Question

write (1+2i) (2+3i) (2+ i)-1
in The form a+bi where i =√-1​

Answer 1

Step-by-step explanation:

(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/13Read more on Sarthaks.com - https://www.sarthaks.com/659694/express-the-following-complex-numbers-in-the-standard-form-a-ib-i-1-i-1-2i-ii-3-2i-2-i-iii-1-2-i-2?show=659732#a659732

Answer 2

Concept

Multiplication or divison of complex numbers is a an operation where two or more complex numbers are multiplied or divided.

Find

(1+2i) (2+3i) (2+i)⁻¹ in a +ib form

Solution

( 1 + 2i )( 2 + 3i )/( 2+i )

( 2 + 4i + 3i +6i² )/( 2 + i )

( 2 + 7i - 6 )/( 2 + i )

( -4 + 7i )/( 2 + i)

Rationalising

((-4 + 7i )*( 2 - i ))/(( 2 + i )*( 2 - i ))

( -8 + 18i - 7i² )/( 4 - i² )

( -1 + 18i )/5

-1/5 + i 18/5

So  (1+2i) (2+3i) (2+i)⁻¹ can be written as -1/5 + i 18/5

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