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Suppose that f(z) = 0 if z is real. Show that f is identically zero.

Any help please?

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- Thread starter arnesmeets
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- #1

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Suppose that f(z) = 0 if z is real. Show that f is identically zero.

Any help please?

- #2

matt grime

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- #3

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[because f is not zero inside the region where f is analytic]

- #4

matt grime

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Can you think of anyway to extend f to be analytic on the region where it is zero?

- #5

HallsofIvy

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Or does the boundary of continuation arise so that the function does not have any limits on it?

- #7

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When n = 1, then every open set is a domain of holomorphy: we can define a holomorphic function which has zeros which accumulate everywhere on the boundary of the domain

But it doesn't give a counter example to the claim in the opening post.

- #8

matt grime

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eg, sin(1/z) is defined for z !=0 and has zeros which accumulate at z=0, but it doesn't go to zero as z->0.

- #10

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eg, sin(1/z) is defined for z !=0 and has zeros which accumulate at z=0, but it doesn't go to zero as z->0.

I had understood this by the time of my previous post, where I said that the function described by the Wikipedia isn't a counter example.

When writing post #6, I thought I had seen something that could be a counter example, but then I found the Wikipedia page, took a closer look, and noticed that I remembered it wrong, and then wrote the post #7.

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- #11

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Define [itex]g_y : \mathbb{R} \rightarrow \mathbb{C}[/itex] so that [itex]g_y(x) = f(x+iy)[/itex]. Cauchy's theorem plus continuity of [itex]f[/itex] at the boundary imply that

[tex]\int_{-a}^a (g_y(x)+g_y(-x))dx = 0[/tex]

(taking a symmetric rectangular contour with base arbitrarily close to the real line). The continuity of [itex]g_y[/itex] gives that

[tex]\frac{1}{2\epsilon} \int_{\epsilon}^{\epsilon} (g_y(x)+g_y(-x))dx \Rightarrow g_y(0)[/tex]

So [itex]f(iy)=0[/itex], and you can apply the identity theorem.

[tex]\int_{-a}^a (g_y(x)+g_y(-x))dx = 0[/tex]

(taking a symmetric rectangular contour with base arbitrarily close to the real line). The continuity of [itex]g_y[/itex] gives that

[tex]\frac{1}{2\epsilon} \int_{\epsilon}^{\epsilon} (g_y(x)+g_y(-x))dx \Rightarrow g_y(0)[/tex]

So [itex]f(iy)=0[/itex], and you can apply the identity theorem.

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