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Let $G(w)=\sin\sqrt w$, where $w=x+iy$.

My initial method was to use the square root of $w$, which is defined as

$(x^2 + y^2)^{\frac{1}{4}}[\cos(\frac{\theta}{2} + k\pi) + i\sin(\frac{\theta}{2} +k\pi)]$ , $k=0,1$

I then set $\theta ' = \frac{\theta}{2} + k\pi$ and I also set

$x'=(x^2 + y^2)^{\frac{1}{4}}\cos(\frac{\theta}{2} + k\pi)$

and

$y'= i(x^2 + y^2)^{\frac{1}{4}}\sin(\frac{\theta}{2} +k\pi)$.

Then I get that $\sin\sqrt w = \sin(x' + iy')$=

=$\sin x'\cos iy' + \cos x'\sin iy'$

=$\sin x'\cosh y' + i\cos x'\sinh y'$

$u= \sin x'\cosh y'$ $v= \cos x'\sinh y'$

so that we get

$\frac{\partial u}{\partial x} = \cos x'\cosh y'$

$\frac{\partial u}{\partial y} = \sin x'\sinh y'$

$\frac{\partial v}{\partial x} = -\sin x'\sinh y'$

$\frac{\partial v}{\partial x} = \cos x'\cosh y'$

Which would mean that $G(w)$ is analytic on the entire complex plane, but then the derivative of $\sin \sqrt w = \frac{\cos\sqrt w}{2\sqrt w}$, and so $w$ definitely can't be equal to $0$. I wanted to know if someone could point out where I went wrong. Also, if a better method could be provided, that would be very helpful.

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  • $\begingroup$ The definition of $\sqrt{\cdot}$ doesn't quite make sense---$k$ takes on two different values, giving two different functions. What is your definition of $\theta$? $\endgroup$ – Travis Mar 8 at 23:24
  • $\begingroup$ $\theta= \arctan\frac {y}{x} $ $\endgroup$ – K.M Mar 8 at 23:31
  • $\begingroup$ $\partial x'/\partial x \ne 1$ $\endgroup$ – eyeballfrog Mar 8 at 23:34
  • $\begingroup$ If I were to use the Cauchy-Riemann equations, would I have to separate the cases where $k=0$ and $k=1$? $\endgroup$ – K.M Mar 8 at 23:36
  • $\begingroup$ @eyeballfrog: this is probably a dumb question, but why does it need to equal $1$? $\endgroup$ – K.M Mar 8 at 23:39

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