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My answer is actually it is not a linear approximiation actually it an approximation of a translation of a linear transformation..whether in complex field if we see complex as a vector space over R and if it is diff at a point then actually it behaves like a translation of a scaled rotation..then why these are called best linear approximiation though they are not linear actually a certain amount of translation of lineae are actually they....

Raju
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    Are you objecting to the word linear, as opposed to (for example) affine? Just say that the thing that $f'(a) , h$ is approximating is the difference $f(a+h)-f(a)$, and then you really get something linear in $h$. See this question, for example. – Hans Lundmark May 27 '21 at 05:17
  • Your question is useful. Please consider it writing in details and omitting short terms. – pmun May 27 '21 at 05:19
  • Sir actually i exactly don't know the defn of affine but one i heard that i recalled that a line is not passing through origin but in this case if it is then yes i meant to it here this by translation of a linear transformation though this words also seems to me not correct.... – Raju May 27 '21 at 06:05
  • Remember that linear isn't linear ... One has to distinguish between $\mathbb R$- and $\mathbb C$-linear maps. Both send $z$ to $a\cdot z$; in the first case $a$ is real, in the latter $a\in\mathbb C$. Now a $\mathbb C$-linear map is, considered as a map $\mathbb R^2\to\mathbb R^2$ a rotation followed by a scaling. Hence not all $\mathbb R$-linear maps are $\mathbb C$-linear. – Michael Hoppe May 27 '21 at 09:29
  • OP are you familiar with the notion of a tangent space from multivariable calculus? You can think of the derivative as a linear transformation of the tangent line at $a$ to the tangent line at $f(a)$, and the origin of the tangent line at $a$ can be taken to be $a$. This makes it literally linear. – paul blart math cop May 27 '21 at 19:46

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