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Confusion with finding the mod-argument for of just '7 + 0i'

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.
I've found the question - I agree with you. Modulus-argument form requires that you give a value for both r and θ.
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,...)
Definitely true - in this instance though answers were required to be in the form −π<θ≤π-\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
Original post by 45.46 Litre Hat
Definitely true - in this instance though answers were required to be in the form −π<θ≤π-\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.
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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