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I have two confluent hypergeometric functions $$A=M(\alpha, 1; e^{i\omega{t}})$$ and $$B=M(\beta, 1; e^{i\omega{t}})$$ Where $$M(a, b, z)=\sum_{k=0}^{\infty}\frac{(a)_{k}}{(b)_{k}k!}z^{k}$$ It seems obvious to me that $$\bar{A}B=A\bar{B}$$ Can someone help me prove this using cauchy formula or something?

Kiryl Pesotski
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  • If I was doing it, I would take A, B apart into sine/cos odd/even parts term by term using Euler's expansion on individual terms. Then play with the results in real terms. An alternative might be to take derivatives and use Cauchy derivative criteria for analytic functions. Bear in mind that I am not sure your hypothesis is true. – rrogers Apr 28 '17 at 13:48

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Counterexample: Let $\alpha = 0$
Then $M(0,1,x)=A=\bar{A}=1 $ but then if $\beta =1 $
Then $M(1,1,e^{i\omega t})=B \neq \bar{B} $

rrogers
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