- #1

buzzmath

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in the reflection principle if f(x) is pure imaginary then the conjugate of f(z)=-f(z*) where z* is the complex conjugate of z.

Any advice on where to start?

thanks

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- #1

buzzmath

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in the reflection principle if f(x) is pure imaginary then the conjugate of f(z)=-f(z*) where z* is the complex conjugate of z.

Any advice on where to start?

thanks

- #2

buzzmath

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sorry I posted in the wrong forum

- #3

Dick

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- #4

rbzima

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Dick

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? I have no idea what that is supposed to mean. The problem is, in fact, about extending the domain of definition of an analytic function. rbzima, aren't you supposed to be working on group theory?

- #6

Wiemster

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in the reflection principle if f(x) is pure imaginary then the conjugate of f(z)=-f(z*) where z* is the complex conjugate of z.

Any advice on where to start?

thanks

How is this a problem (I see no (implicit) question mark)?

- #7

Dick

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He's stating a variant of the Schwarz reflection principle. And trying to prove it.

- #8

rbzima

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? I have no idea what that is supposed to mean. The problem is, in fact, about extending the domain of definition of an analytic function. rbzima, aren't you supposed to be working on group theory?

LOL - This caught my attention, but I managed to finish my group theory stuff. Forgive me if I was wrong, but you don't really have anything to "solve" using what you stated. If f(x) is pure imaginary, it will exist on the imaginary axis, and it's conjugate will then exist on the imaginary axis as well. Maybe I'm just reading the question wrong.

- #9

Dick

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- #10

racland

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Find the Radius of Convergence of the following Power Series:

(a) ∑ as n goes from zero to infinity of Z^n!

(b) ∑ as N goes from zero to infinity of (n + 2^n)Z^n

For (a) I think the radius of convergence is 1 but I'm a bit unsure of that...

- #11

Dick

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Find the Radius of Convergence of the following Power Series:

(a) ∑ as n goes from zero to infinity of Z^n!

(b) ∑ as N goes from zero to infinity of (n + 2^n)Z^n

For (a) I think the radius of convergence is 1 but I'm a bit unsure of that...

You are clearly in the wrong thread.

- #12

buzzmath

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Thanks everyone I think I figured it out

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