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I am trying to prove that $ \overline {\cos(z)} = \cos(\bar z)$

I have tried this myself but I can't see where I have gone wrong. I'd like to continue to try to prove it using the method I will specify and not the double angle formulas.

I'm trying to use the method of using $cos(z)=(e^{iz}+e^{-iz})/2$ and then that $z=x+iy$ however I can't seem to complete the proof.

I'm starting with $ \overline {cos(x+iy)}$= $ \overline {e^{-y}(cos(x)+isin(y))+e^y(cos(x)-isin(x))}$

however I cannot arrive at the correct answer.

Thanks in advance

mathsguy123
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    Easiest would be to use the series expansion of $\cos$ and the fact that the conjugation-map is continuous on $\mathbb C$. – Lukas Betz Dec 02 '16 at 21:58

3 Answers3

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Using the definition

$$ \cos z = \frac{e^{iz} + e^{-iz}}{2}, $$

for $z = x+iy$ we have

$$ \overline{\cos z} = \overline{\frac{e^{i(x+iy)} + e^{-i(x+iy)}}{2}} = \frac{\overline{e^{ix}e^{-y}} + \overline{e^{-ix}e^{y}}}{2} = \frac{e^{-ix}e^{-y} + e^{ix}e^{y}}{2}, $$

and

$$ \cos(\overline {z}) = \frac{e^{i(x-iy)} + e^{-i(x-iy)}}{2} = \frac{e^{ix}e^{y} + e^{-ix}e^{-y}}{2} = \overline{\cos z} $$

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    Is this sufficient proof? Do I not have to expand the e^ix etc..? – mathsguy123 Dec 02 '16 at 22:07
  • no you do not need to expand anything. since $e^{ix} = \cos x + i\sin x$, it follows that $e^{-ix} = \cos(-x) + i\sin(-x) = \cos x - i \sin x = \overline{e^{ix}}$. and $e^y$ is just a real number, so it is equal to its own conjugate. –  Dec 02 '16 at 22:09
  • That makes complete sense. Thank you. I assume the argument is similar for the sin(z) conjugates – mathsguy123 Dec 02 '16 at 22:12
  • Clear and concise. +1 – Mark Viola Dec 02 '16 at 23:40
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You're better off using the addition formula for cosines.

$$\cos(\alpha + i\beta) = \cos\alpha \cos i\beta - \sin \alpha \sin i\beta$$

Now recall that

$$\cos i\beta = \cosh \beta$$

and

$$\sin i\beta = i\sinh\beta$$

and so

$$\cos(\alpha + i\beta) = \cos\alpha \cosh\beta - i\sin \alpha \sinh\beta.$$

Conjugate this and you'll get your result.

Edward Evans
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With heavier machinery: Let $g(z) = \overline{\cos \bar z}$. Then $g$ is entire (see for example this, asked and answered many times). For $z$ real, $g(z) = \cos z$, so by the identity principle, $g(z) = \cos z$ for all $z \in \mathbb{C}$.

The same argument shows that if $f$ is entire and real valued on $\mathbb{R}$, then $f(\bar z) = \overline{f(z)}$

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mrf
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