Complex analysis T/F question

Hey,

I'm just studying for my exam next week and I've found a few of these true or false problems I'm not 100% with,

http://img856.imageshack.us/img856/7685/60423858.jpg

For the first one, I would say its entire since both functions are entire, and also because the product of two entire functions is entire, so e^xcosy is entire.
So f(z) is entire, True.

Is that correct? I just get paranoid with these T/F questions because in some of them are tricky things you can miss and ruin the question.

For the second one I would think that the sequence would converge because the sum diverges if the sequence diverges, so the sum cant converge if the sequence diverges. So True.

For the third one, I not sure at all about the convergence of Laurent series, where would I start with this one?

And for the last, I believe its false because say if it was e^z/cosz then its not entire cause of the singularity at 0, but then i was thinking it would be entire so long as the function doesn't ever = 0 so I'm in a bit of a pickle. In the exam I would say false, because it doesn't specify any two entire functions.

Do you guys think what I have thought is ok?

Thanks in advanced

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Originally Posted by Daniiel

For the first one, I would say its entire since both functions are entire, and also because the product of two entire functions is entire, so e^xcosy is entire. So f(z) is entire, True.

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The statement is false. The Cauchy-Riemann equations are not satisfied.

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For the second one I would think that the sequence would converge because the sum diverges if the sequence diverges, so the sum cant converge if the sequence diverges. So True.

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The reason is: if $\displaystyle \sum_{n=0}^{\infty}z_n$ converges , then $\displaystyle \lim_{n\to \infty}z_n=0$ (by a well known theorem) . So , $\displaystyle z_n$ converges .

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For the third one, I not sure at all about the convergence of Laurent series, where would I start with this one?

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Choose the series $\displaystyle \ldots+\frac{4^3}{(z-2)^3}+\frac{4^2}{(z-2)^2}+\frac{4}{z-2}+1$

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And for the last, I believe its false because say if it was e^z/cosz then its not entire cause of the singularity at 0, but then i was thinking it would be entire so long as the function doesn't ever = 0 so I'm in a bit of a pickle. In the exam I would say false, because it doesn't specify any two entire functions.

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Yes, it is false. Another counterexample: $\displaystyle 1$ and $\displaystyle z$ are entire functions and $\displaystyle 1/z$ is not analytic at $\displaystyle z=0$ , so $\displaystyle 1/z$ is not entire.

Thanks Fernando,

So that series you choose is 4^n(z-2)^-n and it satisfies the question

but the question gave a (z-2)^n, so would to answer it would i let a = 4^-n,

then I get the |z-2|<4 using the ratio test

So would that mean the answer is false?

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Originally Posted by Daniiel

So would that mean the answer is false?

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The series I provided:

$\displaystyle \sum_{n=-\infty}^{\infty}a_n(z-2)^n=\ldots+\frac{4^3}{(z-2)^3}+\frac{4^2}{(z-2)^2}+\frac{4}{z-2}+1+\sum_{n=1}^{\infty}0(z-2)^n$

is convergent if and only if $\displaystyle |z-2|>4$ . So, the answer is: true .

Oh okay,

I was just a little worried, so its alright to split apart the series into two for the case of proving the question true, and saying that the coefficient of the first series work to make the convergence we want and to ensure that we can set the coefficient of the positive degree series to be zero.

So in other words its perfectly fine to say this Laurent series:

$\displaystyle $\sum\limits_{n=-\infty }^{0}{\frac{{{(z-2)}^{n}}}{{{4}^{n}}}=1+\frac{4}{z-2}+\frac{16}{{{(z-2)}^{2}}}}+....$$

Satisfies the question

So that is still a laurent series even though it goes from 0 to - infinity because it doesnt converge within a circle and is not analytic as 2.

Sorry I just want to double check, for some reason laurent series is the only part of the course I havent wrapped my head around properly

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Originally Posted by Daniiel

So in other words its perfectly fine to say this Laurent series: $\displaystyle $\sum\limits_{n=-\infty }^{0}{\frac{{{(z-2)}^{n}}}{{{4}^{n}}}=1+\frac{4}{z-2}+\frac{16}{{{(z-2)}^{2}}}}+....$$ Satisfies the question

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More than fine, extremely fine. :)

wooo,

Thanks for your help Fernando =)