Errors in a false complex proof

Question

This proof tries to show that

$ \mathbb{R} = \mathbb{C} $

$Let z \in \mathbb{C}, \exists r \in \mathbb{R}^{+}, \theta \in [0,2\pi]:z=re^{i\theta}$ $\\z=re^{i\theta\frac{2\pi}{2\pi}} = r(e^{2\pi i})^{\frac{\theta}{2\pi}}= r(1)^{\frac{\theta}{2\pi}}=r \\so z \in \mathbb{R} \\\rightarrow\mathbb{C} \subset \mathbb{R} \\ therefore \mathbb{C}=\mathbb{R}$

This proof is clearly false but I can't seem to find any algebraic mistake in it.

I can't really read it, but it seems that you think that $e^{i\theta\frac{2\pi}{2\pi}}=(e^{i\cdot 2\pi})^{\frac{\theta}{2\pi}}$. This is not the case. — Arthur, Feb 15 '18 at 13:09

score 0 · Accepted Answer · answered Feb 15 '18 at 13:10

0

The equality $$(a^b)^c = a^{bc}$$ is not true for complex numbers (it is only true in general if $a$ is a positive real, and $b,c$ are real), so your proof is false.

answered Feb 15 '18 at 13:10

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Errors in a false complex proof

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