Why is a complex function analytic if it has a non-zero imaginary component?

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The discussion centers on the analyticity of the complex function $log(z^2 + 2)$. It is established that this function is analytic unless $z^2 + 2 \le 0$, which occurs when $z$ is on the real axis or when $z^2$ is negative. The principal logarithm, denoted as $Log\, z$, is defined as $\ln|z| + i\operatorname{arg}(z)$ and is analytic on $\mathbb{C} \setminus (-\infty, 0]$. Participants clarify that complex numbers do not form an ordered field, making expressions like $z > 0$ inappropriate.

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Been working through videos on complex analysis, just when I thought I had a good grasp of the basics, I came across an example that I don't follow.

Where is $ log(z^2 + 2) $ analytic?
The presenter stated, it will be analytic unless $ z^2 + 2 \le 0$. I can't find a property like this anywhere, so where does he get that?

He then expands $ z^2 + 2 $ and states that if both x,y are non-zero, then we will have an imaginary part and the function will be analytic because we are not on the real axis. Again I can't find this property - that a complex function will be analytic if it has a non-zero imaginary component?
 
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Hi ognik,

Start by answering the following questions. How is $\log(z)$ defined? Where is $\log(z)$ analytic?
 
Hi Euge, by definition do you mean from z=$ e^w $, so that w = ln(z)?

A function is analytic at z if it has a derivative there, so with f(z) = ln(z), f'(z) = 1/z. So I can see z cannot = 0. And you can't have log of a negative, so $ z > 0 $

For f(z)=$ ln(z^2 + 2) , f'(z) = \frac{2z}{z^2 + 2} $, so $ z^2 + 2 \:must \:> 0. $ Of course I see it now. But then $ ln(z^2 + 2) $ is analytic for $ z^2 > -2 $ isn't it? Whether on real axis or with imaginary part?
 
Assume $z \neq 0$. The principal logarithm of $z$, typically denoted $Log\, z$, is defined as $\ln|z| + i\operatorname{arg}(z)$ were $-\pi < \arg(z) \le \pi$. It is defined and analytic on $\Bbb C \setminus (-\infty, 0]$. Note that some define $Log\, z$ be to the logarithm with $0 < \arg(z) \le 2\pi$.

Please show more care when discussing order with complex numbers. The complex numbers do not form an ordered field, so expressions like $z > 0$ and $z^2 > -2$ do not make sense. You could say $z$ is positive real number or $z^2 + 2$ lies on the positive real axis instead of these inequalities.

One minor note to add: do not write $ln(z)$ for the complex logarithm; use $log(z)$ instead. Keep in mind that you're not dealing with the natural logarithm of a real number.
 
Thanks as always Euge, all understood. (Just FYI, this is not part of my course, I just decided I'd to explore complex analysis a bit deeper than the text does; I'm doing maths for Physicists, apparently we are not pure :-) ). The text also does not cover ordered fields - but I did a bit more browsing.

I am still coming to grips with notation, for example I thought using log implied base 10, (which is why I wrote ln) - so log(z) implies base e 'cos z is complex?

Back to the main point - is it correct to say "the function will be analytic because we are not on the real axis" - for any complex function?

(I'd like to express again my profound gratitude for this forum, my 'lecturer' does not answer a lot of my emails, so forums are in practice the only interaction I have.)
 

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