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Find the error in following reason

\begin{align*} (-z)^2=z^2 &\implies \log(-z)^2=\log(z)^2\\ &\implies2\log(-z)=2\log(z)\\ &\implies \log(-z)=\log(z) \end{align*}

I think the error is $2\log(-z)=2\log(z)$ because for $z=1$, $2\log(1)=0$, but $2 \log(-1)$ is undefined.

What's bugging me is this problem has a star on it, which mean it's a challenging problem. I don't think it is that easy. So I wonder if anyone could check if I missed or make mistake somewhere.

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The problem is that $\log z$ is multivalued, and so requires a branch cut. $\log z=\log|z|+i\,\text{arg}\,z$. We can choose the branch $-\pi<\text{arg}\,z\le\pi$. Now, let $z=e^{\frac{3\pi i}4}$. $$\log z=\log e^{3\pi i/4}=\frac{3\pi i}4.$$

But $$\log z^2=\log(-i)=-\frac{\pi i}2.$$

Edit: This shows that $\log z^2=2\log z$ does not hold for complex numbers.

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  • $\begingroup$ so you are saying that the second implication is wrong? I'm sorry, but I'm not sure I understand your answer. $\endgroup$ – Diane Vanderwaif Feb 12 '15 at 23:48
  • $\begingroup$ See the edit of my answer. $\endgroup$ – Tim Raczkowski Feb 12 '15 at 23:57
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Let's make it even simpler: If you know that $(-z)^2 = z^2$, does it necessarily follow that $-z = z$? Why not?

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  • $\begingroup$ no, because for $z=a+ib$ the only way to have $-z=z$ is $a=-a$ and $b=-b$ in other word $z=0$ $\endgroup$ – Diane Vanderwaif Feb 12 '15 at 23:08
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The error is in the second implication.

If we work in the real numbers, we have that $\log x^2=2\log|x|$. This follows from the more general notion of Complex Logarithm, which is the one we use when we work with the complex numbers; it is (briefly) defined as follows (once you have choosen the principal branch which allows to work with the so called Principal Logarithm) $$ \log z=\log|z|+i\arg z $$ where $\log|z|$ is the real log.

Clearly $\log|-z|=\log|z|$, but in general $\arg(-z)$ is NOT equal to $\arg z$ (very roughly speaking, if you write $z=re^{i\theta}$ then $\arg z=\theta$).

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