1. ## Complex Analysis questions

I have 3 questions which I have some idea how to solve it, but need confirmations on.

1) Sketch the set of continuity of the function given by the conditions:

f(z) = z if |z| < or = 1
|z|^3 if |z|>1

Is this set open? Is it connected?

So, the sketch should have a circle with radius 1 right? And it is NOT connected since f(z) does not equal at z=1 (or as lim --->1)? And it is open?

2) Let f(z) = z/(z^2 + 1) + e^(1/z)

Find and classify all isolated singularities, and compute the residues at each singularity.

Left polynomial part has removable singularities of +,-i and right has removable singularities at 0, and order of pole is all 1.... is that correct? Just wondering what would be the appropriate justification for it... How should I explain why those are the isolated singularities?

3) What is the order of zero of z^3(1-cos^2(z)) / (e^z-1) at 0

I'm actually not so sure how to do this one. I don't think I should start deriving, so do I just go 1/f and find the order of poles? Would the order of pole at 0 be 4? since we have 3 for z^3 and one for (1-cos^2(z))?

Any help on those questions would be very appreciated. ^^

2. Originally Posted by hohoho00
2) Let f(z) = z/(z^2 + 1) + e^(1/z)

Find and classify all isolated singularities, and compute the residues at each singularity.

Left polynomial part has removable singularities of +,-i and right has removable singularities at 0, and order of pole is all 1.... is that correct? Just wondering what would be the appropriate justification for it... How should I explain why those are the isolated singularities?
Do you know what a removable singularity is ? It's a singularity for which a function is not defined at first sight, but thereafter, you can find that it is extendable by continuity !
$e^{1/z}=\sum_{n=0}^\infty \frac{\left(\frac 1z\right)^n}{n!}=\sum_{n=0}^\infty \frac{1}{z^n n!}$
This is an essential singularity.

i and -i are poles of order 1.

The singularities are the points where the function is not holomorphic (undefined)

3. Originally Posted by hohoho00
3) What is the order of z^3(1-cos^2(z)) / (e^z-1) at 0

I'm actually not so sure how to do this one. I don't think I should start deriving, so do I just go 1/f and find the order of poles? Would the order of pole at 0 be 4? since we have 3 for z^3 and one for (1-cos^2(z))?

Any help on those questions would be very appreciated. ^^
0 is a pole because of $e^z-1$, not because of $z^3$

Get the Taylor series for $e^z-1$ : $1+z+z^2+\dots -1=z+z^2+\dots =z \left(1+\sum_{n=1}^\infty z^n\right)$

Now the order of a pole $z_0$ can be found this way : it's the number n such that $\lim_{z \to z_0} (z-z_0)^n f(z)$ has a nonzero limit.
So here, you can see that it is already simplified with the $z^3$ and hence gives a limit.
So it's a removable singularity.

(I hope I'm not mistaking for this one :s)

4. Originally Posted by Moo
Do you know what a removable singularity is ?
I mistook removable singularity as isolated singularity. My bad.

For 3), question asked for order of zero, not order of poles. For some reason, "zero" part was missing. Sorry about that. So I don't think I should be working with e^z-1

5. The numerator has a zero of order 5 and the denominator, a zero of order 1 and thus the function has a removable singularity at the origin and setting the function equal to zero there produces an analytic function with a zero of order 4 at the origin:

\begin{aligned}\cos^2(z)&=\left(1-\frac{z^2}{2}+\frac{z^4}{4!}+\cdots\right)\left(1-\frac{z^2}{2}+\frac{z^4}{4!}+\cdots\right) \\
&=1-z^2+3 \frac{z^4}{4!}+\cdots\end{aligned}

Thus:

$z^3\left(1-\cos^2(z)\right)=-z^5+\cdots$

and:

$e^z-1=z+z^2/2+\cdots$

Then the order of the zero is 5-1=4.