What is (-sqrt(3)+i)^{-2} ?

The solution to (-sqrt(3)+i)^{-2} is 1/8+(sqrt(3))/8 i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify (-sqrt(3)+i)^{-2}

simplify

(- 3 + i)^{- 2}

\frac{1}{8} + \frac{3}{8} i

Solution steps

(- 3 + i)^{- 2}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

(- 3 + i)^{- 2} = \frac{1}{( - 3 + i ) ^{2}}

= \frac{1}{( - 3 + i ) ^{2}}

Expand

(- 3 + i)^{2} : - 23 i + 2

= \frac{1}{- 2 3 i + 2}

Rewrite

- 23 i + 2

in standard complex form:

2 - 23 i

= \frac{1}{2 - 2 3 i}

Multiply by the conjugate

\frac{2 + 2 3 i}{2 + 2 3 i}

= \frac{1 \cdot ( 2 + 2 3 i )}{( 2 - 2 3 i ) ( 2 + 2 3 i )}

Simplify

\frac{1 \cdot ( 2 + 2 3 i )}{( 2 - 2 3 i ) ( 2 + 2 3 i )} : \frac{2 + 2 3 i}{16}

= \frac{2 + 2 3 i}{16}

Apply the fraction rule:

\frac{a + bi}{c} = \frac{a}{c} + \frac{b}{c} i

= \frac{2}{16} + \frac{2 3}{16} i

Simplify

\frac{2}{16} + \frac{2 3}{16} i : \frac{1}{8} + \frac{3}{8} i

= \frac{1}{8} + \frac{3}{8} i

- 1 1^{2}

\frac{8}{36}

1 + \frac{1}{5}

12 \times 5

- \frac{1}{2} \cdot (- \frac{1}{2})

What is (-sqrt(3)+i)^{-2} ?
The solution to (-sqrt(3)+i)^{-2} is 1/8+(sqrt(3))/8 i