Sum

Express the following in the form a + ib, a, b ∈ R, using De Moivre's theorem:

(1 + i)^{6}

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#### Solution

Let z = 1 + i

∴ a = 1, b = 1 , i.e. a > 0, b > 0

∴ |z| = `sqrt("a"^2 + "b"^2) = sqrt(1^2 + 1^2) = sqrt(2)`

Here (1, 1) lies in 1^{st} quadrant.

∴ amp (z) = `tan^-1("b"/"a")`

= `tan^1(1/1)`

= `pi/4`

z^{6} = (1 + i)^{6}

= `[sqrt(2)(cos pi/4 + "i" sin pi/4)]^6`

= `8[cos (6pi)/4 + "i"sin (6pi)/4]` ...[∵ (cos θ + i sin θ)^{n} = (cos n θ + i sin n θ)]

= `8[cos (3pi)/2 + "i"sin (3pi)/2]`

= `8[0 + "i" (-1)]`

= – 8i

Concept: De Moivres Theorem

Is there an error in this question or solution?

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