Is [math]i^i[/math] a complex number, and what is its standard form?

Let [math]y = i^i \,\,\,\, ———(1)[/math]

[math]\implies \ln(y) = i\ln(i)[/math]

[math]\implies y = e^{i\ln(i)} \,\,\,\, ———(2)[/math]

As, [math]i = \cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2}) = e^{i\frac{\pi}{2}}[/math]

[math]\implies \ln(i) = \ln(e^{i\frac{\pi}{2}})[/math]

[math]\implies \ln(i) = i\frac{\pi}{2}[/math]

So, putting the value of [math]\ln(i)[/math] into equation [math](2)[/math], we have,

[math]y = e^{i(i\frac{\pi}{2})}[/math]

[math]\implies i^i = e^{-\frac{\pi}{2}}[/math]

One Reply to “Is [math]i^i[/math] a complex number, and what is its standard form?”

  1. It is a complex number as well, as all real numbers are. Funnily enough [math]i^i[/math] is a real number.

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