Are all analytical function differentiable? I have a few confusion in complex analysis. Like what is the difference between a function being differentiable and a function being complex differentiable?
Also, A function being analytic implies:
$f'\left( z \right) = \mathop {\lim }\limits_{\Delta z \to 0} \frac{{f\left( {z + \Delta z} \right) - f\left( z \right)}}{{\Delta z}}$
must exist. (2)
Also  the function is differentiable at some point z if
$f'\left( z \right) = \mathop {\lim }\limits_{\Delta z \to 0} \frac{{f\left( {z + \Delta z} \right) - f\left( z \right)}}{{\Delta z}}$
exists. (3)
I am sure the third equation is correct but not sure if the second one is justified or not.
Also,I know that every differentiable function is not analytic that is why I think equation 3 is not correct.
Also, is it necessary for a function to be complex differentiable at a point if it satisfies CR equations at that point? I found it somehwere on quora where people said that  **We know that a function is complex differentiable if and only if it satysfies the Cauchy-Riemann equations at the point. So, lets think as your function as a function from R 2 to R ** but, I found a question...
question
where the function is not complex differentiable at a point but satisfies the CR equations.Kindly help me..Different sources have left me confused.
 A: tl; dr: There are two subtleties:

*

*Whether $f$ is approximated near a point by a real-linear mapping ($f$ is "real-differentiable", or simply "differentiable"), or by a complex-linear  mapping ($f$ is "complex-differentiable");

*Whether $f$ is complex-differentiable at every point of a non-empty open set ($f$ is "holomorphic") or not.


By definition, if $f$ is holomorphic in a plane region $\Omega$, then $f$ is complex-differentiable at each point of $\Omega$, and therefore real-differentiable at each point of $\Omega$.
Real-differentiability at a point does not imply complex-differentiability at that point, and complex-differentiability does not imply holomorphicity at that point (complex-differentiability in a neighborhood). Writing $z = x + iy$ with $x$, $y$ real:

*

*$f(z) = \bar{z} = x - iy$ is real-differentiable everywhere in the plane, but complex-differentiable nowhere.

*$f(z) = z\bar{z} = x^{2} + y^{2}$ is real-differentiable everywhere in the plane, but complex-differentiable at $0$ and nowhere else.

*$f(z) = (\operatorname{Re} z)^{2} = x^{2}$ is real-differentiable everywhere in the plane, but complex-differentiable only along the imaginary axis $\{x = 0\}$, and therefore holomorphic nowhere.

*$f(z) = z = x + iy$ is holomorphic everywhere.

Notes:

*

*All these examples are real polynomials, hence real-analytic (represented by convergent real power series in a neighborhood of each point). The first three fail to be holomorphic anywhere. The fourth is a complex polynomial, a polynomial in $z$.


*With respect to the Cartesian standard basis in the real plane, multiplication by $i$, or by a complex number $a + bi$ with $a$ and $b$ real, are given by the matrices
$$
J = \left[\begin{array}{@{}rr@{}}
    0 & -1 \\
    1 &  0 \\
  \end{array}\right],\qquad
aI + bJ = a\left[\begin{array}{@{}rr@{}}
    1 & 0 \\
    0 & 1 \\
  \end{array}\right]
+ b\left[\begin{array}{@{}rr@{}}
    0 & -1 \\
    1 &  0 \\
  \end{array}\right]
= \left[\begin{array}{@{}rr@{}}
    a & -b \\
    b &  a \\
  \end{array}\right].
$$
Writing a complex-valued function $f = u + iv$ in terms of real and imaginary parts, and treating the complex variable $z = x + iy$ as an ordered pair $(x, y)$ of real variables, we have
$$
Df = \left[\begin{array}{@{}cc@{}}
    u_{x} & u_{y} \\
    v_{x} & v_{y} \\
  \end{array}\right].
$$
Comparing the real derivative matrix with the real representation of a general complex matrix, we find $Df$ is complex-linear if and only if the Cauchy-Riemann equations $u_{x} = v_{y}$ and $u_{y} = -v_{x}$ hold.


*The sets of complex-valued (!) functions $\{x, y\}$ and $\{z, \bar{z}\}$ span the same two-dimensional complex vector space because $z = x + iy$, $\bar{z} = x - iy$ and
$$
x = \tfrac{1}{2}(z + \bar{z}),\qquad
y = -\tfrac{1}{2}i(z - \bar{z}).
$$
A short formal chain rule calculation shows the Cauchy-Riemann equations are equivalent to $\partial f/\partial\bar{z} = 0$.


*It turns out that if $f$ is holomorphic in some plane region $\Omega$, then $f$ is complex-analytic in that region: For each point $z_{0}$ of $\Omega$, there is a convergent complex power series centered at $z_{0}$ that converges to $f$ in some open disk. (In fact, we may pick the "largest" open disk about $z_{0}$ that is contained in $\Omega$, namely the union of all such open disks.)
