Question

1)Express the complex number i⁹ + i¹⁹ in the form of a + ib
2) Find multiplicative Inverse of 4 - 3i​

Answer 1

$\underline{\underline{\blue{\textbf{Step - By - Step - Explanation : -}}}}$

(i) Express the complex number i⁹ + i¹⁹ in the form of a + ib .

Solution :

$\rm = {i}^{9} + {i}^{19}$

$\rm = ( {i}^{2} {)}^{4} \times i + ({i}^{2} {)}^{9} \times i$

we know that,

• i² = -1

$\rm = ( - 1) {}^{4} \times i + ( - 1) ^{9} \times i$

$\rm = 1 \times i + ( - 1) \times i$

$\rm = i - i$

$\rm = 0$

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(ii) Find multiplicative Inverse of 4 - 3i

= 1/(4 - 3i)

• Rationalising the denominator :

= 1/(4 - 3i) × (4 + 3i)/(4 + 3i)

= (4 + 3i)/(4 - 3i)(4 + 3i)

• By using Identity (a + b)(a - b) = a² - b²

= (4 + 3i)/(4² - (3i)²)

= (4 + 3i)/(16 + 9)

= (4 + 3i)/25

= 4/25 + 3i/25

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Answer 2

the correct answer is 0

jehd