TITLE: Does Euler's formula give $e^{-ix}=\cos(x) -i\sin(x)$?
QUESTION [7 upvotes]: Does Eulers formula give $$e^{-ix}=\cos(x) -i\sin(x)$$
I know that $$e^{ix}=\cos(x)+i\sin(x)$$ But how does it work when we have a $-$ in front

REPLY [3 votes]: From the property that for any $z_1,z_2\in\mathbb{C}$, $$e^{z_1}\cdot e^{z_2}=e^{z_1+z_2}$$ we have that $$e^{ix}\cdot e^{-ix}=e^{ix-ix}=e^0=1$$ so $$e^{-ix}=\frac1{e^{ix}}=\frac{\overline{e^{ix}}}{e^{ix}\cdot\overline{e^{ix}}}=\frac{\overline{e^{ix}}}{|e^{ix}|^2}=\overline{e^{ix}}$$ using that $\cos^2x+\sin^2x=1$.