MHB G3popstar's question at Yahoo Answers (Principal value of i^i)

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The principal value of i^i can be calculated using the formula z^w = e^(w log z). For i, the principal argument is π/2, leading to log(i) = log(1) + i(π/2) = i(π/2). Thus, i^i = e^(i log i) = e^(i * i(π/2)) = e^(-π/2). This result shows that the principal value of i^i is e^(-π/2). Further questions can be directed to the math help boards for additional assistance.
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Hello g3popstar,

For $z,w$ complex nombers and $z\neq 0$ we define $z^w=e^{w\log z}$. In our case, $i^i=e^{i\log i}$. On the other hand, $\log z=log |z|+i\arg z$. The principal argument of $i$ is $\frac{\pi}{2}$, so the principal value of $\log i$ is $\log i=\log 1+i\frac{\pi}{2}=i\frac{\pi}{2}$. As a consequence $$i^{i}=e^{i\log i}=e^{i\cdot i\frac{\pi}{2}}=e^{-\frac{\pi}{2}}$$ If you have further questions, you can post them in the http://www.mathhelpboards.com/f50/ section.
 
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