ g = 1 

2f = -4 ⇒ f = -2 

c = 4 

(a) Centre (-g, -f) = (-1, 2) = -1 + 2i 

(b) Radius = √(g² + f² - c) = √(1 + 4 - 4) = 1

Aliter: 2|(z + 1 – 2i)| 

= 2 |z – (-1 + 2i)| = 1 

Centre = -1 + 2i 

Radius = 1

(iii) |3(x + iy) – 6 + 12i| = 8 

⇒ |3x + i3y – 6 + 12i| = 8 

⇒ |3(x – 2) + i3 (y + 4)| = 8 

⇒ 3|(x – 2) + i (y + 4)| = 8

= 3√((x - 2)² + (y + 4)²) = 8

Squaring on both sides, 9[(x – 2)² + (y + 4)²] = 64 

⇒ x² – 4x + 4 + y² + 8y + 16 = 64/9

⇒ x² + y² – 4x + 8y + 20 – (64/9) = 0

x² + y² – 4x + 8y + (116/9) = 0 represents a circle. 

2g = -4 ⇒ g = -2 

2f = 8 ⇒ f = 4 

c = 116/9

(a) Centre (-g, -f) = (2, -4) = 2 – 4i 

(b) Radius = √(g² + f² - c) = √(4 + 16 - (116/9)) = √(180 - 116)/9 = 8/3

Aliter: |z – 2 + 4i| = 8/3

⇒ |z – (2 – 4i)| = 8/3

Centre = 2 – 4i, Radius = 8/3

Z₁ = 1 + i, 

z₂ = 4 – 3i, 

z₃ = 2 + 5i 

Z₁ (z₂ – z₃) = 1 + i[(4 – 3i) – (2 + 5i)] = 1 + i[2 – 8i] 

= 2 – 8i + 2i + 8 

= 10 – 6i …(1) 

Z₁ z₂ = (1 + i) (4 – 3i) = 4 – 3i + 4i + 3 = 7 + i 

Z₁ Z₃ = (1 + i) (2 + 5i) = 2 + 5i + 2i – 5 = 7i – 3 

Z₁ Z₂ – Z₁ z₃ = 7 + i – (7i – 3) = 7 + i – 7i + 3 

= 10 – 6i …(2) 

From (1) and (2) we get, 

Z₁ (Z₂ Z₃) = Z₁ Z₂ – z₁ z₃

(i) Z₁ + (z₂ + z₃) 

= 4 – 7i + (2 + 3i + 1 + i) 

= 4 – 7i + (3 + 4i) 

= 7 – 3i 

(z₁ + z₂) + z₃ = (4 – 7i + 2 + 3i) + (1 + i) 

= (6 – 4i) + (1 + i) 

= 7 – 3i …(2) 

From (1) and (2) we get, z₁ + (z₂ + z₃) = (z₁ + z₂) + z₃

(ii) (z₁ z₂) z₃ = (4 – 7i) + (2 + 3i) (1 + i) 

= (8 + 12i – 14 i + 21) (1 + i) 

= (29 – 2i)(1 + i) = 29 + 29i – 2i + 2 

= 31 + 27i ….(1) 

Z₁ (z₂ z₃) = (4 – 7i) [(2 + 3i) (1 + i)] 

= 4 – 7i [2 + 2i + 3i – 3] 

= 4 – 7i[5i – 1] = 20i – 4 + 35 + 7i 

= 31 + 27i …(2) 

From (1) and (2) we get, (Z₁ Z₂) z₃ = z₁ (z₂ z₃)

(i) z = x + iy 

|z – 4| = 16

⇒ |x + iy – 4| = 16 

⇒ |(x – 4) + iy| = 16

= √((x - 4)² + y²) = 16

Squaring on both sides 

(x – 4)² + y² = 256 

⇒ x² – 8x + 16 + y² – 256 = 0 

⇒ x² + y² – 8x – 240 = 0 represents the equation of circle

(ii) |x + iy – 4|² – |x + iy – 1|² = 16 

⇒ |(x – 4) + iy|² – |(x – 1) + iy|² = 16 

⇒ [(x – 4)² + y²] – [(x – 1)² + y²] = 16 

⇒ (x² – 8x + 16 + y²) – (x² – 2x + 1 + y²) = 16

⇒ x² + y² – 8x + 16 – x² + 2x – 1 – y² = 16 

⇒ -6x + 15 = 16 

⇒ 6x + 1 = 0

Given: 6i⁵⁰ + 5i³³ – 2i¹⁵ + 6i⁴⁸

To prove: 6i⁵⁰ + 5i³³ – 2i¹⁵ + 6i⁴⁸ = 7i

⇒ 6i^4×12+2 + 5i^4×8+1 – 2i^4×3+3 + 6i^4×12

⇒ 6i² + 5i¹ – 2i³ + 6i⁰

⇒ -6+5i+2i+6

⇒ 7i

⇒ L.H.S. = R.H.S.

