complex numbers. help in proof that $e^{-it}=1/e^{it}$
I need to show that $e^{-it}=\frac{1}{e^{it}}$. but I don't understand what needs to be proven, it seems trivial to me. If anyone could help me. Is the claim true even if t is not real? Thank you
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$\begingroup$You have$$e^{it}\cdot e^{-it}=e^{it-it}=e^0=1$$and therefore$$e^{-it}=\frac1{e^{it}}.$$
$\endgroup$ 2 $\begingroup$From Euler's formula,$$e^{-it} = \cos(t)+i\sin(-t) = \cos(t)-i\sin(t).$$Also,$$e^{it} = \cos(t)+i\sin(t).$$Now, compute the product$$e^{it} \cdot e^{-it}.$$You find that the answer is $1$. This shows that $e^{it}$ is the reciprocal of $e^{-it}$.
$\endgroup$ $\begingroup$Using Eulers formula,
$$\cos(x) + i\sin(x) = e^{ix}$$
$$\cos(x) - i\sin(x) = e^{-ix}$$
Multiply the two equations and you’re done (work out the left hand side containing trigonometric quantities which should turn out to be $1$, while the right hand side leave it as $e^{ix}e^{-ix}$
$\endgroup$ $\begingroup$The inverse of the complex number $e^{t\cdot i}$ is, by definition, another complex number $e^{u+v\cdot i}$ such that $$ e^{t\cdot i}\cdot e^{u+v\cdot i}=1 $$On the other hand we know that$$ e^{t\cdot i}\cdot e^{u+v\cdot i}=e^{u+(t+v)\cdot i} \quad \mbox{ and } \quad 1=e^{0+0\cdot i} $$Equating the right sides of these two equations we have$$ e^{u+(t+v)\cdot i}=e^{0+0\cdot i}\implies u=0 \quad \mbox{ and } \quad t+v=0 $$Consequently we have $ v = -t $ and therefore $$ \frac{1}{e^{t\cdot i}}=e^{-t\cdot i} $$
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