Find Complex Conjugate of 1/(1+e^(ix))

[nicodemus1]

Good Day,

I would like to know how to find the complex conjugate of the complex number 1/(1+e^(ix)).

I got (1+e^(-(ix)))/(2+2 cos x) but the solution is 0.5 sec (x/2) e^(i(x/2)).

Any help will be greatly appreciated.

Thanks & Regards

P.S. Apologies for not using LATEX as it was formatting the expressions wrongly

 

[Ackbach]

Well, the first step is to actually conjugate, which is simply to replace all $i$'s with $-i$'s:
$$ \frac{1}{1+e^{ix}} \to \frac{1}{1+e^{-ix}}.$$
Next, one thing we could do is to rationalize the denominator to make the result have a real number in the denominator:
$$ \frac{1}{1+e^{-ix}} \cdot \frac{1+e^{ix}}{1+e^{ix}}
=\frac{1+e^{ix}}{1+e^{ix}+e^{-ix}+1}=\frac{1+e^{ix}}{2+2 \cos(x)}.$$
That's slightly different from your result.

Another approach would be to create symmetry where there isn't any, by multiplying top and bottom by $e^{ix/2}$:
$$\frac{1}{1+e^{-ix}} \cdot \frac{e^{ix/2}}{e^{ix/2}}=\frac{e^{ix/2}}{e^{ix/2}+e^{-ix/2}}= \frac{e^{ix/2}}{2 \cos(x/2)},$$
which is where your book's answer comes from.

 

[Prove It]

Good Day,

I would like to know how to find the complex conjugate of the complex number 1/(1+e^(ix)).

I got (1+e^(-(ix)))/(2+2 cos x) but the solution is 0.5 sec (x/2) e^(i(x/2)).

Any help will be greatly appreciated.

Thanks & Regards

P.S. Apologies for not using LATEX as it was formatting the expressions wrongly

I would lean towards trying to write your complex number in terms of its real and imaginary parts, then the conjugation is easy...

[tex]\displaystyle \begin{align*} \frac{1}{1 + e^{i\,x}} &= \frac{1}{1 + \cos{(x)} + i\sin{(x)}} \\ &= \frac{1 \left[ 1 + \cos{(x)} - i\sin{(x)} \right] }{\left[ 1 + \cos{(x)} + i\sin{(x)} \right] \left[ 1 + \cos{(x)} - i\sin{(x)} \right] } \\ &= \frac{1 + \cos{(x)} - i\sin{(x)} }{ \left[ 1 + \cos{(x)} \right] ^2 + \sin^2{(x)} } \\ &= \frac{1 + \cos{(x)} - i\sin{(x)}}{ 1 + 2\cos{(x)} + \cos^2{(x)} + \sin^2{(x)}} \\ &= \frac{1 + \cos{(x)} - i\sin{(x)} }{ 2 + 2\cos{(x)} } \\ &= \frac{1 + \cos{(x)}}{2\left[ 1 + \cos{(x)} \right] } - i\left\{ \frac{\sin{(x)}}{2 \left[ 1 + \cos{(x)} \right] } \right\} \\ &= \frac{1}{2} - i\left\{ \frac{\sin{(x)}}{2 \left[ 1 + \cos{(x)} \right] } \right\} \end{align*}[/tex]

So the conjugate is [tex]\displaystyle \begin{align*} \frac{1}{2} + i\left\{ \frac{\sin{(x)}}{2 \left[ 1 + \cos{(x)} \right] } \right\} \end{align*}[/tex]

 

[nicodemus1]

Yes, it is quite simple actually. I used the approach to express the given complex number in x+iy but I made a careless mistake there.

Thank you very much for all your help and advice.

 

[CellCoree]

.Hello,

Thank you for your question. To find the complex conjugate of a complex number, we simply need to change the sign of the imaginary part. In this case, the complex number is 1/(1+e^(ix)), so its complex conjugate would be 1/(1+e^(-ix)).

To verify this, we can multiply the original complex number by its conjugate:

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

Therefore, we can see that the complex conjugate of 1/(1+e^(ix)) is indeed 1/(1+e^(-ix)).

I hope this helps. Let me know if you have any further questions.

Best regards,