Prof says that $z^{a+b}=z^a z^b$ is not always true

Asked Jun 05 '23 at 12:02

Active Jun 05 '23 at 12:02

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Our prof said that for $z, a,b \in \mathbb C$ we not always have $z^{a+b}=z^a z^b$.

I don't understand why. We have:

$$z^{a+b}=\exp(\ln(z)(a+b))=\exp(a \ln(z)+ b \ln(z))$$

According to our lecture we can write this as: $$\exp(a \ln(z)) \cdot \exp(b \ln(z)) = z^a \cdot z^b$$

Can someone enlighten me?

asked Jun 05 '23 at 12:02

1

$\ln$ is not defined for all complex numbers! You have to choose a branch cut, and that's exactly when the problems begin. – Andromeda Jun 05 '23 at 12:05
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Consider Why $\sqrt{-1\cdot-1}\neq \sqrt{(-1)}^2$ – JMoravitz Jun 05 '23 at 12:06
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Well, no, $z^{a+b}=z^a z^b$ is true if you have fixed a choice of logarithm and $z$ is nonzero. What is not true in general is $z^a\cdot w^a=(zw)^a$ (for fixed choice of logarithm). Similarly $(z^a)^b=z^{ab}$ is not true in general – FShrike Jun 05 '23 at 12:17
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It is likely you have misinterpreted your professor. $\exp(a+b)=\exp(a)\exp(b)$ is the only index law that carries over from the real numbers – FShrike Jun 05 '23 at 12:22

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