Complex conjugate

nicodemus

New member
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
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 $e^{ix/2}$:
$$\frac{1}{1+e^{-ix}} \cdot \frac{e^{ix/2}}{e^{ix/2}}=\frac{e^{ix/2}}{e^{ix/2}+e^{-ix/2}}= \frac{e^{ix/2}}{2 \cos(x/2)},$$

Prove It

Well-known member
MHB Math Helper
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...

\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*}

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

nicodemus

New member
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.