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Use the Cauchy-Riemann equations to show that the function $f(z) = \exp(\bar{z})$

Here's my attempt:

we have $e^{\bar{z}} = e^{x-iy} = e^{x}e^{-iy} = e^x(\cos(y)-i\sin(y))$

we can then define:

  • $u(x,y)= e^x\cos(y)$
  • $v(x,y)= -e^x\sin(y)$

Then we can easily see that the Cauchy-Riemann equations aren't satisfied.

What I find weird is that I am asked to show that $f(z)$ is not defined everywhere whereas it's defined almost nowhere.

Is this the right way to go about this problem?

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    $\begingroup$ $f(z)$ is defined everywhere! $\endgroup$ – Nosrati Oct 2 '17 at 18:40
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First, we want to find the partial derivative of $u(x, y)$ and $v(x, y)$ with respect to both $x$ and $y$.

$$u_x = e^x \cos(y)$$ $$u_y = -e^x \sin(y)$$ $$v_x = -e^x \sin(y)$$ $$v_y = -e^x \cos(y)$$

The Cauchy-Riemann equations are satisfied when:

$$v_x = v_y$$ $$v_y = -v_x$$

As you pointed out, these equations are clearly not satisfied. A function $f(z)$ can only be analytic at a point if the Cauchy-Riemann equations are satisfied at that point. If they are not, then the function is not analytic there. Thus, because there are points where $f(z)$ is not analytic, it is not analytic everywhere. Hope that answers your question.

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  • $\begingroup$ (it meant everywhere non-analytic instead of non-analytic somewhere) $\endgroup$ – reuns Oct 2 '17 at 18:45
  • $\begingroup$ Well, if you instead define z conjugate = x - iy, and then take the partial derivatives with respect to that, you get $u_x = 1$ and $v_y = -1$, which is true nowhere. $\endgroup$ – user484876 Oct 2 '17 at 18:53

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