How do you take this complex conjugate?

[AxiomOfChoice]

What's the complex conjugate of

<br /> \frac{1}{\sqrt{1+it}}, \quad t \geq 0.<br />

 

[pmsrw3]

<br /> \frac{1}{\sqrt{1-it}}<br />

 

[SteveL27]

What's the complex conjugate of

<br /> \frac{1}{\sqrt{1+it}}, \quad t \geq 0.<br />

The key is that for all complex z, z \overline{z} = |z|^2 so that \overline{z} = \frac{|z|^2}{z}

You can see this since (a + bi)(a - bi) = a^2 + b^2Now |\frac{1}{\sqrt{1+it}}|^2 <br /> <br /> = \frac{1}{|1+it|}<br /> <br /> = \frac{1}{\sqrt{1 + t^2}}

since abs commutes with multiplication, division, and exponentiation.

and so \overline{\frac{1}{\sqrt{1+it}}} <br /> <br /> = \frac{\frac{1}{\sqrt{1 + t^2}}} {\frac{1}{\sqrt{1+it}}} <br /> <br /> = \frac {\sqrt{1+it}} {\sqrt{1 + t^2}}

(edit)
Oops of course pmsrw3 is correct, and that's the same answer as mine but a lot easier! I'd delete my response but I don't want to waste all that TeX!

 

[Hurkyl]

(Don't forget about branch cuts! A little bit of care must be used to ensure that the function and its proposed conjugate make consistent choices of principal value)

 

[pmsrw3]

(Don't forget about branch cuts! A little bit of care must be used to ensure that the function and its proposed conjugate make consistent choices of principal value)

It works out OK in this case :-) I did actually think about that. If you picture the operations on the complex plane, it's pretty easy to see.