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This exercise is from Arfken mathematical methods for physicists): "The function $f(z)=u(x,y)+iv(x,y)$ is analytic. Show that $f^*(z^*)$ is also analytic."

There must be some simple proof (and not related to series), because there is little said about complex analysis in the book before this exercise (The only important thing said is Cauchy-Riemann conditions). Not sure, but I think if $f(z)=u(x,y)+iv(x,y)$ then $f^*(z^*)=u(x,-y)-iv(x,-y)=g(x,y)+ih(x,y)$. Now $g$ and $h$ must satisfy the Cauchy-Reimann conditions and their first partial derivatives with respect to $x$ and $y$ must be continuous. Yet I can't show any of them. Any suggestions for developing this idea?

Simorq
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    @mrf This is not the same. He wants to show it from the Cauchy-Riemann equations and the question you are saying is a duplicate explicitly say in the answer "I think it will be easiest not to split into real and imaginary parts -- so no Cauchy-Riemann " – JKnecht Mar 26 '16 at 20:02
  • It's been asked a dozen times or so. Here's another duplicate: http://math.stackexchange.com/questions/769099 (but I guess I wasn't able to reopen and close again.) And another: http://math.stackexchange.com/questions/809145 – mrf Mar 26 '16 at 20:26
  • @mrf Got it. Did not know that. – JKnecht Mar 26 '16 at 20:31
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    @mrf I asked a new question because none of those questions say how real and imaginary parts of f(z) satisfy the Cauchy-Riemann conditions which is my main problem. – Simorq Mar 27 '16 at 07:46

3 Answers3

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$f(z) = u(x,y) + iv(x,y)$

You want to show that the function

$f^*(z^*) = u(x,-y) - iv(x,-y)$

is also analytic.

Denote $f^*(z^*)$ by

$f^*(z^*) = u_1 + iv_1$

with

$u_1(x,y) = u(x,-y)$

$v_1(x,y) = -v(x,-y)$

The partial derivatives with respect to $x$ are related by

$$\frac{\partial{u_1}}{\partial{x}}(x,y) =\frac{\partial{u}}{\partial{x}}(x,-y), \:\:\:\: \frac{\partial{v_1}}{\partial{x}}(x,y) =-\frac{\partial{v}}{\partial{x}}(x,-y) $$

The partial derivatives with respect to $y$ are related by

$$\frac{\partial{u_1}}{\partial{y}}(x,y) =-\frac{\partial{u}}{\partial{y}}(x,y), \:\:\:\: \frac{\partial{v_1}}{\partial{y}}(x,y) =\frac{\partial{v}}{\partial{y}}(x,y) $$

You know that the Cauchy Riemann equations holds for $f(z) = u + iv$. The above relations shows that they also hold for $f^*(z^*) = u_1 + iv_1$ and thus the function $f^*(z^*)$ is also analytic.

jg mr chapb
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JKnecht
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  • Thanks, but I don't understand any of these. For example why \frac{\partial{u_1}}{\partial{x}} =\frac{\partial{u}}{\partial{x}} – Simorq Mar 26 '16 at 19:15
  • @Simorq I have edited my answer. Do you understand now? – JKnecht Mar 26 '16 at 19:57
  • @Simorq You just plug the above expressions for the partial derivatives into the Cauchy Riemann equations. – JKnecht Mar 26 '16 at 20:05
  • How did you find those relations between partial derivatives of u and u1? This is what I don't understand. – Simorq Mar 27 '16 at 07:50
  • @Simorq I used these $$u_1(x,y) = u(x,-y), :: ::

    v_1(x,y) = -v(x,-y)$$

     – JKnecht Mar 27 '16 at 08:28
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    I already know it. The thing I don't understand is why partial derivative of u1 with respect to x is the same as partial derivative of u with respect to x. And why partial derivative of v1 with respect to x is minus partial derivative of v with respect to x. (And partial derivatives with respect to y as well.) – Simorq Mar 27 '16 at 09:08
  • @Simorq The relations in my previous comment is a definition. Take the partial derivatives on both sides and you are done. – JKnecht Mar 27 '16 at 09:19
  • No I don't understand it yet :| – Simorq Mar 28 '16 at 13:50
  • @Simorq Do you know what the Cauchy-Riemann conditions are? Have you written them down for $u$ and $v$ in $f(z) = u + iv$? Have you then written down the relations between the partial derivatives? And then finally written down the Cauchy-Riemann conditionis for $u_1$ and $v_1$ in $f^(z^) = u_1 + iv_1$? – JKnecht Mar 28 '16 at 14:09
  • @Simorq The first C-R equation is