Hence proved.

(b) imaginary axis

|z + 2| = | z – 2| 

⇒ |x + iy + 2| = |x + iy – 2| 

⇒ |x + 2 + iy|² = |x – 2 + iy|²

⇒ (x + 2)² + y² = (x – 2)² + y² 

⇒ x² + 4 + 4x = x² + 4 – 4x 

⇒ 8x = 0 

⇒ x = 0

(c) 2 + i

Complex roots occur in pairs when -i + 2 is one root, i.e., when 2 – i is a root, the other root is 2 + i

(a) The answer is |z| = 1

(d) 10

α = -i + 3 = 3 – 7 

⇒ The other root is β = 3 + i 

Product of the roots = αβ = (3 – i)(3 + i) = 10 

Product of the roots of x² – 6x + k = 0 is k ⇒ k 

= 10

We have, (2 – 3i)³

= 2³ – 3 × 2² × 3i – 3 × 2 × (3i)² – (3i)³

= 8 – 36i + 54 + 27i

= 46 - 9i.

Let, z = -3i

Let 0 = r cosθ and -3 = r sinθ

By squaring and adding, we get

(0)² + (-3)² = (r cosθ)² + (r sinθ)²

⇒ 0+9 = r²(cos²θ + sin²θ)

⇒ 9 = r²

⇒ r = 3

∴ cosθ = 0 and sinθ = -1

Since, θ lies in fourth quadrant, we have

θ = 3π/2

Thus, the required polar form is \(3[cos(3\frac{3\pi}{2})+i\,sin(\frac{3\pi}{2})]\)

(i) ω is a cube root of unity ω³ = 1; 

1 + ω + ω² = 0 

(1 – ω + ω²)⁶ + (1 + ω – ω²)⁶ 

= (-ω – ω)⁶ + (-ω²– ω ²)⁶ 

= (-2ω)⁶ + (-2ω²)⁶ 

= (-2)⁶ (ω⁶ + ω¹²) 

= (64)(1 + 1) 

= 128

(ii) (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸) …… (1 + ω²ⁿ) 

= (1 + ω)(1 + ω²)(1 + ω⁴)(1 + ω⁸) ……. 2n factors 

= (-ω²)(-ω)(-ω²)(-ω) …… 2n factors 

= ω³ . ω³ 

= 1

Let z = 1 + i√3

a = 1, b = √3, i.e. a > 0, b > 0

|z| = \(\sqrt{a^2+b^2}=\sqrt{162+(\sqrt3)^2}=\sqrt{1+3}=2\)

Here, (1, √3) lies in 1st quadrant.

amp(z) = tan^-1\((\frac{b}a)\) = tan^-1√3 = \(\frac{\pi}3\)

We have, z = (–1 + i√3)

Let -1 = r cosθ and √3 = r sinθ

By squaring and adding, we get

(-1)² + (√3)² = (r cosθ)² + (r sinθ)²

⇒ 1+3 = r² (cos²θ + sin²θ)

⇒ 4 = r²

⇒ r = 2

∴ cosθ = -1/2 and sinθ = √3/2

Since, θ lies in second quadrant, we have

θ = π - π/3 = 2π/3

Thus, the required polar form is 2[cos\(\frac{2\pi}{3}\)+i sin\(\frac{2\pi}{3}\)]

False

Explanation:

Given z ≠ 0 and Re (z) = 0

Let z = x + iy

Then x = 0

∴ z = iy

⇒ Im (z²) = i²y² = -y² ≠ 0

\(\frac{1-i}{1+i} = \frac{(1-i)}{(1+i)}\times\frac{(1-i)}{(1-i)}\) 

=0 + i(-1)

arg = tan^-1 \(\frac{-1}{0}\) 

= \(\frac{-\pi}{2}\) 

\(\frac{1}{i}=\frac{1}{i}\times\frac{i}{i}\)

= \(\frac{i}{-1}\) 

= 0+i(-1)

amp = tan^-1\(\frac{-1}{0}\) ​

= \(\frac{-\pi}{2}\)​​​​​​

(x + 2y) + (2x – 3y)i + 4i = 5

(x + 2y) + (2x – 3y)i = 5 – 4i

Equating real and imaginary parts, we get

x + 2y = 5 ……(i)

and 2x – 3y = -4 …..(ii)

Equation (i) x 2 – equation (ii) gives

7y = 14

∴ y = 2

Substituting y = 2 in (i), we get

x + 2(2) = 5

x + 4 = 5

∴ x = 1

∴ x = 1 and y = 2

Check:

For x = 1 and y = 2

Consider, L.H.S. = (x + 2y) + (2x – 3y)i + 4i

= (1 + 4) + (2 – 6)i + 4i

= 5 – 4i + 4i

= 5

= R.H.S.

