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Write each expression in the standard form a+bi

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(3 + 4i)^4? using De Moivre's theorem.

 

asked Oct 30, 2014 in PRECALCULUS by anonymous

1 Answer

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The expression is (3+4i)4.

Write the complex number z = x + iy in polar from z = r[cos(θ) + i sin(θ)].

r = √(x² + y²) = √(3² + 4²) = √(9 + 16) = √25 = 5.

Here, x = 3 >0, so θ = tan-1(y/x) = tan-1(4/3) = tan-1(4/3) = 53.13o.

Therefore 3 + 4i = 5[cos(53.13o) + i sin(53.13o)].

(3 + 4i)4 = {5[cos(53.13o) + i sin(53.13o)]}4

Apply De Moivre's theorem: If z = r[cos(θ) + i sin(θ)] ⇒ zn = rn[cos(nθ) + i sin(nθ)].

(3 + 4i)4 = 54[cos(4*53.13o) + i sin(4*53.13o)]

(3 + 4i)4 = 625[cos(212.52o) + i sin(212.52o)]

(3 + 4i)4 = 625[-0.8432 + i (-0.5376)]

(3 + 4i)4 = - 527 - i 336.

answered Oct 30, 2014 by casacop Expert

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