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Cauchy-Riemann equations

To check that the derivative is well-defined, separate the complex function into the sum of two real functions, just as z can be written as the sum of a real and an imaginary part:

z = x + i y (42)
f = u(x,y) + i v(x,y). (43)

By first taking $\Delta z$ real,

$\displaystyle \frac{df}{dz}$ = $\displaystyle \lim_{\Delta z \rightarrow 0} \frac{f(z+\Delta x)}{\Delta x}$ (44)
  = $\displaystyle \lim_{\Delta x \rightarrow 0} \frac{
u(x+\Delta x)+iv(x+\Delta x) - (u(x,y) + i v(x,y))}{\Delta x}$ (45)
  = $\displaystyle \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x},$ (46)

and then imaginary,
$\displaystyle \frac{df}{dz}$ = $\displaystyle \lim_{\Delta z \rightarrow 0} \frac{f(z+i\Delta y)}{i\Delta y}$ (47)
  = $\displaystyle \lim_{\Delta y \rightarrow 0}
\frac{u(x,y+\Delta y)+iv(x,y+\Delta y)-(u(x,y) + i v(x,y))}{i\Delta y}$ (48)
  = $\displaystyle -i\frac{\partial u}{\partial y} + \frac{\partial v}{\partial y}.$ (49)

Supposing that the derivative is independent of phase, at least in the context taken here, comparison shows

$\displaystyle \frac{\partial u}{\partial x}$ = $\displaystyle \frac{\partial v}{\partial y}$ (50)
$\displaystyle \frac{\partial v}{\partial x}$ = $\displaystyle -\frac{\partial u}{\partial y}$ (51)

The members of this pair of equations are known as the Cauchy - Riemann equations, relating the real and imaginary parts of a complex function which is supposed to have a derivative. Ignoring the relationship is equivalent to failing to check whether or not a supposed derivative is well-defined. Strictly the relation should be checked for all other phases, but in the end all of them depend on just these two.

Consider the Jacobian matrix for the real functions u(x,y) and v(x,y) of the real variables x and y,

$\displaystyle \frac{\partial(u,v)}{\partial(x,y)}$ = $\displaystyle \left[ \begin{array}{cc}
\frac{\partial u}{\partial x} & \frac{\p...
...rac{\partial u}{\partial y} & \frac{\partial v}{\partial y}
\end{array} \right]$ (52)
  = $\displaystyle \left[ \begin{array}{cc}
\frac{\partial u}{\partial x} & \frac{\p...
...rac{\partial v}{\partial x} & \frac{\partial u}{\partial y}
\end{array} \right]$ (53)
  = $\displaystyle \frac{\partial u}{\partial x} {\bf 1}+\frac{\partial v}{\partial x} {\bf i}$ (54)

The impact of the Cauchy-Riemann equations is to give the Jacobian matrix the form of a complex number in quaternion disguise; none other will suffice. Writing the matrix as an exponential shows how the derivative is a complex number with absolute value and a phase.

The conclusion is not only that not any old pair of functions could be joined to get an analytic functionanalytic function (as differentiable functions of a complex variable are called), but that by knowing one of them, one effectively knows the other. That is,

v(x,y) - v(a,y) = $\displaystyle \int_a^x\frac{\partial v(\sigma, y)}{\partial \sigma} d\sigma$ (55)
  = $\displaystyle -\int_a^x\frac{\partial u(\sigma, y)}{\partial y} d\sigma$ (56)

Knowing u, take its derivative and integrate to get v. For example, according to Euler's formula,

ez = $\displaystyle e^x(\cos y + i \sin y).$ (57)

u = $\displaystyle e^x \cos y$ (58)
$\displaystyle \frac{\partial u}{\partial y}$ = $\displaystyle -e^x\sin y$ (59)
$\displaystyle - \int_a^be^\sigma\sin y d\sigma$ = $\displaystyle e^\sigma\sin y\vert _a^x$ (60)
v(x,y) - v(a,y) = $\displaystyle e^x\sin y - e^a\sin a$ (61)

which is consistent with
v(x,y) = $\displaystyle e^x\sin y$ (62)

So trying to think of something like $e^x(\cos y + i \tan y)$ as an analytic function just wouldn't work.

The reason there aren't analytic quaternion functions is twofold. First, they anticommute (complicating division by $\Delta q$), and besides, there is getting their Jacobian matrices to act like numbers.

Consider anticommutativity and the derivative of q2. Having written

$\displaystyle (q + \Delta q)^2 - q^2$ = $\displaystyle q^2 +q\ \Delta q +\Delta q\ q +{\Delta q}^2 - q^2,$ (63)

should we divide by $\Delta q$ on the left, on the right, take half the sum of the foregoing, or divide by $\surd(\Delta q)$ on both sides. No matter what, it is questionable whether the result would be 2q, or anything else free of $\Delta q$

Historically, some progress has been made by requiring quaternion functions to satisfy linear partial differential equations similar to the Cauchy-Riemann equations. However, examining the possibilities in more detail would be a distraction from our concern with complex functions,

Remaining with functions of a complex numbers, it seems that all the manipulations which work for real variables seem to work for complex variables. That is because they are confined to polynomials and perhaps their limits, where there is always a term free of $\Delta z$ along with others having $\Delta z$ as a factor which can vanish in the limit.

But those are not the only functions of two real variables which can take complex values. Consider the complex conjugate,

$\displaystyle \overline{x+iy}$ = x - i y. (64)

Its Jacobian matrix is
J(x,y) = $\displaystyle \left[ \begin{array}{cc}
1 & 0 \\  0 & -1
\end{array} \right],$ (65)

which certainly doesn't satisfy the Cauchy-Riemann equation. Neither does the squared norm
$\displaystyle z{\bar{z}}$ = x2 + y 2, (66)

whose Jacobian matrix is
J(x,y) = $\displaystyle \left[ \begin{array}{cc}
2x & 0 \\  2y & 0
\end{array} \right].$ (67)

Therefore any polynomial which mixes z and $\bar{z}$, or depends on $\bar{z}$ alone, will fail to be analytic unless it is purely zero.

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Next: harmonic functions of real Up: The derivative of a Previous: The derivative of a