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QUESTION: 1

Solution:

(x+iy)(x−iy) = (a+ib)(a−ib)/(c+id)(c−id)

⇒x^{2}−iy^{2} = √[(a^{2}−i^{2}b^{2})/(c^{2}−i^{2}d^{2})]

⇒x^{2}+y^{2} = √[(a^{2}+b^{2})/(c^{2}+d^{2})] [1i^{2 }= -1]

(x^{2}+y^{2})^{2} = (a^{2}+b^{2})/(c^{2}+d^{2})

QUESTION: 2

Solution:

(a + ib)^{1/2} = (x + iy)

Squaring both sides,

a + ib = (x + iy)^{2}

a + ib = x^{2} - y^{2} + 2ixy

Equating real and imaginary

a = x^{2} - y^{2} b = 2xy............(1)

Using (x^{2} + y^{2})^{2} = (x^{2} - y^{2})^{2} + 4xy

(x^{2} + y^{2})^{2} = a^{2} + b^{2}

(x^{2} + y^{2}) = (a^{2} + b^{2})^{1/2}.......(2)

Adding (1) and (2)

2x^{2} = (a^{2} + b^{2})^{1/2} + a

x = +-{1/2(a^{2} + b^{2})^{1/2} + a}^{1/2}

Subtract (2) from (1)

2y^{2} = (a^{2} + b^{2})^{1/2} - a

y = x = +-{1/2(a^{2} + b^{2})^{1/2} - a}^{1/2}

Therefore, (a+ib)^{1/2} = x+iy

=> +-{1/2(a^{2} + b^{2})1/2 + a}^{1/2} + i+-{1/2(a^{2} + b^{2})^{1/2} - a}^{1/2}

QUESTION: 3

Square root of 5 + 12i is

Solution:

(a + ib)^{2} = 5 + 12i

use (a + b)^{2} = a^{2} + b^{2} + 2*a*b

=> a^{2} + b^{2}*i^{2} + 2*a*b*i = 5 + 12i

i^{2} = -1

⇒ a^{2} - b^{2} + 2*a*b*i = 5 + 12i

equate the real and complex coefficients

⇒ a^{2} - b^{2} = 5 and ab = 6

a = 6/b

substitute in a^{2} - b^{2} = 5

⇒ 36/b^{2} - b^{2} = 5

⇒ 36 - b^{4} = 5b^{2}

⇒ b^{4} + 5b^{2} - 36 = 0

⇒ b^{4} + 9b^{2} - 4b^{2} - 36 = 0

⇒ b^{2}(b^{2} + 9) - 4(b^{2} + 9) = 0

⇒ (b^{2} - 4)(b^{2} + 9) = 0

⇒ b^{2} = 4 and b^{2} = -9

b is a real number, so we eliminate b^{2} = -9

b^{2 }= 4

⇒ b = 2 and b = -2

a = 3 and a = -3

The required number can be 3 + 2i and -3 -2i

QUESTION: 4

The square root of i is

Solution:

QUESTION: 5

Solution:

Let √(5 – 12i) = x + iy

Squaring both sides, we get

5 – 12i = x^{2} + 2ixy +(iy)^{2} = x^{2} – y^{2} + 2xyi.

Comparing real and imaginary parts , we get

5 = x^{2} – y^{2} ———– (1) and xy = – 6 ———— (2)

Squaring (1), we get

25 = (x^{2} – y^{2})2 = (x^{2} + y^{2})^{2} – 4x^{2}y^{2}

⇒ 25 = (x^{2} + y^{2})^{2} – 4(– 6)^{2}

⇒ (x^{2 }+ y^{2})^{2} = 169

⇒ x^{2} + y^{2} = 13 ———- (3)

Adding (1) and (3) we get

2x^{2} = 18

⇒ x = ± 3.

Subtracting (1) from (3) we get

2y^{2} = 8

⇒ y = ± 2.

Hence, square root of √(5 – 12i) is (3 – 2i)

Similarly, √(5 + 12i) is (3 + 2i)

√(5 + 12i) + √(5 – 12i)

⇒ (3 + 2i) + (3 - 2i)

⇒ 6

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