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I am studying for my (introductory) complex analysis final exam tomorrow. I am practicing an old final exam, which unfortunately has no answer key. Here is a link:

http://www.math.ubc.ca/Ugrad/pastExams/Math_300_December_2008.pdf

I just want to know if I am on the right track for Problem 1, which is collection of 10 (kind of tricky!) true/false questions. Here are my attempts:

1) If $f(z)$ satisfies Cauchy-Riemann equations at $z_0$, then $f(z)$ is differentiable at $z_0$.

False. For differentiability, one also needs continuity of partial derivatives.

2) If $f(z)$ has a pole at $z_0$, then $\lim_{z\to z_0}|f(z)|=\infty$.

True. Though I am not sure if I know how to prove it rigorously.

3) If $f(z)$ is analytic in a domain $D$ containing a simple closed contour $\Gamma$, then $\int f(z)dz=0$.

False. For conclusion to be true in general, one needs simply-connected domain.

4) If the two power series $\sum_{k=0}^\infty a_k (z-z_0)^k$ and $\sum_{k=0}^\infty b_k (z-z_0)^k$ converge to the same function in the disk $\{|z-z_0|\}$, then $a_k=b_k$ for all $k$.

I think the answer is True for this one. I believe this follows from Taylor theorem. The coefficients $a_k$ and $b_k$ are determined from derivatives of $f$, so they must be equal. Is this correct?

5) There does not exist any function $f(z)$ which is analytic at the point $0$ and nonanalytic everywhere else.

I think the answer is False. I can't think of a counterexample. Maybe the function $|z|^2$ ?

6) The function $\text{Log}(z^2)$ is analytic for all values of $z$ except those on the negative real axis.

False. The function $\text{Log}(z^2)$ is analytic for all values of $z$ except those on upper half of imaginary axis.

7) Any entire function is the complex derivative of another entire function

True. This follows from Cauchy Integral Formula.

8) If $f(z)$ has an essential singularity at $z_0$, then Res$((z-z_0)f(z); z_0)=0$.

I am pretty sure this is false. For example, $f(z)=e^{1/z}$ would be a counter-example.

9) If $f(z)$ and $g(z)$ have a simple poles at $0$, then $(fg)(z)$ has a simple pole at $0$.

I think it is true. By the way, I am not sure if the notation above stands for product, or composition. The question doesn't indicate the usage. But I think in either case, the statement would be true. Any comments on this one?

10) If the disk of convergence of the Taylor series of a function $f(z)$ is $\{|z|=2\}$, then the disk of convergence for the Taylor series of $f(z^2)$ is $\{|z|=4\}$.

Frankly, I am completely struck at this one. I have not the slightest idea of relating radii of convergence for the functions $f(z)$ and $f(z^2)$.

Any help and feedback on any of the questions is much appreciated. I suspect I have made couple mistakes above, apart from the ones I am stuck on. Confirming one of the my answers as correct would be just as awesome! :)

Thanks.

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2 Answers

up vote 4 down vote accepted

For 2) Lets say that $z_0$ is a pole of order $n \geq 1$. Then there exists an analytic function $g(z)$ with $g(z_0)\neq 0$ so that $f(z)=\frac{g(z)}{(z-z_0)^n} $. From here it is easy.

For 5): The definition of analiticity is that the power series must converge in a neighborhood of $0$. But then $f(z)$ is analityc in that neighborhood.

Note that $f(z)=|z|^2$ is differentiable at $z=0$ but not analytic...

For 9) That is the product, and you can actually show that $fg(z)$ has a pole of order $2$.

For 10)

Let $f(z)=\sum_{n=0}^\infty a_nz^n \,.$ Then $f(z^2)=\sum_{n=0}^\infty a_nz^{2n}$. It is easy to prove that the first series converges for all $|z| \leq a$ if and only if the second series converges for all $|z| \leq \sqrt{a}$.

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These are great explanations. Thank you! By the way, I will come back and upvote both yours and Tony's answer once I pass 15 point reputation limit. I don't know which answer to select, since both are very helpful. :( –  Prism Jan 5 '13 at 21:46
    
There we go, got my 15 reputation! Upvoting answers is gonna be fun! :) I will choose your answer, since I particularly liked the explanation for 5). –  Prism Jan 5 '13 at 21:54
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I agree with most of your answers. Here are some comments on the others:

2) To prove it, one can forget about the analytic part, since analytic functions are bounded in a neighborhood of any point. By translating, you can also assume that the point is $0$, so the function is a finite linear combination of $z^{-n}$. From here, it should be pretty straightforward.

4) Yes.

9) This is false for either interpretation. Take $f(z)=g(z)=1/z$. The product is $1/z^2$, which has no simple pole. The composition is $z$, which also has no simple pole.

10) This seems like a bizarre question to me; if $f(z^2)$ converged on all $z$ of distance $\leq 4$ from the origin, wouldn't we expect $f(z)$ to converge on points of distance $\leq 16$? Here's maybe the intended meaning of the problem. It's a property of complex-analytic functions that the radius of convergence is the distance to the nearest singularity. Therefore, we know that the nearest singularity of $f(z)$ has distance $2$ from the origin, so the nearest singularity of $f(z^2)$ has distance $\sqrt{2}$ from the origin.

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Thanks, Tony! Ah, for 9) I totally overlooked the word "simple". Oh I see, so roughly speaking, f(z^2) will behave "less analytic" than $f(z)$ in general. So I guess 10) is False then. –  Prism Jan 5 '13 at 21:40
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