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## Main Question or Discussion Point

Here's an interesting way to look at CR I feel is often overlooked:

Let:

[tex]z = x + i y[/tex]

[tex]z^{\ast} = x - i y[/tex]

One common form for the CR condition is to say that if some function [itex]f[/itex] is analytic then it does not depend on [itex]z^{\ast}\;[/itex]. That is,

[tex]\frac{\partial f}{\partial z^{\ast}} = 0[/tex]

But is there anything special about [itex]z^{\ast}\;[/itex]? It turns out there isn't. In fact, any complex variable [itex]w[/itex] that is linearly independent with [itex]z[/itex] will do. To illustrate this, let

[tex]w = a x - i b y[/tex]

Then,

[tex]b z + w = (a+b)x, \, a z - w = i(a+b)y[/tex]

So,

[tex]x = \frac{b z + w}{a + b}, \, y = \frac{a z - w}{i(a + b)}[/tex]

[tex]\frac{\partial x}{\partial w} = \frac{1}{a+b}, \, \frac{\partial y}{\partial w} = -\frac{1}{i(a+b)}[/tex]

We then have

[tex]\frac{\partial}{\partial w} = \frac{\partial x}{\partial w} \frac{\partial}{\partial x} + \frac{\partial y}{\partial w} \frac{\partial}{\partial y} = \frac{1}{a+b}[\frac{\partial}{\partial x} - \frac{1}{i} \frac{\partial}{\partial y}][/tex]

We then confirm that effectively,

[tex]\frac{\partial z}{\partial w} = \frac{1}{a+b}[\frac{\partial}{\partial x}(x + i y) - \frac{1}{i} \frac{\partial}{\partial y}(x + i y)] = \frac{1}{a+b}[1 - \frac{i}{i}] = 0[/tex]

which holds as long as [itex]a+b \neq 0[/itex] which is to say that [itex]z[/itex] and [itex]w[/itex] are linearly independent. Other than this constraint, our choice of [itex]a[/itex] and [itex]b[/itex] is totally arbitrary. The CR condition is simply:

[tex]\frac{\partial f}{\partial w} = \frac{1}{a+b}[\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}]f = 0[/tex]

Or simply,

[tex][\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}]f = 0[/tex]

Let:

[tex]z = x + i y[/tex]

[tex]z^{\ast} = x - i y[/tex]

One common form for the CR condition is to say that if some function [itex]f[/itex] is analytic then it does not depend on [itex]z^{\ast}\;[/itex]. That is,

[tex]\frac{\partial f}{\partial z^{\ast}} = 0[/tex]

But is there anything special about [itex]z^{\ast}\;[/itex]? It turns out there isn't. In fact, any complex variable [itex]w[/itex] that is linearly independent with [itex]z[/itex] will do. To illustrate this, let

[tex]w = a x - i b y[/tex]

Then,

[tex]b z + w = (a+b)x, \, a z - w = i(a+b)y[/tex]

So,

[tex]x = \frac{b z + w}{a + b}, \, y = \frac{a z - w}{i(a + b)}[/tex]

[tex]\frac{\partial x}{\partial w} = \frac{1}{a+b}, \, \frac{\partial y}{\partial w} = -\frac{1}{i(a+b)}[/tex]

We then have

[tex]\frac{\partial}{\partial w} = \frac{\partial x}{\partial w} \frac{\partial}{\partial x} + \frac{\partial y}{\partial w} \frac{\partial}{\partial y} = \frac{1}{a+b}[\frac{\partial}{\partial x} - \frac{1}{i} \frac{\partial}{\partial y}][/tex]

We then confirm that effectively,

[tex]\frac{\partial z}{\partial w} = \frac{1}{a+b}[\frac{\partial}{\partial x}(x + i y) - \frac{1}{i} \frac{\partial}{\partial y}(x + i y)] = \frac{1}{a+b}[1 - \frac{i}{i}] = 0[/tex]

which holds as long as [itex]a+b \neq 0[/itex] which is to say that [itex]z[/itex] and [itex]w[/itex] are linearly independent. Other than this constraint, our choice of [itex]a[/itex] and [itex]b[/itex] is totally arbitrary. The CR condition is simply:

[tex]\frac{\partial f}{\partial w} = \frac{1}{a+b}[\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}]f = 0[/tex]

Or simply,

[tex][\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}]f = 0[/tex]

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