Complex numbers involving modulus

[xenonp]

Hi, please could you help me solve these two, and find the real and imaginary parts of both. Thanks.

1 / (1+e^(-ix))

and

modulus ( e^z)^2 where z is complex.

Thanks so much.

 

[Deleted member 4993]

Hi, please could you help me solve these two, and find the real and imaginary parts of both. Thanks.

1 / (1+e^(-ix))

and

modulus ( e^z)^2 where z is complex.

Thanks so much.

1/ [1 + e^(-ix)] = 1/[{1+cos(x)} - i*sin(x)]

now continue...

Please share your work with us ...

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

 

[xenonp]

1/ [1 + e^(-ix)] = 1/[{1+cos(x)} - i*sin(x)]

now continue...

Please share your work with us ...

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

I can get to this point but then don't know how to work out the real and imaginary parts of it. For the first one, I know how to solve it if there are no modulus signs but I'm not sure how it changes WITH the modulus signs there.

 

[pka]

find the real and imaginary parts of both. Thanks.
1 / (1+e^(-ix))
and
modulus ( e^z)^2 where z is complex.

Use this fact \(\displaystyle \dfrac{1}{z} = \dfrac{{\overline z }}{{{{\left| z \right|}^2}}}\) to write \(\displaystyle Z = \dfrac{1}{{1 + {e^{ - ix}}}} = \dfrac{{1 + {e^{ix}}}}{{\left| {1 + {e^{ - ix}}} \right|^2}}\)

\(\displaystyle \text{Re}(Z)=\dfrac{{1 + \cos(x)}}{{{{\left| {1 + {e^{ - ix}}} \right|}^2}}}~\&~\text{Im}(Z)=~?\)


#2
\(\displaystyle {\left( {{e^z}} \right)^2} = {e^{2z}} = {e^{2x}}{e^{i2y}} = {e^{2x}}\cos (2y) + i{e^{2x}}\sin (2y)\)

 

[Deleted member 4993]

I can get to this point but then don't know how to work out the real and imaginary parts of it. For the first one, I know how to solve it if there are no modulus signs but I'm not sure how it changes WITH the modulus signs there.

Really! Why didn't you show that beginning work in your original post?

To continue:

1/ [1 + e^(-ix)] = 1/[{1+cos(x)} - i*sin(x)] = 1/[{1+cos(x)} - i*sin(x)] * [{1+cos(x)} + i*sin(x)] /[{1+cos(x)} + i*sin(x)]

Now continue....

 

[xenonp]

Really! Why didn't you show that beginning work in your original post?

To continue:

1/ [1 + e^(-ix)] = 1/[{1+cos(x)} - i*sin(x)] = 1/[{1+cos(x)} - i*sin(x)] * [{1+cos(x)} + i*sin(x)] /[{1+cos(x)} + i*sin(x)]

Now continue....

In your solution to #2, you haven't used the modulus signs...aren't they relevant?

OK so for this one, we expand out the brackets top and bottom and I'm left with:

1+cosx+isinx / 2+2cosx. Now?

 

[pka]

In your solution to #2, you haven't used the modulus signs...aren't they relevant?

OK so for this one, we expand out the brackets top and bottom and I'm left with:

1+cosx+isinx / 2+2cosx. Now?

Look at reply #5.

 

[xenonp]

Took out a factor of 2 in the denominator and split it, simplifed.

Got out: 1/2 + i(sinx/(2+2cosx)).

Is this right?

 

[Deleted member 4993]

Took out a factor of 2 in the denominator and split it, simplifed.

Got out: 1/2 + i(sinx/(2+2cosx)).

Is this right?

Yes.