What is (1+2i)^{(-2)} ?

The solution to (1+2i)^{(-2)} is -3/25-4/25 i

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simplify (1+2i)^{(-2)}

simplify

(1 + 2 i)^{(- 2)}

- \frac{3}{25} - \frac{4}{25} i

Solution steps

(1 + 2 i)^{- 2}

Apply exponent rule:

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

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

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

Expand

(1 + 2 i)^{2} : 4 i - 3

= \frac{1}{4 i - 3}

Rewrite

4 i - 3

in standard complex form:

- 3 + 4 i

= \frac{1}{- 3 + 4 i}

Multiply by the conjugate

\frac{- 3 - 4 i}{- 3 - 4 i}

= \frac{1 \cdot ( - 3 - 4 i )}{( - 3 + 4 i ) ( - 3 - 4 i )}

Simplify

\frac{1 \cdot ( - 3 - 4 i )}{( - 3 + 4 i ) ( - 3 - 4 i )} : \frac{- 3 - 4 i}{25}

= \frac{- 3 - 4 i}{25}

Apply the fraction rule:

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

= \frac{- 3}{25} + \frac{- 4}{25} i

Simplify

\frac{- 3}{25} + \frac{- 4}{25} i : - \frac{3}{25} - \frac{4}{25} i

= - \frac{3}{25} - \frac{4}{25} i

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