# Conjugate of a complex number

Can you guys please tell that what's the conjugate of $i^i$? I tried to solve on the basis that multiplying conjugate with it's respective complex no. yields real no. So my try gives its answer as $i^{-i}$. Is it correct?

• Welcome to math.SE: I have tried to improve the readability of your question by introducing Tex. – AlexR Apr 25 '18 at 16:33
• Can you use $i = e^{\frac{i\pi}{2}}$? – Paul Apr 25 '18 at 16:34
• Do you realize that there are infinitely many values of $i^i?$ – saulspatz Apr 25 '18 at 16:35
• Because $i = e^{i(\frac{\pi}{2}+2k\pi)}$ for any integer k. – Paul Apr 25 '18 at 16:37

$$\overline{i^i}={\overline i}^{\overline i}=(1/i)^{-i}=i^i$$ so the conjugate of $i^i$ is itself. I should mention that I got this trick from a different answer on this site, but I do not remember which one.
The definition of $z^w$ is $\exp(w\log z)=\exp(w(\log |z|+i\arg(z))$. This is a multivalued funciton, because $\text{arg}$ is multivalued, since it is only well defined up to a multiple of $2\pi$. Therefore, $$i^i=\exp(i(\log |i|+i\arg i))=\exp(-(\pi/2+2\pi k)),$$ where the above equation is valid for any $k\in \mathbb Z$. No matter which $k$ you choose, the result is real, so the conjugate of $i^i$ is always itself.
We can compute $i^i=e^{\frac{-\pi}{2}+2k\pi}, k \in \Bbb Z$ .
What is worth noting is that, despite it being a highly irrational number, all values of $i^i$ are real - they don't have a complex conjugate, or at the very least, its conjugate is itself (I don't know for certain which is the better definition).