# Complex conjugate of a simple function

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 ...

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.
• Are there functions for which $\overline{f(x+ij)}=f(\overline{x+ij})$ ? May 5, 2019 at 10:55
• Yes, the exponential function and logarithm functions both have such a property - $e^{\overline{z}}=\overline{e^z}$ and $\ln{(\overline{z})}=\overline{\ln{(z)}}$ May 5, 2019 at 10:57
• @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. May 6, 2019 at 0:28
• @AleQuercia an obvious example where this breaks is $\sqrt{z}$ since such a function is not even single valued on the real line. May 6, 2019 at 0:30
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.