Question

Select the correct answer. What is the standard form of this complex number?

Answer 1

$\dfrac{1}{3 - \sqrt{-4}} = \dfrac{1}{3 - i \sqrt{4}} = \dfrac{1}{3 - 2i} \times \dfrac{3+2i}{3+2i} = \dfrac{3+2i}{3^2 + 2^2} = \dfrac{3}{13} + \dfrac{2}{13}i$

Answer: B

Answer 2

The standard form of the complex number 1 / [ 3 - √(-4)] will be 3/13 + 2/13i so option (B) will be correct.

What is a complex number?

Complex numbers are helpful in finding the square root of negative numbers.

If we solve x² + 1 = 0 ⇒ x = √(-1) which is called as iota(i).

The general form of a complex number is a + ib where the first part a called real and the ib is called imaginary.

Given that number  1 / [ 3 - √(-4)]

1 / [ 3 - √(-4)] = 1 / [ 3 - √(-1)√4]

Since i = √(-1) so

1 / [ 3 - √(-4)] = 1 / [ 3 - 2i ]

By rationalization,

⇒ 1 / [ 3 - 2i ] × (3 + 2i)/(3 + 2i)

⇒ (3 + 2i)/(9 - 6 i²)

⇒ (2/13) + (2/13)i

Hence (2/13) + (2/13) I will be the standard form of the given complex number.

To learn more about complex numbers,

brainly.com/question/10251853

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