Find all values of $i^\sqrt3$.
I am trying to apply de Moivre's formula here but cannot find a way to do so. I am not sure if i am approaching this wrong.
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Hint: by definition, $b^c = \exp(c \log b)$. Use all values of $\log b$.
answered Jan 17 at 3:24
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$\begingroup$ thanks, and b here does not need to a real number? (need a little explanation if possible) $\endgroup$ – james black Jan 17 at 3:25
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$\begingroup$ But then: $i^\sqrt{3}=\exp(\sqrt{3}\ln{i})=\exp(\sqrt{3}i(\pi/2+2\pi n)) [\forall n \in Z]$, which has infinit results? $\endgroup$ – Roger Jan 17 at 3:35
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$\begingroup$ Infinitely many results, @Roger. Yes. In fact, raising a nonpositive complex number to an exponent that’s not an integer will always be ill-defined, with perhaps infinitely many values. $\endgroup$ – Lubin Jan 17 at 3:43
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Consider: from deMoivre's
$e^{i \theta} = \cos \theta + i \sin \theta \tag 1$
with
$\theta = 2n \pi + \dfrac{\pi}{2}, \; n \in \Bbb Z, \tag 2$
we have
$e^{i(2n\pi + \pi/2)} = i; \tag 3$
thus
$i^{\sqrt 3} = e^{i(2n\pi + \pi/2)\sqrt 3}$ $= \cos ((2n\pi + \pi/2)\sqrt 3) + i \sin ((2n\pi + \pi/2)\sqrt 3), \ n \in \Bbb Z. \tag 4$
We may check for consistency: (4) yields
$-i = i^3 = (i^{\sqrt 3})^{\sqrt 3} $ $= (e^{i(2n\pi + \pi/2)\sqrt 3})^{\sqrt 3} = e^{i(2n\pi + \pi/2)(3)}$ $= e^{6\pi i}e^{3\pi i /2} = -i. ✓ \tag 5$
answered Jan 17 at 20:42
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1$\begingroup$ Great! I appreciate often your answer....my sincere and cordial greetings. $\endgroup$ – Sebastiano Jan 17 at 20:49
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1$\begingroup$ @Sebastiano: greetings to you as well, my friend . . . and thanks for the kind words! $\endgroup$ – Robert Lewis Jan 17 at 20:51
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The de Moivre's formula is useful when you deal with a complex number of the form $e^{ix}$ ($x\in\mathbb{R}$).
When you have exponentiation of a complex number, what you should look for first is the definition, particularly when you have $z^w$ where $z$ is a complex number and $w$ is not an integer.
In your case, $$ i^{\sqrt{3}}:=\exp(\sqrt{3}\log(i)) $$ where $\log$ denotes the complex logarithm, which is a multi-valued function.
