math135-s11-2011

# math135-s11-2011 - Assignment 11 – Solutions 1 Express...

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Unformatted text preview: Assignment 11 – Solutions 1. Express the complex number (1- √ 3 i ) 8 in standard form. Solution: | 1- √ 3 i | = 2 and the argument of (1- √ 3 i ) is θ = 5 π/ 3. So (1- √ 3 i ) = 2(cos(5 π/ 3)+ i sin(5 π/ 3)). By De Moivre’s Theorem, we have (1- √ 3 i ) 8 = 2 8 (cos(40 π/ 3) + i sin(40 π/ 3)) = 2 8 (cos(4 π/ 3) + i sin(4 π/ 3)) = 128(- 1- √ 3 i ) 2. (a) Use De Moivre’s Theorem to prove that cos4 θ = 8cos 4 θ- 8cos 2 θ + 1 sin4 θ = 4cos θ (sin θ- 2sin 3 θ ) . Solution: Using De Moivre’s Theorem, cos4 θ + i sin4 θ = (cos θ + i sin θ ) 4 and by the Binomial theorem, (cos θ + i sin θ ) 4 = cos θ 4 + 4 i cos 3 θ sin θ- 6cos 2 θ sin 2 θ- 4 i cos θ sin 3 θ + sin 4 θ . Hence, cos4 θ = cos 4 θ- 6cos 2 θ sin 2 +sin 4 θ sin4 θ = 4cos 3 θ sin θ- 4 i cos θ sin 3 θ = 4cos θ (cos 2 θ sin θ- sin 2 θ ) Using sin 2 θ = 1- cos 2 θ we obtain cos4 θ = 8cos 4 θ- 8cos 2 θ + 1 sin4 θ = 4cos θ (sin θ- 2sin 3 θ ) ....
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math135-s11-2011 - Assignment 11 – Solutions 1 Express...

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