# Argument and modulus

• Jul 24th 2009, 09:23 AM
dhiab
Argument and modulus
• Jul 24th 2009, 09:49 AM
J.R
you can write:

$\displaystyle 1+e^{i\theta}=e^{i\frac{\theta}{2}}(2cos(\frac{\th eta}{2}))$

don't forget that the modulus is positive ...
• Jul 24th 2009, 11:53 AM
dhiab
Quote:

Originally Posted by J.R
you can write:

$\displaystyle 1+e^{i\theta}=e^{i\frac{\theta}{2}}(2cos(\frac{\th eta}{2}))$

don't forget that the modulus is positive ...

Hello: Thank you four your solution.
Now student sign of : 2cos(teta/2)(Itwasntme)
• Jul 24th 2009, 10:07 PM
J.R
you know

$\displaystyle 0< \frac{\theta}{2} <\pi$

so,

$\displaystyle if \ \ \pi<\theta <2\pi$

then $\displaystyle 2cos(\frac{\theta}{2}+\pi)>0$

$\displaystyle (1+e^{i\theta})^n=(2cos(\frac{\theta}{2}+\pi))^ne^ {ni(\frac{\theta}{2}+\pi)}$

$\displaystyle if \ \ 0<\theta <\pi$ (it's easier than the precedent ;) )

...
• Jul 25th 2009, 11:48 PM
dhiab
Quote:

Originally Posted by J.R
you know

$\displaystyle 0< \frac{\theta}{2} <\pi$

so,

$\displaystyle if \ \ \pi<\theta <2\pi$

then $\displaystyle 2cos(\frac{\theta}{2}+\pi)>0$

$\displaystyle (1+e^{i\theta})^n=(2cos(\frac{\theta}{2}+\pi))^ne^ {ni(\frac{\theta}{2}+\pi)}$

$\displaystyle if \ \ 0<\theta <\pi$ (it's easier than the precedent ;) )

...

Hello : THANK YOU
YOU HAVE 3 CASES :
http://www.mathramz.com/xyz/latexren...c28f8c02eb.png
• Jul 25th 2009, 11:59 PM
J.R
wath's the point of the topic ?

it's a challenge or you don't know the solution ?

would you like help ?
• Jul 26th 2009, 01:16 AM
dhiab
Quote:

Originally Posted by J.R
wath's the point of the topic ?

it's a challenge or you don't know the solution ?

would you like help ?

Good morning, I thank you for your observation, nevertheless, that the objective of the forum, to help to solve every exercices, as well as exchanges it of views and ideas to be of benefit to everybody, it is the most important in my view..... Thank you