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    Hi, i would have thought the modulus-argument form of 7 would = 7(cos0 + isin0), which would give the right answer but i look in the answer section of my FP2 book and i'm seeing that they have 7 = 7(cosθ + isinθ), maybe a mistake by the authors thinking they were zeros but just want to make sure i'm not missing anything here.
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    I've found the question - I agree with you. Modulus-argument form requires that you give a value for both r and θ.
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    (Original post by Dylex)
    Hi, i would have thought the modulus-argument form of 7 would = 7(cos0 + isin0), which would give the right answer but i look in the answer section of my FP2 book and i'm seeing that they have 7 = 7(cosθ + isinθ), maybe a mistake by the authors thinking they were zeros but just want to make sure i'm not missing anything here.
    My answer to this question would be neither. It'd be 7 = 7(cos(n*2pi)+isin(n*pi))
    where n is a natural number (0,1,2,3,...)
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    Definitely true - in this instance though answers were required to be in the form -\pi < \theta \leq \pi.

    Attwood, G. and Pledger, K. (2009) Edexcel AS and A Level Modular Mathematics Further Pure Mathematics 2. Harlow: Edexcel.
    pg. 23 ex. 3A qu. 1a
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    (Original post by 45.46 Litre Hat)
    Definitely true - in this instance though answers were required to be in the form -\pi < \theta \leq \pi.

    Attwood, G. and Pledger, K. (2009) Edexcel AS and A Level Modular Mathematics Further Pure Mathematics 2. Harlow: Edexcel.
    pg. 23 ex. 3A qu. 1a
    Theta is inferior or equal to pi. sin is 0 both at theta = 0 and theta = pi.
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    (Original post by candyaljamila)
    Theta is inferior or equal to pi. sin is 0 both at theta = 0 and theta = pi.
    Ah I missed that, thanks.
 
 
 
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