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I have to prove that the function $f:\mathbb{C}\to\mathbb{C}$ defined by $f(z)=\sin(\overline{z})$ is not holomorphic at any point of $\mathbb{C}$.

Now, I want to show that $f$ does not satisfy the Cauchy-Riemann equations, but first I must write $f(z)$ in the form $u(z)+iv(z)$.

How can I find $u(z)$ and $v(z)$?

Thanks.

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    $\begingroup$ $\sin (a+b) = \sin a \cos b + \sin b \cos a$. Let $a = x$, $b = -iy$. But it's probably better to use the Wirtinger derivatives and compute $\frac{\partial}{\partial\overline{z}}\sin \overline{z}$. $\endgroup$ – Daniel Fischer Sep 18 '13 at 11:20
  • $\begingroup$ $\sin(x-iy)=\sin(x)\cos(-iy)+\sin(-iy)\cos(x)=\sin(x)\cos(iy)-\sin(iy)\cos(x)$... What can we do now? Btw, I don't know what it is a Wirtinger derivative. $\endgroup$ – Talexius Sep 18 '13 at 11:29
  • $\begingroup$ $\cos (iy) = \cosh y$, and $\sin (iy) = i\sinh y$. If you haven't yet learned about the Wirtinger derivatives, ignore that for the moment, the real form of the CR equations is simple enough here. $\endgroup$ – Daniel Fischer Sep 18 '13 at 11:34
  • $\begingroup$ ... even if the C-R equations are satisfied at an isolated point, still the function is not holomorphic there ... $\endgroup$ – GEdgar Sep 18 '13 at 12:19
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By definition

$$\sin\overline z:=\frac{e^{\overline z}-e^{-\overline z}}{2i}=\frac{e^xe^{-iy}-e^{-x}e^{iy}}{2i}=\frac1{2i}\left(e^x(\cos y-i\sin y)-e^{-x}(\cos y+i\sin y)\right)=$$

$$=\frac1{2i}\left[\cos y(e^x-e^{-x})-i\sin y(e^x+e^{-x})\right]=\frac12\left[-\sin y(e^x+e^{-x})-\cos y\left(e^x-e^{-x}\right)i\right]=$$

$$=-\sin y\cosh x-i\cos y\sinh x=u(x,y)+iv(x,y)$$

Well, now check the Riemann-Cauchy equations, for example

$$u_x=\sin y\sinh x\;,\;\;v_y=\sin y\sinh x\;\ldots$$

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