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I saw the theorem that Julia set of $f(z)=z^2+c$ is symmetric about the origin and I'm wondering why and how to prove it!

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Look at your favourite definition of the Julia set, and note that (since $f(z) = f(-z)$) $z$ satisfies the definition if and only if $-z$ does.

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    Isn't in this way it will be a even function rather an odd function symmetric about the origin? because I suppose odd function should be -f(z)=f(-z)? – 2017-02-19
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    Yes, $f$ is an even function. – 2017-02-20
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    So therefore it's not symmetric about origin? Because odd function should be the one which is symmetric about origin – 2017-02-20
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    The graph of the function $f$ would be symmetric about the origin if $f(z) = - f(z)$. But we're talking about the Julia set, not the graph. – 2017-02-20