Given: x + 4yi = ix + y + 3

or x + 4yi = ix + (y + 3)

Comparing the real parts, we get

x = y + 3 Or x – y = 3 …(i)

Comparing the imaginary parts, we get

4y = x …(ii)

Putting the value of x = 4y in eq. (i), we get

4y – y = 3

⇒ 3y = 3

⇒ y = 1

Putting the value of y = 1 in eq. (ii), we get

x = 4(1) = 4

Hence, the value of x = 4 and y = 1

x(1 + 3i) + y(2 – i) – 5 + i³ = 0

x + 3xi + 2y – yi – 5 – i = 0 ……[∵ i³ = -i]

(x + 2y – 5) + (3x – y – 1)i = 0 + 0i

Equating real and imaginary parts, we get

x + 2y – 5 = 0 …..(i)

and 3x – y – 1 = 0 ……(ii)

Equation (i) + equation (ii) × 2 gives

7x – 7 = 0

7x = 1

∴ x = 1

Substituting x = 1 in (i), we get

1 + 2y – 5 = 0

2y = 4

y = 2

∴ x = 1 and y = 2

∴ x + y = 1 + 2 = 3

Let, (a + ib)² = - 7 + 24i

Now using, (a + b)² = a² + b² + 2ab

a² + (bi)² + 2abi = -7 + 24i

Since i² = -1

a² - b² + 2abi = -7 + 24i

Now, separating real and complex parts, we get

⇒ a² - b² = - 7…………..eq.1

⇒ 2ab = 24…….. eq.2

⇒ a = 12/b

Now, using the value of a in eq.1, we get

⇒ (12/b)² – b² = - 7

⇒ 144 – b⁴ = -7b²

⇒ b⁴ - 7b² - 144 = 0

Simplify and get the value of b², we get,

⇒ b² = - 9 or b² = 16

As b is real no. so, b² = 16

b = 4 or b = - 4

Therefore, a = 3 or a = - 3

Hence the square root of the complex no. is 3 + 4i and - 3 - 4i.

Let z = x + iy

|z – 3| = 2

|x + iy – 3| = 2

|(x – 3) + iy| = 2

\(\sqrt{(x-3)^2+y^2}=2\) 

\(\therefore\) (x - 3)² + y² = 4

Let z = x + iy

|z| = 10

|x + iy| = 10

\(\sqrt{x^2+y^2}\) = 10

\(\therefore\) x² + y² = 100

Let z = x + iy

|z – 2 – 2i| = |z + 2 + 2i|

|x + iy – 2 – 2i | = |x + iy + 2 + 2i |

|(x – 2) + i(y – 2)| = |(x + 2) + i(y + 2)|

\(\sqrt{(x-2)^2+(y-2)^2}=\sqrt{(x+2)^2 + (y+2)^2}\) 

(x – 2)² + (y – 2)² = (x + 2)² + (y + 2)²

x² – 4x + 4 + y² – 4y + 4

= x² + 4x + 4 + y² + 4y + 4

- 4x – 4y = 4x + 4y

8x + 8y = 0

x + y = 0 

y = -x

Let z = x + iy

|z + 8| = |z – 4|

|x + iy + 8| = |x + iy – 4|

|(x + 8) + iy | = |(x – 4) + iy|

\(\sqrt{(x+8)^2+y^2}=\sqrt{(x-4)^2+y^2}\)

(x + 8)² + y² = (x - 4)² + y²

x² + 16x + 64 + y² = x² – 8x + 16 + y²

16x + 64 = -8x + 16

24x + 48 = 0

∴ x + 2 = 0

Let z = x + iy

|z – 5 + 6i| = 5

|x + iy – 5 + 6i| = 5

|(x – 5) + i(y + 6)| = 5

\(\sqrt{(x-5)^2+(y+6)^2}=5\) 

\(\therefore\) (x - 5)² + (y + 6)² = 25

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

= (2i)² + (-2i)² 

= -4 + -4 

= -8

Given |1 - i|^x = 2^x

⇒ √(1 + 1)^x = 2^x

2^x/2 = 2^x

∴ x/2 = x ⇒ x = 0

∴ There is no non-zero integral solution

We know that 

i= √(-1) 

i²= i × i 

=\(\sqrt{-1}\times\sqrt{-1}\)

= -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

Correct option (a) 2

Explanation:

