complex numbers. help in proof that $e^{-it}=1/e^{it}$

Question

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

Answer 1

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 $e^{-it}$.

Answer 2

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 $e^{ix}e^{-ix}$

Answer 3

You have$$e^{it}\cdot e^{-it}=e^{it-it}=e^0=1$$and therefore$$e^{-it}=\frac1{e^{it}}.$$

Answer 4

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 $$ \frac{1}{e^{t\cdot i}}=e^{-t\cdot i} $$