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If i take the complex number $e^{i(3+2i)}$, it's conjugate is $e^{i(-3+2i)}$.

However, the conjugate of the function f, defined as $f(x+iy)=e^{i(x+iy)}$, is, according to my book: $\overline{f(x+iy)}=e^{i(x-iy)}$.

I can't understand this difference ...

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2 Answers 2

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I think the book meant that $$f(\overline{x+iy})=f(x-iy)=e^{i(x-iy)}$$ but otherwise the book is in fact wrong as you say.

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  • $\begingroup$ Are there functions for which $\overline{f(x+ij)}=f(\overline{x+ij})$ ? $\endgroup$
    – AleWolf
    May 5, 2019 at 10:55
  • $\begingroup$ Yes, the exponential function and logarithm functions both have such a property - $e^{\overline{z}}=\overline{e^z}$ and $\ln{(\overline{z})}=\overline{\ln{(z)}}$ $\endgroup$ May 5, 2019 at 10:57
  • $\begingroup$ Mhm and the sin and cos functions ? How can I understand why some functions have this property ? $\endgroup$
    – AleWolf
    May 5, 2019 at 10:59
  • $\begingroup$ @AleQuercia Note this occurs iff $\overline{f(\bar z)} = f(z)$, which if you consider the power series expansion of $f$ implies that, for each coefficient $a_k$ in the series, we must have $\bar a_k = a_k,$ i.e. each $a_k$ is real. Thus $f$ must take real values along the real line. It is easy to check this is both necessary and sufficient. Thus both the $\sin$ and $\cos$ functions will have this property, as will all polynomials, etc. $\endgroup$ May 6, 2019 at 0:28
  • $\begingroup$ @AleQuercia an obvious example where this breaks is $\sqrt{z}$ since such a function is not even single valued on the real line. $\endgroup$ May 6, 2019 at 0:30
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Well write it out: \begin{align} \overline{ e^{i(x+iy)} } &= \overline{e^{ix-y}} = e^{-y}\overline{e^{ix}} = e^{-y}\overline{(\cos x+ i \sin x)} = e^{-y}(\cos x-i\sin x)\\ &= e^{-y}(\cos(-x) + i \sin (-x)) = e^{-y}e^{-ix} = e^{i(-x+iy)}, \end{align} so the book likely has a typo.

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