Question

The variable complex number ß is given by ß=1+cos 2 θ
+isin2
θ ,
where
θ
takes all values in the interval−1 2
π
<
θ
<1 2
π .
(i) Show that the modulus of ß is 2 cos
θ
and the argument of ß is
θ

Answer 1

is beta like this:
\[\Large \beta = 1 + \cos \left( {2\theta } \right) + i\sin \left( {2\theta } \right)\]

Answer 2

yes

Answer 3

ok! Then by definition, the modulus of \beta is given by the subsequent formula:
\[\Large \left| \beta \right| = \sqrt {{{\left\{ {1 + \cos \left( {2\theta } \right)} \right\}}^2} + {{\left\{ {\sin \left( {2\theta } \right)} \right\}}^2}} = ...?\]

Answer 4

i am not being able to solve... help pls ?

Answer 5

ok!

Answer 6

here is your steps:
\[ \begin{gathered}
{\left\{ {1 + \cos \left( {2\theta } \right)} \right\}^2} + {\left\{ {\sin \left( {2\theta } \right)} \right\}^2} = 1 + {\left( {\cos \left( {2\theta } \right)} \right)^2} + 2\cos \left( {2\theta } \right) + {\left( {\sin \left( {2\theta } \right)} \right)^2} = \hfill \\
= 2 + 2\cos \left( {2\theta } \right) = 2\left( {1 + \cos \left( {2\theta } \right)} \right) = 2 \times 2{\left( {\cos \theta } \right)^2} = 4{\left( {\cos \theta } \right)^2} \hfill \\
\end{gathered} \]

Answer 7

so we can write:
\[\Large \left| \beta \right| = \sqrt {4{{\left( {\cos \theta } \right)}^2}} = 2\left| {\cos \theta } \right|\]

Answer 8

thank you :)
and the second part ?

Answer 9

for second part, we can write, I call with \alpha the principal argument of \beta, so I can write:
\[\Large \begin{gathered}
\cos \alpha = \frac{{1 + \cos \left( {2\theta } \right)}}{{2\cos \theta }}, \hfill \\ \\
\sin \alpha = \frac{{\sin \left( {2\theta } \right)}}{{2\cos \theta }} \hfill \\
\end{gathered} \]

Answer 10

then we get:
\[\Large \tan \alpha = \frac{{\sin \left( {2\theta } \right)}}{{2\cos \theta }} \times \frac{{2\cos \theta }}{{1 + \cos \left( {2\theta } \right)}} = \frac{{\sin \left( {2\theta } \right)}}{{1 + \cos \left( {2\theta } \right)}}\]

Answer 11

and finally:
\[\Large \begin{gathered}
\tan \alpha = \frac{{\sin \left( {2\theta } \right)}}{{1 + \cos \left( {2\theta } \right)}} = \frac{{2\sin \theta \cos \theta }}{{1 + {{\left( {\cos \theta } \right)}^2} - {{\left( {\sin \theta } \right)}^2}}} = \hfill \\
\hfill \\
= \frac{{2\sin \theta \cos \theta }}{{2{{\left( {\cos \theta } \right)}^2}}} = ...? \hfill \\
\end{gathered} \]

Answer 12

thanks

Answer 13

thanks!

Answer 14

Prove that the real part of
1 ß is constant.

Answer 15

the real part of (1+\beta) ?

Answer 16

yes

Answer 17

we have:
\[\Large \begin{gathered}
\Re \left( {1 + \beta } \right) = \frac{{1 + \beta + 1 + {\beta ^*}}}{2} = \frac{{2 + \beta + {\beta ^*}}}{2} = \hfill \\
\hfill \\
= 1 + \Re \beta \hfill \\
\end{gathered} \]

Answer 18

and:
\[\Large \begin{gathered}
\Re \beta = \hfill \\
= \frac{{1 + \cos \left( {2\theta } \right) + i\sin \left( {2\theta } \right) + 1 + \cos \left( {2\theta } \right) - i\sin \left( {2\theta } \right)}}{2} = \hfill \\
\hfill \\
= 1 + \cos \left( {2\theta } \right) \hfill \\
\end{gathered} \]

Answer 19

so:
\[\Large \Re \left( {1 + \beta } \right) = 1 + \Re \beta = 2 + \cos \left( {2\theta } \right)\]

Answer 20

which is not constant, since it depends on \theta