    $$\frac{\partial{u}}{\partial{x}} =\frac{\partial{v}}{\partial{y}}$$

    You know this holds for $f(z)$ $(*)$

    Now

    $$\frac{\partial{u_1}}{\partial{x}} =\frac{\partial{u}}{\partial{x}}$$

    and

    $$\frac{\partial{v_1}}{\partial{y}} =\frac{\partial{v}}{\partial{y}}$$

    The first C-R equation:

    $$\frac{\partial{u_1}}{\partial{x}}= \frac{\partial{u}}{\partial{x}}=(*)= \frac{\partial{v}}{\partial{y}}= \frac{\partial{v_1}}{\partial{y}}$$

    Hence

    $$\frac{\partial{u_1}}{\partial{x}}= \frac{\partial{v_1}}{\partial{y}}$$

     – JKnecht Mar 28 '16 at 14:21
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    I know these all. Forget about Cauchy-Riemann conditions. Suppose I have a function f(x,y). Then I define another function g(x,y)=f(x,-y). Why partial derivatives of these two functions with respect to x are the same? This is what I don't get. – Simorq Mar 29 '16 at 12:19
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    But they're not the same! In the second we've replaced y with -y.. Suppose u=xy so partial derivative with respect to x is y. Now I put -y. Partial derivative of -xy is -y which is not the sames as y :| – Simorq Mar 29 '16 at 12:55
  • @Simorq good question. one thing to note is that the partial of g(x,y) wrt x is the same as the partial of f with respect to x AT x,-y – jg mr chapb Mar 03 '22 at 21:09
  • You have to evaluate the partial derivatives of $f, g$ at different points so that they coincide. One is evaluated at $(x, y) $ and the other one at $(x, -y) $. – Compacto Mar 03 '22 at 23:52
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JKnecht's answer is partially wrong, it is true that

$$\frac{\partial u_1}{\partial x}(x, y) =\frac{\partial u}{\partial x}(x, -y)$$

$$\frac{\partial v_1}{\partial x}(x, y) =-\frac{\partial v}{\partial x}(x, -y)$$

Let's prove the first one, for example. Observe that $u_1(x, y) = u(c(x, y)) $, with $c(x, y) = (c_1(x, y),c_2(x, y))= (x, -y) $ is the complex conjugate.

Applying chain rule, we get:

$$\frac{\partial u_1}{\partial x}(x, y) = \frac{\partial(u\circ c) }{\partial x}(x, y) = \frac{\partial u}{\partial x}(c(x, y)) \frac{\partial c_1}{\partial x}(x, y) + \frac{\partial u}{\partial y}(c(x, y)) \frac{\partial c_2}{\partial x}(x, y) =\frac{\partial u}{\partial x}(x, -y) .$$

We have used that $\frac{\partial c_1}{\partial x} = 1, \frac{\partial c_2}{\partial x} = 0$.

But the partial derivatives w. r. t. $y$ are related by another formula:

$$\frac{\partial u_1}{\partial y}(x, y) =\frac{\partial (u\circ c)}{\partial y}(x, y)=\frac{\partial u}{\partial x}(c(x, y)) \frac{\partial c_1}{\partial y}(x, y) + \frac{\partial u}{\partial y}(c(x, y)) \frac{\partial c_2}{\partial y}(x, y) =-\frac{\partial u}{\partial y}(x, -y) . $$

We have used that $\frac{\partial c_1}{\partial y} = 0,\frac{\partial c_2}{\partial y} = -1$.

