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?