# Log of negative numbers I know that log of

Log of negative numbers
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?
$\mathrm{log}\left(-a\right)=\frac{2×\mathrm{log}\left(-a\right)}{2}=\frac{1}{2}\mathrm{log}\left({a}^{2}\right)=\frac{2}{2}\mathrm{log}\left(a\right)=\mathrm{log}\left(a\right)$
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Sawyer Anthony
Because the property is actually
$\mathrm{ln}\left({x}^{a}\right)=a\mathrm{ln}\left(|x|\right)$
(for even integer a), and not
$\mathrm{ln}\left({x}^{a}\right)=a\mathrm{ln}\left(x\right)$
and since no number has a negative absolute value, the second equality is incorrect.
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Alannah Farmer
Let's cut the complex plane along the positive real axis and work on the Riemann sheet for which
$0\le argz<2\pi$ (1)
For $a$ real-valued with $a>0$, we have
$\mathrm{log}\left(-a\right)=\mathrm{log}a+i\pi$ (2)
Now, we have
$\mathrm{log}\left(-a\right)=\frac{2\mathrm{log}\left(-a\right)}{2}$
$=\frac{1}{2}{\mathrm{log}a}^{2}+i\pi$
$=\mathrm{log}a+i\pi$
$\ne \mathrm{log}a$
In fact, if we multiply Equation (2) by a factor of $2$, we get
$2\mathrm{log}\left(-a\right)=\mathrm{log}a+i2\pi \ne \mathrm{log}a$
That is, in multiplying by 2, we actually move to the subsequent Riemann surface since Equation (1) restricts the argument to be less than $2\pi$.