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I have a question involving the evaluation of $3^i$, but I am unsure how to do this. I know how to solve such questions involving $e^{i\theta}$, but how does this work with a different base? (I understand that the angle is 1 radian).

I have attempted to convert $3$ to Euler's form giving $3 \ e^{0}$ but this does not get me anywhere. What is the general way to solve questions in the form $a^i, a\in \mathbb{R}$?

Anixx
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Ruben
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2 Answers2

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From http://www.math.toronto.edu/mathnet/questionCorner/complexexp.html:

Finally, for a real number $a$, you can define $a^{(b+ic)}$ by writing $a = e^{\ln a}$: \begin{align} a^{b+ic} &= e^{(\ln a)(b+ic)} = e^{(b\ln a) + i(c\ln a)}\\ &= e^{(b\ln a)}\left(\cos(c\ln a)+i\sin(c\ln a)\right)\\ &= a^b\left(\cos(c\ln a)+i\sin(c\ln a)\right) \end{align}

That simplifies to this form:

$$ \cos (\ln(a))+i \sin (\ln(a)) $$

Hence,

$$ 3^i=\cos (\ln (3))+i \sin (\ln (3)) $$

Also, interesting side-note: any problem of form $a^i, a\in \mathbb{R}$ can be expressed in polar coordinates as $r=1, \theta=\frac{180 \ln (a)}{\pi }{}^{\circ}$

Carter Pape
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  • Should it not be $3^i=\cos{\ln{3}} + i\sin{\ln{3}}$ then? That's is what the other answers brought to my understanding. – Ruben Jan 30 '14 at 23:37
  • Yes, Ruben. Thanks for pointing that out. I'm used to just writing $\log$ because in my math class, $\log$ is implied to be $\ln$. – Carter Pape Jan 30 '14 at 23:39
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    I'm in CS, where there are many $\log$s. But a math prof of mine once said that "there's only one $\log$." :-) – Josephine Moeller Jan 30 '14 at 23:42
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    @JohnMoeller here, in fact, that are many different $\log$'s ;) (Although the interesting distinction is their domain and range rather than base.) – Jonathan Y. Jan 30 '14 at 23:55
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Note that $ e^{i\theta}=(e^{\theta})^i.$ To find $3^i $, first solve $e^{\theta}=3$.

Unwisdom
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