1. ## Argument and modulus

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Calculate the argument and modulus for the number complex A:

2. you can write:

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

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

3. Originally Posted by J.R
you can write:

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

don't forget that the modulus is positive ...
Hello: Thank you four your solution.
Now student sign of : 2cos(teta/2)

4. you know

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

so,

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

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

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

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

...

5. Originally Posted by J.R
you know

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

so,

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

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

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

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

...
Hello : THANK YOU
YOU HAVE 3 CASES :

6. wath's the point of the topic ?

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

would you like help ?

7. 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