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Complex conjugate

nicodemus

New member
Jul 22, 2012
16
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

Indicium Physicus
Staff member
Jan 26, 2012
4,193
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

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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]
 

nicodemus

New member
Jul 22, 2012
16
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.