Show that
$\displaystyle f(z) = \left\{\begin{array}{ll} \cos (z^{1/2}),& z\neq 0\\1,&z=0\end{array}$
is single-valued and analytic in $\displaystyle C$. Calculate $\displaystyle f'(0)$.
I don't even know where to begin with this one.
Show that
$\displaystyle f(z) = \left\{\begin{array}{ll} \cos (z^{1/2}),& z\neq 0\\1,&z=0\end{array}$
is single-valued and analytic in $\displaystyle C$. Calculate $\displaystyle f'(0)$.
I don't even know where to begin with this one.
I don't understand why this is listed as a hybrid function, since $\displaystyle \cos{0} = 1$ anyway... So I'm going to assume we can just write the function as $\displaystyle f(z) = \cos{(z^{\frac{1}{2}})}$ for all $\displaystyle z$.
$\displaystyle f(z) = \cos{(z^{\frac{1}{2}})}$
$\displaystyle = \cos{[(r\,e^{i\theta})^{\frac{1}{2}}]}$
$\displaystyle = \cos{(r^{\frac{1}{2}}e^{\frac{\theta}{2}i})}$
$\displaystyle = \cos{\left[r^{\frac{1}{2}}\left(\cos{\frac{\theta}{2}} + i\sin{\frac{\theta}{2}}\right)\right]}$
$\displaystyle = \cos{\left(r^{\frac{1}{2}}\cos{\frac{\theta}{2}} + i\,r^{\frac{1}{2}}\sin{\frac{\theta}{2}}\right)}$
$\displaystyle = \cos{\left(r^{\frac{1}{2}}\cos{\frac{\theta}{2}}\r ight)}\cosh{\left(r^{\frac{1}{2}}\sin{\frac{\theta }{2}}\right)} - i\sin{\left(r^{\frac{1}{2}}\cos{\frac{\theta}{2}}\ right)}\sinh{\left(r^{\frac{1}{2}}\sin{\frac{\thet a}{2}}\right)}$
Now convert it back to Cartesians and use the Cauchy Riemann Equations.
i.e. For $\displaystyle f(z) = u(x, y) + i\,v(x, y)$, the function is analytic if $\displaystyle \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\displaystyle \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$
The first order partial derivatives must be continuous as well in order for the function to be analytic, which I'm guessing is a part of the problem. The reason the function looks like that probably has something to do with $\displaystyle z^{1/2} = e^{\frac{1}{2}log(z)}$ when $\displaystyle z \neq 0$.
Well, we don't know exactly what you are allowed to use - can you use the fact that the cosine function is entire and square root is analytic in $\displaystyle \mathbb{C} - \{ 0 \}$?
Also, it is not defined at 0 since, as you said, $\displaystyle \sqrt{z}$ is defined as $\displaystyle e^{\frac{1}{2} log(z)}$, and $\displaystyle logz$ has a branch point at 0.
I don't think it is 1:1, though -- $\displaystyle f(4 \pi ^2) = f(0)$. Regarding the calculation of the derivative - you should probably do it by definition.
One-to-one might be the wrong wording here. You're supposed to show that it's not multivalued. You're allowed to use that cos is entire, but like I wrote previously, I think you need to show that f and the partial derivatives are continuous. Anyways, this is just an exercise, but I would really like to know how to solve it to further my knowledge.
Actually, the Cauchy-Riemann Equations in Polar Coordinates is more useful in this case.
You need to show that $\displaystyle \frac{\partial u}{\partial r} = \frac{1}{r}\,\frac{\partial v}{\partial \theta}$ and $\displaystyle \frac{\partial v}{\partial r} = -\frac{1}{r}\,\frac{\partial u}{\partial \theta}$.
So for your function...
$\displaystyle f(z) = \cos{\left(r^{\frac{1}{2}}\cos{\frac{\theta}{2}}\r ight)}\cosh{\left(r^{\frac{1}{2}}\sin{\frac{\theta }{2}}\right)} - i\sin{\left(r^{\frac{1}{2}}\cos{\frac{\theta}{2}}\ right)}\sinh{\left(r^{\frac{1}{2}}\sin{\frac{\thet a}{2}}\right)}$
$\displaystyle u = \cos{\left(r^{\frac{1}{2}}\cos{\frac{\theta}{2}}\r ight)}\cosh{\left(r^{\frac{1}{2}}\sin{\frac{\theta }{2}}\right)}$ and $\displaystyle v = -\sin{\left(r^{\frac{1}{2}}\cos{\frac{\theta}{2}}\r ight)}\sinh{\left(r^{\frac{1}{2}}\sin{\frac{\theta }{2}}\right)}$.
Can you go from here?
Remembering the Taylor expansion of $\displaystyle cos(*)$ You can write...
$\displaystyle \displaystyle \cos \sqrt{z} = \sum_{n=0}^{\infty} (-1)^{n} \frac{z^{n}}{(2n)!}$ (1)
... so that $\displaystyle \cos \sqrt{z}$ is analytic on the whole z plane. Its derivative can be computed deriving (1) 'term by term' and is $\displaystyle f' (0)= - \frac{1}{2}$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
The Cauchy-Riemann equations is only a necessary condition in order for the function to be analytic. In order to prove that the function is analytic you also need to show that the partial derivatives are continuous. I don't know how to get the partial derivatives for the function though since it's a split function. I would think that the derivative of the constant part is 0, but the answer is $\displaystyle f'(0)=-1/2$