(1 + 2i)^{-3}

Let us simplify and express in the standard form of (a + ib),

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

= \( \dfrac{1}{(1+6i+4i^2+8i^3)}\)

=\( \dfrac{1}{(1+6i+4(-1)+8i^2.i)}\) [since, i^{2 }= -1]

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

[Multiply and divide with (3-2i)]

=\( \dfrac{-1}{(3+2i)}× \dfrac{(3-2i)}{(3-2i)}\)

= \(\dfrac{ (-3+2i) }{ (9-4(-1))}\)

=\(\dfrac{ (-3+2i) }{13}\)

The values of a, b are \(\dfrac{ -3}{13}, \dfrac{2}{13}\)

Answered by Aaryan | 1 year agoShow that 1 + i^{10} + i^{20} + i^{30} is a real number?

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