2

My question is a bit straightforward. How can I solve $i^i$? Do I have to work it out based on polar form of complex number? Even that doesn't seem to help!!

Avery
  • 127
  • If you start with the definition of exponentiation, it's not too difficult: $$i^i = e^{i \log i}$$ where $\log$ refers to whatever branch of the logarithm you're using at the moment. –  Jan 13 '14 at 02:01
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    $i=e^{\frac\pi2i}$ so $i^i=e^{\frac\pi2i\cdot i}$. – Berci Jan 13 '14 at 02:04

2 Answers2

5

$i^i = e^{\pi /2 \cdot i \cdot i} = e^ {-\pi/2}$

lennon310
  • 628
2

$i^i=e^{i \, log \,i}=exp\left(i\left(ln(1)+i\frac{\pi}{2}\right)\right)=\cdots$

emka
  • 6,494