Why is Euler's formula defined for non-integer values?

Question

Say that for some complex number $w$

$$e^{wi} = a$$

Now raise both sides to $1/4$.

$$e^{wi/4} = a^{1/4}$$

Now $e^{wi/4}$ has a single defined value. Yet $a^{1/4}$ can have multiple values. So why is Euler's formula well defined for non-integer powers, since non-integer powers can yield multiple values?

Answer 1

The exponential function $\exp$ is single-valued. Non-integer powers are multi-valued functions defined by $a^b = \exp(b \log a)$, using any branch of the logarithm. In particular, $\exp(iwb)$ is just one possible value of $(\exp(iw))^b$. Since $\log(\exp(iw)) = iw + 2 \pi i n$ for arbitrary integer $n$, the full result is $$ \left(\exp(iw)\right)^b = \exp((iw + 2 \pi i n) b) = \exp(iwb + 2 \pi i n b) = \exp(iwb) \exp(2\pi i n b)$$ In particular, when $b=1/4$ there are four values, corresponding to $e^{2\pi i n/4} = 1, i, -1, -i$.

Answer 2

If $e^{wi} = a$, then $e^{(w+2\pi n)i} = a$ for any integer $n$.

Raising to the $1/4$ power, $a^{1/4} =(e^{(w+2\pi n)i})^{1/4} =e^{i(w/4+\pi n/2)} $.

Answer 3

Depending on the pedagogical order, this phenomenon of multivaluedness is really a consequence of periodicity. That is to say, there is an underlying periodicity in the formula which is not immediately apparent, and is inherited from the periodicity of the trigonometric functions (that is, if you learn about the complex exponential through DeMoivre's formula--which is not a given, considering Walter Rudin's prologue to Real and Complex Analysis).

Recall Euler's formula $$e^{i\theta} = \cos \theta + i \sin \theta.$$ From this, we can immediately deduce that the complex-valued mapping $f : \mathbb C \to \mathbb C$, $f(z) = e^z$, is periodic with period $2\pi i$: $$f(z + 2\pi i k) = f(z)$$ for all $k \in \mathbb Z$. So this periodicity is what leads to the inverse mapping, the complex logarithm $$f^{-1}(z) = \log z$$ to be "multivalued" (i.e., "one-to-many"). As there are infinitely many complex numbers whose exponentials are equal--explicitly, the set $S = \{z + 2 \pi i k \mid k \in \mathbb Z\}$ all map to the same value $w = e^z$ under $f$--then, for a general complex number $w$, there are infinitely many preimages whose exponential equals $w$.

Now, the same thing happens with (complex) powers of complex numbers, because we have $$z^w = e^{w \log z}.$$ When $w$ is an integer there are no issues: since $$\log z = \log|z| + i \arg(z) + 2 \pi i k,$$ multiplying by an integer $w$ causes the $2\pi i k$ term to remain an integer multiple of $2\pi i$, and the multivaluedness introduced by the logarithm is canceled by exponentiation. But when $w$ is not an integer (or is non-real), then this does not occur and we have an essentially multivalued mapping.