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I know that log of negative numbers is complex numbers. But I just got over this little proof and wondering what is wrong with this? $$\log(-a)=\frac{2\times\log(-a)}{2}=\frac{1}{2}\log(a^2)=\frac{2}{2}\log(a)=\log(a)$$

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2 Answers

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Because the property is actually $$\ln(x^a)=a\ln(\vert x\vert)$$ (for even integer a), and not$$\ln(x^a)=a\ln(x)$$ and since no number has a negative absolute value, the second equality is incorrect.

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Let's cut the complex plane along the positive real axis and work on the Riemann sheet for which

$$0\le \arg{z}<2\pi \tag 1$$

For $a$ real-valued with $a>0$, we have

$$\log (-a) =\log a +i\pi \tag 2$$

Now, we have

$$\begin{align} \log (-a)&=\frac{2\log(-a)}{2}\\\\ & =\frac12 \log a^2 +i\pi \\\\ &=\log a +i \pi \\\\ &\ne \log a \end{align}$$

In fact, if we multiply Equation $(2)$ by a factor of $2$, we get

$$2\log(-a)=\log a +i2\pi \ne \log a$$

That is, in multiplying by $2$, we actually move to the subsequent Riemann surface since Equation $(1)$ restricts the argument to be less than $2\pi$.

CONCLUSION:

$n\log z \ne \log z^n$ in general.

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