# Math Help - Complex function

1. ## Complex function

Show that

$f(z) = \left\{\begin{array}{ll} \cos (z^{1/2}),& z\neq 0\\1,&z=0\end{array}$

is single-valued and analytic in $C$. Calculate $f'(0)$.

I don't even know where to begin with this one.

2. I don't understand why this is listed as a hybrid function, since $\cos{0} = 1$ anyway... So I'm going to assume we can just write the function as $f(z) = \cos{(z^{\frac{1}{2}})}$ for all $z$.

$f(z) = \cos{(z^{\frac{1}{2}})}$

$= \cos{[(r\,e^{i\theta})^{\frac{1}{2}}]}$

$= \cos{(r^{\frac{1}{2}}e^{\frac{\theta}{2}i})}$

$= \cos{\left[r^{\frac{1}{2}}\left(\cos{\frac{\theta}{2}} + i\sin{\frac{\theta}{2}}\right)\right]}$

$= \cos{\left(r^{\frac{1}{2}}\cos{\frac{\theta}{2}} + i\,r^{\frac{1}{2}}\sin{\frac{\theta}{2}}\right)}$

$= \cos{\left(r^{\frac{1}{2}}\cos{\frac{\theta}{2}}\r ight)}\cosh{\left(r^{\frac{1}{2}}\sin{\frac{\theta }{2}}\right)} - i\sin{\left(r^{\frac{1}{2}}\cos{\frac{\theta}{2}}\ right)}\sinh{\left(r^{\frac{1}{2}}\sin{\frac{\thet a}{2}}\right)}$

Now convert it back to Cartesians and use the Cauchy Riemann Equations.

i.e. For $f(z) = u(x, y) + i\,v(x, y)$, the function is analytic if $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$

3. 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 $z^{1/2} = e^{\frac{1}{2}log(z)}$ when $z \neq 0$.

4. 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 $\mathbb{C} - \{ 0 \}$?

Also, it is not defined at 0 since, as you said, $\sqrt{z}$ is defined as $e^{\frac{1}{2} log(z)}$, and $logz$ has a branch point at 0.

I don't think it is 1:1, though -- $f(4 \pi ^2) = f(0)$. Regarding the calculation of the derivative - you should probably do it by definition.

5. 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.

6. Actually, the Cauchy-Riemann Equations in Polar Coordinates is more useful in this case.

You need to show that $\frac{\partial u}{\partial r} = \frac{1}{r}\,\frac{\partial v}{\partial \theta}$ and $\frac{\partial v}{\partial r} = -\frac{1}{r}\,\frac{\partial u}{\partial \theta}$.

$f(z) = \cos{\left(r^{\frac{1}{2}}\cos{\frac{\theta}{2}}\r ight)}\cosh{\left(r^{\frac{1}{2}}\sin{\frac{\theta }{2}}\right)} - i\sin{\left(r^{\frac{1}{2}}\cos{\frac{\theta}{2}}\ right)}\sinh{\left(r^{\frac{1}{2}}\sin{\frac{\thet a}{2}}\right)}$

$u = \cos{\left(r^{\frac{1}{2}}\cos{\frac{\theta}{2}}\r ight)}\cosh{\left(r^{\frac{1}{2}}\sin{\frac{\theta }{2}}\right)}$ and $v = -\sin{\left(r^{\frac{1}{2}}\cos{\frac{\theta}{2}}\r ight)}\sinh{\left(r^{\frac{1}{2}}\sin{\frac{\theta }{2}}\right)}$.

Can you go from here?

7. Remembering the Taylor expansion of $cos(*)$ You can write...

$\displaystyle \cos \sqrt{z} = \sum_{n=0}^{\infty} (-1)^{n} \frac{z^{n}}{(2n)!}$ (1)

... so that $\cos \sqrt{z}$ is analytic on the whole z plane. Its derivative can be computed deriving (1) 'term by term' and is $f' (0)= - \frac{1}{2}$...

Kind regards

$\chi$ $\sigma$

8. 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 $f'(0)=-1/2$

9. To be continuous at that point, you just need to show that

$\lim_{z \to 0 + 0i}f(z) = 1$.