# Is $i^i$ a complex number, and what is its standard form?

Let $y = i^i \,\,\,\, ———(1)$

$\implies \ln(y) = i\ln(i)$

$\implies y = e^{i\ln(i)} \,\,\,\, ———(2)$

As, $i = \cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2}) = e^{i\frac{\pi}{2}}$

$\implies \ln(i) = \ln(e^{i\frac{\pi}{2}})$

$\implies \ln(i) = i\frac{\pi}{2}$

So, putting the value of $\ln(i)$ into equation $(2)$, we have,

$y = e^{i(i\frac{\pi}{2})}$

$\implies i^i = e^{-\frac{\pi}{2}}$

## One Reply to “Is $i^i$ a complex number, and what is its standard form?”

1. Ravi Ranjan Kumar Singh says:

It is a complex number as well, as all real numbers are. Funnily enough $i^i$ is a real number.