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I need to use theorem $2$ in this paper and in it they have the term

$$\frac{\langle b-a,f(z_1)-f(a)\rangle}{\langle b-a,z_1-a\rangle}$$

which I can't seem to understand. What are the entities on the numerator and denominator? Are they vectors, intervals? And how does one carry out division on them. For context, the theorem is one of complex analysis and $f$ is a holomorphic function defined on the open convex subset $D\in\mathbb{C}$ where $a,b\in D$ and $z_1\in (a,b)$.

Sorry for the dull question, I just can't seem to figure it out, nor do I know what to google because I don't know the name of the entities being used.

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    $\begingroup$ In the line above the theorem, they define $\langle u,v\rangle=\Re(u\bar{v})$. $\endgroup$ – carmichael561 Jul 29 '16 at 17:24
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    $\begingroup$ More generally, $\langle x,y\rangle$ is common notation for "inner product of $x$ and $y$" (which are vectors in some inner product vector space). This inner product is usually clear, or defined before. $\endgroup$ – Clement C. Jul 29 '16 at 17:27
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The line above the statement of the theorem has the definition $\langle u, v\rangle = \operatorname{Re}(u\overline{v})$. This is a real inner product on $\mathbb{C}$ - the notation $\langle\cdot, \cdot\rangle$ is usually used to denote inner products.

More generally, $\langle u, v\rangle = \operatorname{Re}(u^T\overline{v})$ is a real inner product on $\mathbb{C}^n$.

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  • $\begingroup$ This answer would gain by specifying that this is actually an inner product that is defined, not just an arbitrary binary operation. (The notation $\langle \cdot,\cdot\rangle$ is common/most often used for this.) $\endgroup$ – Clement C. Jul 29 '16 at 17:30
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    $\begingroup$ @ClementC. Good point. I've edited my answer accordingly. $\endgroup$ – Michael Albanese Jul 29 '16 at 17:44

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