# what's the conjugate of $i^{-\frac{1}{2}}$?

If a complex number is $A=a+bi$, then its conjugate is $\bar{A}=a-bi$. What's more, the conjugate of $e^{i\theta}$ is $e^{-i\theta}$. Well, it is known to us.

Now, if a complex number is $i^{-\frac{1}{2}}$, what's its conjugate? Thanks in advance!

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## 2 Answers

HINT: $i=e^{i\pi/2}\implies i^{-1/2}=e^{-i\pi/4}$

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The complex exponential is usually defined by the equation $a^b = \exp(b \log a)$. But you already know how to take the complex conjugate of a complex exponential.

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Thanks, I forgot too much things. –  user39843 Mar 7 '13 at 10:44