How do you take this complex conjugate?

[AxiomOfChoice]

What's the complex conjugate of

$$\frac{1}{\sqrt{1+it}}, \quad t \geq 0.$$

 

[pmsrw3]

$$\frac{1}{\sqrt{1-it}}$$

 

[SteveL27]

What's the complex conjugate of

$$\frac{1}{\sqrt{1+it}}, \quad t \geq 0.$$

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

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.