Question

If a+bi=c+i/c-i, prove that a squar +b square=1 and b/a=2c/c square -1

Answer 1

Answer:

Given:

• a + bi = (c + i) / (c - i)

Solution:

》a + bi = (c + i) / (c - i)...(1) (Given)

》a - bi = (c - i) / (c + i)...(2) (Conjugate pair)

Multiply equation (1) by equation (2), we get

》a² - b²i² = 1

》a² + b² = 1

Divide equation (1) from equation (2), we get

》(a + bi) / (a - bi) = (c + i)² / (c - i)²

By componendo and dividendo

》[(a + bi) + (a - bi)] / [(a + bi) - (a - bi)]

⠀⠀⠀= [(c + i)² + (c - i)²] / [(c + i)² - (c - i)²]

》2a/2bi = (2c² + 2i²) / (4ci)

》2a/2b = 2 (c² - 1) / 4c

》a/b = (c² - 1) / 2c

By invertendo, we get

》b/a = 2c / (c² - 1)

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Note;

• (a + b)² + (a - b)² = 2a² + 2 b²

• (a + b)² - (a - b)² = 4ab

Answer 2

Yes, we can prove that a² + b² = 1 and b/a = 2c/c²- 1.

Given,

a + bi = c + i/c - i

To Find,

We have to prove that a² + b² = 1 and b/a = 2c/c²- 1.

Solution,

We have, a + bi = c + i/c - i                          ------------------------------------(1)

On conjugating (1),

⇒ a - bi = c - i/c + i                                       ------------------------------------(2)

On multiplying (1) and (2), we get,

⇒ a² - b²i² = 1

⇒ a² + b² = 1                                        (∵ i² = -1)

On dividing (1) and (2), we get,

⇒ a + bi/a - bi = (c + i)²/(c - i)²

Using componendo and dividendo, we get,

⇒ a + bi + a - bi/a + bi - a - bi = (c + i)² + (c - i)²/(c + i)² - (c - i)²

⇒ 2a/2bi = 2c² + 2i²/4ci

⇒ a/b = c² - 1/2c

⇒ b/a = 2c/c² - 1

Hence proved, a² + b² = 1 and b/a = 2c/c²- 1.

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