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- Thread starter shen07
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- #1

- Jan 17, 2013

- 1,667

For clarification you mean by \(\displaystyle u_g=\text{Re}(g) \) and \(\displaystyle v_g=\text{Im}(g)\) using that \(\displaystyle g(x,y) = u(x,y)+iv(x,y)\) , right?

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yeas rightFor clarification you mean by \(\displaystyle u_g=\text{Re}(g) \) and \(\displaystyle v_g=\text{Im}(g)\) using that \(\displaystyle g(x,y) = u(x,y)+iv(x,y)\) , right?

- Jan 17, 2013

- 1,667

I would suggest starting by

\(\displaystyle u_f = \frac{f(z)+\overline{f(z)}}{2}\)

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- #5

I would suggest starting by

\(\displaystyle u_f = \frac{f(z)+\overline{f(z)}}{2}\)

One more question what is \(\displaystyle \overline{f(\overline{z})}\) actually?? i dont quite understand this!

- Aug 30, 2012

- 1,353

Consider a simple example:One more question what is \(\displaystyle \overline{f(\overline{z})}\) actually?? i dont quite understand this!

[tex]f(z) = u(z) + i v(z)[/tex] with z = x + iy.

Then

[tex]f(z) = u(x + iy) + i v(x + iy)[/tex]

[tex]f( \overline{z} ) = u(x - iy) + i v(x - iy)[/tex]

[tex]\overline{f( \overline{z} ) } = u(x - iy) - i v(x - iy)[/tex]

Is this what you are looking for? Or something more conceptual?

-Dan

Last edited:

- Jan 29, 2012

- 661

$$f(z)=z^2+i \Rightarrow \overline{f\left({\bar{z}}\right)}=\overline{(\bar{z})^2+i}=\overline{ \overline{z^2}+i}=\overline{\overline{z^2}}+\bar{i}=z^2-i$$