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Use the Cauchy-Riemann equations to show that the function $$g(z) = \sin (\bar z)$$ is not analytic at any point of $\mathbb{C}$.

Here's as far as I got -

$$\sin \left(\frac{\bar z}{1}.\frac{z}{z}\right) =\sin \left(\frac{|z|^2}{z}\right) =\sin \left(\frac{x^2 + y^2}{x+iy}\right)$$

I can't see how to separate the real and imaginary parts so that I can apply the Cauchy-Riemann equations.

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1  
And the complex $\sin(z)$ function is defined as..... –  N. S. Apr 18 '12 at 0:42
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You could have just straight-up written $z=x-iy$ if you wanted explicit real and imaginary parts. Are you using the definition $$\sin z = \frac{e^{iz}-e^{-iz}}{2i}~?$$ –  anon Apr 18 '12 at 0:43
    
I need a break, I cant believe the crap I am missing tonight :( –  Jim_CS Apr 18 '12 at 0:46
    
(That should be $\bar{z}$ in my comment, obviously.) No worries Jim. –  anon Apr 18 '12 at 0:50
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You can prove that: CR equations $\Longleftrightarrow \frac{\partial f}{\partial \bar{z}}=0$. –  Riemann Apr 18 '12 at 4:30

2 Answers 2

Writing $f(z) = \sin\bar{z} = \sin(x - iy) = \sin x \cosh y - i \cos x \sinh y$, we have

$f(z) = u(x,y) + i v(x,y)$

where

$u(x,y) = \sin x \cosh y $ and $v(x,y) = - \cos x \sinh y $

If the Cauchy - Riemann equations $u_x = v_y$ , $u_y = -v_x$ are to hold , it is easy to see that it will hold only at the points $z = \frac{\pi}{2} + n\pi$, $n \in\mathbb{Z}$.

Clearly, then, there is no neighborhood of any point throughout which $f$ is analytic. Also note that these are the isolated points. A function is analytic when Cauchy-Riemann equations hold in an open set. So, we may conclude that $\sin\bar{z}$ is nowhere analytic. Similarly we can show that $\cos\bar{z}$ is nowhere analytic.

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Note that $\sin(\overline z)=\overline{\sin(z)}$. Since $\sin$ is analytic, if $g$ is analytic on some open set, then so is the real-valued function $\sin+g$. But a nonconstant real-valued function cannot be analytic (e.g., use the CR-equations or the open mapping theorem).


More generally, if $f$ is a nonconstant analytic function*, then $g(z)=f(\overline z)$ is not analytic on any open set. It is true that $\overline g$ is analytic, so if $g$ were analytic then the real-valued functions $\mathrm{Re}(g)=\frac{1}{2}(g+\overline g)$ and $\mathrm{Im}(g)=\frac{1}{2i}(g-\overline g)$ would be analytic, hence constant, which means that $g$ (and hence $f$) would be constant.

Furthermore, the only points at which the Cauchy-Riemann equations can be satisfied for $g$, being the complex conjugate of the analytic function $\overline g$, are at the zeros of the derivative of $\overline g$, which are isolated.

*(I am assuming that the domain of $f$ is symmetric about the real axis for the arguments used here.)

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+1 for your answer. –  srijan Jun 4 '12 at 8:03
    
Thanks, srijan, I had up-arrowed your answer, too. –  Jonas Meyer Jun 4 '12 at 8:14

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