# 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 $e^{ix/2}$:
$$\frac{1}{1+e^{-ix}} \cdot \frac{e^{ix/2}}{e^{ix/2}}=\frac{e^{ix/2}}{e^{ix/2}+e^{-ix/2}}= \frac{e^{ix/2}}{2 \cos(x/2)},$$
which is where your book's answer comes from.

#### 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.

Thank you very much for all your help and advice.