Is the Function Analytic and Where?

[squaremeplz]

Homework Statement

Determine if the function is analytic and sketch where it's analytic.

a) $$f(x) = \frac {e^z}{z^2 + 4}$$

b) $$f(z) = \frac {conj(z)}{|z|^2}$$

c) $$\sum_{n=0}^\infty \frac {e^z}{3^n} (2z-4)^n$$

Homework Equations

The Attempt at a Solution

a) $$e^z$$ is analytic everywhere so $$f(x) = \frac {e^z}{z^2 + 4}$$ is analytic everywhere except at

$$z^2 = (x + yi)^2 = -4$$

I tried separating the function into $$f(x,y) = u(x,y) + i*v(x,y)$$ but get a very complex polynomial when I try to get rid of the imaginary part in the denominator for example:

$$\frac {e^x (cos(y) + i*sin(y)}{(x+yi)^2 + 4} * \frac{(x-yi)^2 + 4}{(x-yi)^2 + 4}$$

does not work for me

b)

$$f(z) = \frac{conj(z)}{ |z|^2}$$

$$f(z) = \frac{1}{z}$$

$$\frac{1}{z} = \frac{1}{ x+yi}$$

$$\frac{x - yi} {x^2 + y ^2}$$

$$= \frac{x}{x^2 + y ^2} - \frac {yi}{x^2 + y ^2}$$

$$since \frac{du}{dx} u(x,y) = \frac{dv}{dy} u(x,y)$$

and $$\frac{dv}{dx} u(x,y) = - \frac{dv}{dx} u(x,y)$$

the function is analytic everywhere except at the origin.

d) After using the ratio test, the result I get is

$$|\frac{1}{3} (2z - 4) |$$

$$\frac {1}{3} |2z - 4|$$

so the function is analytic on the disk $$0 < |2z - 4| < 6$$ for all values n

 

[gabbagabbahey]

a) $$e^z$$ is analytic everywhere so $$f(x) = \frac {e^z}{z^2 + 4}$$ is analytic everywhere except at

$$z^2 = (x + yi)^2 = -4$$

Right, so $f(z)$ is analytic everywhere except at $z=\pm2i$

I tried separating the function into $$f(x,y) = u(x,y) + i*v(x,y)$$ but get a very complex polynomial when I try to get rid of the imaginary part in the denominator for example:

$$\frac {e^x (cos(y) + i*sin(y)}{(x+yi)^2 + 4} * \frac{(x-yi)^2 + 4}{(x-yi)^2 + 4}$$

does not work for me

I'm not sure why you'd want to do it this way, since you've already found the answer using a much easier method, but if you want to be a masochist about it, this method should work fine as well.

$$\left[(x+yi)^2 + 4\right]\left[(x-yi)^2 + 4\right]=(x+iy)^2(x-iy)^2+4(x+iy)^2+4(x-iy)^2+16=(x^2+y^2)^2+8(x^2-y^2)+16$$

and so on...

$$since \frac{du}{dx} u(x,y) = \frac{dv}{dy} u(x,y)$$

and $$\frac{dv}{dx} u(x,y) = - \frac{dv}{dx} u(x,y)$$

the function is analytic everywhere except at the origin.

You mean, "since $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$ except at the origin, where $u$, $v$ and their partial derivatives do not exist, the function is analytic everywhere except at the origin", right?

d) After using the ratio test, the result I get is

$$|\frac{1}{3} (2z - 4) |$$

$$\frac {1}{3} |2z - 4|$$

so the function is analytic on the disk $$0 < |2z - 4| < 6$$ for all values n

Where is the '6' coming from?