7

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.

Joe
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Diane Vanderwaif
  • 4,292

3 Answers3

8

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.

Tim Raczkowski
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3

Let's make it even simpler: If you know that $(-z)^2 = z^2$, does it necessarily follow that $-z = z$? Why not?

heropup
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1

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 $\arg z=\theta$).

Joe
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