# What is the domain of analyticity for $G(w) = \sin\sqrt w$?

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.

• 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 $\theta$? – Travis Mar 8 at 23:24
• $\theta= \arctan\frac {y}{x}$ – K.M Mar 8 at 23:31
• $\partial x'/\partial x \ne 1$ – eyeballfrog Mar 8 at 23:34
• If I were to use the Cauchy-Riemann equations, would I have to separate the cases where $k=0$ and $k=1$? – K.M Mar 8 at 23:36
• @eyeballfrog: this is probably a dumb question, but why does it need to equal $1$? – K.M Mar 8 at 23:39