Finding modulus and argument of a complex number

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To find the modulus and argument of the complex number z=1-cos(a)-i*sin(a), the modulus is calculated as mod(z)=sqrt((1-cos(a))^2+(-sin(a))^2), simplifying to sqrt(2-2cos(a)). The argument is expressed as arg(z)=arctan((-sin(a))/(1-cos(a))). Trigonometric identities are suggested to simplify the expressions further, specifically using sin and cos half-angle formulas. The discussion highlights the need for familiarity with trigonometric identities to complete the solution effectively.
choob
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Homework Statement


Find the modulus and argument of z=1-cos(a)-i*sin(a)

Homework Equations


mod(z)=sqrt(a^2+b^2)

The Attempt at a Solution


mod(z)=sqrt((1-cos(a))^2+(-sin(a))^2)
=sqrt(2-2cos(a))

arg(z)=arctan((-sin(a))/(1-cos(a)))

This is as far as I can get, I have asked my math teacher but he is not very familiar with thise.

my textbook gives mod(z) as 2sin(2a) and arg(z) as (a-pi)2

BTW this question is from the IB math textbook, there exists a solutions manual but I do not have it... the question is 11.2, 19 a)

Thanks in advance!
 
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Hi choob! :smile:

You need to learn your trigonometric identities …

in this case, sin = 2 sin1/2 cos1/2

and 1 - cos = 2 sin21/2 :wink:
 
i can get arg(z) to arctan(-cos(a/2)/(sin a/2)), how do i finish this?
 
This will help you.

tan(\frac{\pi}{2}-\theta)= \frac{1}{tan\theta}
 
wow thanks a lot, lol.
 

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