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(2i³)²

= 4i⁶

= 4(i²)³

= 4(-1)³

= -4 …..[∵ i² = -1]

= -4 + 0i

(i) (1 + i) (1 + 2i)

Now let us simplify and express in the standard form of (a + ib),

(1 + i) (1 + 2i) = (1 + i)(1 + 2i)

= 1(1 + 2i) + i(1 + 2i)

= 1 + 2i + i + 2i²

= 1 + 3i + 2(-1) [since, i² = -1]

= 1 + 3i - 2

= -1 + 3i

Thus the values of a, b are -1, 3.

(ii) (3 + 2i)/(-2 + i)

Now let us simplify and express in the standard form of (a + ib),

(3 + 2i)/(-2 + i) = [(3 + 2i)/(-2 + i)] × (-2 - i)/(-2 - i) [multiply and divide with (-2 - i)]

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

= [-6 - 3i – 4i - 2i²]/(4 - i²)

= [-6 - 7i - 2(-1)]/(4 – (-1)) [since, i² = -1]

= [-4 - 7i]/5

∴ The values of a, b are -4/5, -7i/5

(iii) 1/(2 + i)²

Now let us simplify and express in the standard form of (a + ib),

1/(2 + i)² = 1/(2² + i² + 2(2) (i))

= 1/ (4 – 1 + 4i) [since, i² = -1]

= 1/(3 + 4i) [By multiply and divide with (3 – 4i)]

= 1/(3 + 4i) × (3 – 4i)/(3 – 4i)]

= (3-4i)/ (3² – (4i)²)

= (3-4i)/ (9 – 16i²)

= (3-4i)/ (9 – 16(-1)) [since, i² = -1]

= (3-4i)/25

∴ The values of a, b are 3/25, -4i/25

(iv) (1 – i) / (1 + i)

Now let us simplify and express in the standard form of (a + ib),

(1 – i) / (1 + i) = (1 – i) / (1 + i) × (1 - i)/(1 - i) [multiply and divide with (1 - i)]

= (1² + i² – 2(1)(i)) / (1² – i²)

= (1 + (-1) -2i) / (1 – (-1))

= -2i/2

= -i

∴ The values of a, b are 0, -i

(v) (2 + i)³/(2 + 3i)

Now let us simplify and express in the standard form of (a + ib),

(2 + i)³/(2 + 3i) = (2³ + i³ + 3(2)²(i) + 3(i)²(2))/(2 + 3i)

= (8 + (i².i) + 3(4)(i) + 6i²)/(2 + 3i)

= (8 + (-1)i + 12i + 6(-1))/(2 + 3i)

= (2 + 11i)/(2 + 3i)

[by multiply and divide with (2 - 3i)]

= (2 + 11i)/(2 + 3i) × (2 - 3i)/(2 - 3i)

= [2(2 - 3i) + 11i(2 - 3i)]/(2² – (3i)²)

= (4 – 6i + 22i – 33i²)/(4 – 9i²)

= (4 + 16i – 33(-1))/(4 – 9(-1)) [since, i² = -1]

= (37 + 16i)/13

Thus the values of a, b are 37/13, 16i/13

Given: 

⇒ a + ib = \(\frac{5+\sqrt2i}{1-\sqrt2i}\)

Multiplying and dividing with 1+√2i

⇒ a + ib =  \(\frac{5+\sqrt2i}{1-\sqrt2i} \times \frac{1+\sqrt2i}{1+\sqrt2i}\) 

⇒ a + ib =  \(\frac{5(1+\sqrt2i)+\sqrt2i(1+\sqrt2i)}{1^2-(\sqrt2i)^2}\) 

⇒ a + ib =  \(\frac{5+5\sqrt2i+\sqrt2i+2i^2}{1-2i^2}\)

We know that i²=-1

⇒ a + ib = \(\frac{5+6\sqrt2i+2(-1)}{1-2(-1)}\) 

⇒ a + ib =  \(\frac{3+6\sqrt2i}{3}\) 

⇒ a + ib = 1 + 2√2i

∴ The values of a, b are 1, 2√2 .