# Value of $i^\sqrt3$

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.

Hint: by definition, $$b^c = \exp(c \log b)$$. Use all values of $$\log b$$.

• thanks, and b here does not need to a real number? (need a little explanation if possible) – james black Jan 17 at 3:25
• 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? – Roger Jan 17 at 3:35
• 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. – Lubin Jan 17 at 3:43

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

$$-i = i^3 = (i^{\sqrt 3})^{\sqrt 3}$$ $$= (e^{i(2n\pi + \pi/2)\sqrt 3})^{\sqrt 3} = e^{i(2n\pi + \pi/2)(3)}$$ $$= e^{6\pi i}e^{3\pi i /2} = -i. ✓ \tag 5$$

• Great! I appreciate often your answer....my sincere and cordial greetings. – Sebastiano Jan 17 at 20:49
• @Sebastiano: greetings to you as well, my friend . . . and thanks for the kind words! – Robert Lewis Jan 17 at 20:51

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.