# Two holomorphic functions

#### StefanM

##### New member
_z=conjugate
p_g=partial diff
_p_g=conjugate pde
p_g_x=partial diff with respect to x
Suppose that f : D → C . Let g(z) : D → C be defined by
g(z) = f(_z). Calculate_p_g and p_g where _p_g=(p_g_x+i(p_g_y))/2, p_g=(p_g_x+i(p_g_y))/2;
Conclude that f is holomorphic on D if
and only if p_g = 0 on D.
I've calculated the pde and observed that I get different signs that supposed to for a regular exercise.Also there is a property that states that f is holomorphic if _p_g=0.....given the nature of the function g(z) by intuition am assuming that in this case is has to be the other way around.
Any help will be great Thank you.

#### ThePerfectHacker

##### Well-known member
_z=conjugate
p_g=partial diff
_p_g=conjugate pde
p_g_x=partial diff with respect to x
Suppose that f : D → C . Let g(z) : D → C be defined by
g(z) = f(_z). Calculate_p_g and p_g where _p_g=(p_g_x+i(p_g_y))/2, p_g=(p_g_x+i(p_g_y))/2;
Conclude that f is holomorphic on D if
and only if p_g = 0 on D.
I've calculated the pde and observed that I get different signs that supposed to for a regular exercise.Also there is a property that states that f is holomorphic if _p_g=0.....given the nature of the function g(z) by intuition am assuming that in this case is has to be the other way around.
Any help will be great Thank you.
The way you wrote it is a mess. I have no idea what you are asking.
Maybe you are asking the following.

Write $f(x+iy) = u(x,y) + iv(x,y)$. Then we define $f_x = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}$ and $f_y = \frac{\partial u}{\partial y} + i\frac{\partial v}{\partial y}$.
We also define $f_z = \frac{1}{2} \left( f_x - i f_y \right)$ and $f_{\bar z} = \frac{1}{2} \left( f_x + i f_y \right)$.

If $f$ is holomorphic on $D$ then by the substituting Cauchy-Riemann equations we get that $f_{\bar z} = 0$ (and conversely also).
All you have to do is write everything down in the definition and then substitute CR-equations to get it equal to 0.

#### hmmm16

##### Member
I think the question should read as follows:

Suppose that $f:\mathbb{D}\rightarrow\mathbb{C}$ is of class $\mathcal{C}^1$. Let $g(z):\mathbb{D}\righatrrow\mathbb{C}$ be defined by $g(z)=f(\bar{z})$. Calculate $\partial g$ and $\bar{\partial} g$. Conclude that f is holomorphic on $\mathbb{D}$ iff $\partial g = 0$ on $\mathbbh{D}$

where if $z=x+iy$ and : $$\bar{\partial} = \frac{1}{2}\left ( \frac{\partial}{\partial x}+i*\frac{\partial}{\partial y} \right )$$

$$\partial = \frac{1}{2}\left ( \frac{\partial}{\partial x}-i*\frac{\partial}{\partial y} \right )$$

So that we have that f is holomorphic iff $\bar{\partial} f = 0$

To solve we write $f(z)=u(x,y)+iv(x,y)$ then $g(z)=u(x,-y)+iv(x,-y)$ and then:

$\partial g=\frac{1}{2}(g_x-ig_y)=\frac{1}{2} ((u_x+iv_x++i(u_y+iv_y))(x,-y)=\bar{\partial}f(\bar{z})$

So $\partial g=\bar{\partial} f(\bar{z})$ and so as f is holomorphic iff $\bar{\partial}f=\partial g = 0$ on $\mathbb{D}$

I think this is what is being asked for and the answer

(I don't know why it won't save my edits on the first few lines I will try again latter sorry)

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