Complex exponentials.

Definition 4.6.1       

For any $ z \in \mathbb{C}^*$ and every $ \alpha \in \mathbb{C}$

$ z^{\alpha} = e^ {\alpha \log z}$

Example 4.6.2       

$ 16^{\frac 12}$ $ = e^{\frac 12 \log 16}$ $ = e^{\frac 12 (\ln 16 +
2k \pi i )}$
     $ = e^{\frac 12 (4 \ln 2 + 2k \pi i )}$ $ =
e^{2 \ln 2 + k \pi i }$
     $ = e^{2 \ln 2} \cdot e^{k \pi i }$ $ = 4 e^{k \pi i }$
     $ = \pm 4$ .  
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Proposition 4.6.3   $ \frac {d}{dz}(\alpha ^z)= \alpha ^z \; \log \alpha$ .

For the proof, we use the definition $ \alpha ^z=e^{z \log \alpha}$ . Then we have:

$\displaystyle \frac {d}{dz}(\alpha ^z)= \frac {d}{dz}(e^{z \log \alpha}) =\log \alpha \; e^{z \log \alpha} = \alpha ^z \; \log \alpha.$    

Proposition 4.6.4   $ \frac {d}{dz}(z^{\alpha })= \alpha z^ {\alpha -1}$ .

For the proof, we use the definition $ z^{\alpha} = e^ {\alpha \log z}$ . Then we have:

$\displaystyle \frac {d}{dz}(z^{\alpha })= \frac {d}{dz}(\alpha \log z) \; e^{\a...
...{z} e^{\alpha \log z} = \frac {\alpha }{z} z^ {\alpha } =\alpha z^ {\alpha -1}.$    

Noah Dana-Picard 2007-12-24