(x – 1)⁴ = 16

x – 1 = (16)^1/4

x – 1 = ± 2, ± 2i

x = 1 ± 2, 1 ± 2i

Sum of non-real roots is (1 + 2i) + (1 – 2i) = 2

Correct option (b) –1

Explanation:

z = 7√ –1 

z⁷ = –1

∴ z⁸⁶  +  z¹⁷⁵  +  z²⁸⁹ = (z⁷ )¹² z² + (z⁷ )²⁵ + (z⁷ ) ⁴¹.z² 

 = z² – 1 – z² = –1

According to the question,

We have,

z = x + iy

⇒ z̅ = x – iy

Now, we also have,

z z̅ + 2 (z + z̅) + b = 0

⇒ (x + iy) (x – iy) + 2 (x + iy + x – iy) + b = 0

⇒ x² + y² + 4x + b = 0

The equation obtained represents the equation of a circle.

Let, (a + ib)² = -11 - 60i

Now using, (a + b)² = a² + b² + 2ab

⇒ a² + (bi)² + 2abi = -11 - 60i

Since i² = -1

⇒ a² - b² + 2abi = -11 - 60i

Now, separating real and complex parts, we get

⇒ a² - b² = -11…………..eq.1

⇒ 2ab = - 60…….. eq.2

⇒ a = -30/b

Now, using the value of a in eq.1, we get

⇒ \((-\frac{30}{b})^2\) – b² = -11

⇒ 900 – b⁴ = -11b²

⇒ b⁴ - 11b² - 900 = 0

Simplify and get the value of b² , we get,

⇒ b² = 36 or b² = -25

as b is real no. so, b² = 36

b = 6 or b = -6

Therefore , a = - 5 or a = 5

Hence the square root of the complex no. is - 5 + 6i and 5 – 6i.

Let, (a + ib)² = - 4 - 3i

Now using, (a + b)² = a² + b² + 2ab

⇒ a² + (bi)² + 2abi = -4 -3i

Since i² = -1

⇒ a² - b² + 2abi = - 4 - 3i

Now, separating real and complex parts, we get

⇒ a² - b² = - 4…………..eq.1

⇒ 2ab = - 3…….. eq.2

⇒ a = - 3/2b

Now, using the value of a in eq.1, we get

⇒ \((-\frac{3}{2b})^2\) – b² = - 4

⇒ 9 – 4b⁴ = -16b²

⇒ 4b⁴ - 16b² - 9 = 0

Simplify and get the value of b², we get,

⇒ b² = 9/2 or b² = - 2

As b is real no. so, b² = 9/2

b = \(\frac{3}{\sqrt2}\) or b = -\(\frac{3}{\sqrt2}\)

Therefore, a = -\(\frac{1}{\sqrt2}\) or a = \(\frac{1}{\sqrt2}\)

Hence the square root of the complex no. is -\(\frac{1}{\sqrt2}\) + \(\frac{3}{\sqrt2}\)i and \(\frac{1}{\sqrt2}\) - \(\frac{3}{\sqrt2}\)i.

Let, (a + ib)² = -15 - 8i

Now using, (a + b)² = a² + b² + 2ab

⇒ a² + (bi)² + 2abi = -15 - 8i

Since i² = -1

⇒ a² - b² + 2abi = -15 - 8i

Now, separating real and complex parts, we get

⇒ a² - b² = -15…………..eq.1

⇒ 2ab = -8…….. eq.2

⇒ a = - 4/b

Now, using the value of a in eq.1, we get

⇒ \((-\frac{4}{b})^2\) – b² = -15

⇒ 16 – b⁴ = -15b²

⇒ b⁴ - 15b² - 16 = 0

Simplify and get the value of b², we get,

⇒ b² = 16 or b² = -1

As b is real no. so, b² = 16

b = 4 or b = - 4

Therefore, a = -1 or a = 1

Hence the square root of the complex no. is -1 + 4i and 1 - 4i.

Let 

\(\sqrt{i}\)  = \(\sqrt{a+ib}\) .........….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 

a = b = \(\frac{1}{\sqrt{2}}\)   

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

\(\sqrt{i}\)  = ±  \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\)i    

\(\sqrt{i}\) = ±  \(\frac{1}{\sqrt{2}}\) (1+i)

False

Explanation:

We can compare two complex numbers when they are purely real. Otherwise comparison of complex numbers is not possible or has no meaning.

(1 + 3i)² (3 + i)

= (1 + 6i + 9i²)(3 + i)

= (1 + 6i – 9)(3 + i) ……[∵ i² = -1]

= (-8 + 6i)(3 + i)

= -24 – 8i + 18i + 6i²

= -24 + 10i + 6(-1)

= -24 + 10i – 6

= -30 + 10i