I am reading a proof of a theorem and I see this:
"if we put $F(x)$ in this form: $F(x)= r e^{i\theta}$"
$F(x)$ takes values in $\mathbb{C}$, so the question is:
In $\mathbb{C}$, every element can be put in that form?
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It is something like polar system? where $r$ is the modul and $e^{i \theta}$ is the angle? – Maths Student Jan 15 '14 at 18:48
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Yes, exactly.${}$ – David Mitra Jan 15 '14 at 18:50
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Yes. for any complex number, $x, y\in\Bbb R\not{0},; z=x+iy$ we have $r=|z|=\sqrt{x^2+y^2}$ and $\theta = \text{arg} z$ arg can be computed using arctangent. – Ali Caglayan Jan 15 '14 at 18:55
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Yes, but not uniquely represented in this way. Even if you ignore the $2\pi$ periodicity of $\theta$, there's still the ambiguity when $z=0$ because $0=0e^{i\theta}$ for all $\theta$. – Disintegrating By Parts Jan 15 '14 at 19:20
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It holds: $x=|x|e^{i\theta}$ where $\theta:= arcsin(\Im(x) /|x|) \ \forall x \in \mathbb{C}\setminus \{0\}$ or something like that.
Edoardo Lanari
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It depends on the argument of the function, I think that is correct – Edoardo Lanari Jan 15 '14 at 18:56