Similarly, one proves that:

$$\frac{\partial v_1}{\partial y}(x, y) =\frac{\partial v}{\partial y}(x, -y) . $$

In conclusion, the partial derivatives must be compared at different points. This still implies the CR equations.

$$ \frac{\partial u_1}{\partial x}(x, y) =\frac{\partial u}{\partial x}(x, -y)= \frac{\partial v}{\partial y}(x, -y)= \frac{\partial v_1}{\partial y}(x, y). $$

We have used that $u, v$ verify Cauchy-Riemann at the point $(x, -y) $, and the previous equalities.

Same goes for the other equation:

$$ \frac{\partial u_1}{\partial y}(x, y) =-\frac{\partial u}{\partial y}(x, -y)= \frac{\partial v}{\partial x}(x, -y)= -\frac{\partial v_1}{\partial x}(x, y). $$

EDIT: Another easy way of showing this goes by seeing that the complex conjugate preserves angles (but not orientations), so clearly $c(f(c(x, y))$ preserves angles (because $f$ also does) AND orientations (because $f$ preserves orientations and $c$ reverses them twice, so the composition preserves them). Every map preserving orientations and angles is holomorphic.

EDIT2: Proof of the equality of derivatives w. r. t. $y$. Recall that $v_1= -v\circ c$:

$$\frac{\partial v_1}{\partial y}(x, y) =-\frac{\partial (v\circ c)}{\partial y}(x, y)=-\frac{\partial v}{\partial x}(c(x, y)) \frac{\partial c_1}{\partial y}(x, y) - \frac{\partial v}{\partial y}(c(x, y)) \frac{\partial c_2}{\partial y}(x, y) =\frac{\partial v}{\partial y}(x, -y) . $$

We have used that $\frac{\partial c_1}{\partial y} = 0,\frac{\partial c_2}{\partial y} = -1$.

Compacto
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Let's start with JKnecht's answer:

$f(z) = u(x,y) + iv(x,y)$

You want to show that the function

$f^*(z^*) = u(x,-y) - iv(x,-y)$

is also analytic

Denote $f^*(z^*)$ by

$f^*(z^*) = u_1 + iv_1$

with

$u_1(x,y) = u(x,-y)$

$v_1(x,y) = -v(x,-y)$

Let $-y = w$

Then we have

$$\frac{\partial{u_1}}{\partial{x}}(x,y) =\frac{\partial{u}}{\partial{x}}(x,w), \:\:\:\: \frac{\partial{v_1}}{\partial{x}}(x,y) =-\frac{\partial{v}}{\partial{x}}(x,w) $$

The partial derivatives with respect to $y$ are related by

$$\frac{\partial{u_1}}{\partial{y}}(x,y) =\frac{\partial{u(x,-y)}}{\partial{y}} = -\frac{\partial{u}}{\partial{w}}(x,w), \:\:\:\: \frac{\partial{v_1}}{\partial{y}}(x,y) =-\frac{\partial{v(x,-y)}}{\partial{y}} = \frac{\partial{v}}{\partial{w}}(x,w) $$

the chain rules that produce these results can be intuitively seen if we treat u and v as multiple different single variable functions.

The key here is that we are assuming that f is holomorphic at the complex conjugate of every point that f is holomorphic on. we are assuming that the domain is symmetrical about the real axis

jg mr chapb
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  • This is wrong, $dv_1/dy(x, y) = dv/dy(x, -y) $. No minus sign involved. – Compacto Mar 04 '22 at 18:30
  • It's true that the domain is symmetric about real axis, otherwise the question makes no sense: $f^(z^) $ would not be defined. – Compacto Mar 04 '22 at 18:34
  • no this is correct, it follows directly from the definition of v1. And let me clarify, the function f must be analytic on the conjugate of z – jg mr chapb Mar 04 '22 at 20:25
  • You don't know how to differentiate man. Need to study chain rule again. – Compacto Mar 04 '22 at 22:00
  • @Compacto. Listen. it's the exact same answer. only your notation is confusing. when you say dv/dy. you are differentiating v wrt to the SECOND variable of v(x,w). AT the point w = -y. it's the same thing. MY notation does NOT mean to differentiate v wrt the SECOND variable. It is the variable y itself. So I am right. and my answer includes the further detail that we must have complex differentiability of f at z*. – jg mr chapb Mar 05 '22 at 00:56