Let $X$ and $Y$ be independent identical distributions with characteristic function $\varphi$. Express the characteristic functions of $-X$ and $X-Y$ with $\varphi$
My attempt:
I know that $\varphi_X(t)=\mathbb{E}(e^{itX})$ and that $\mathbb{E}(X_1X_2)=\mathbb{E}(X_1)\mathbb{E}(X_2)$ if $X_1, X_2$ are independent.
Knowing that, I can compute $$\varphi_{X-Y}(t) = \mathbb{E}(e^{it(X-Y)})=\mathbb{E}(e^{itX}e^{-itY})\overset{?}{=}\mathbb{E}(e^{itX})\mathbb{E}(e^{-itY})=\varphi_X(t)\varphi_{-Y}(t)$$
I am not 100% sure about the third step (marked with a "?"). I know that measurable functions of independent variables are also independent and that $exp$ is borel-measurable, but maybe the $i$ makes a difference?
Moving on to $\varphi_{-X}$ $$\varphi_{-X}(t)=\mathbb{E}(e^{it(-X)})=\mathbb{E}(e^{i(-t)X})=\varphi_{X}(-t)$$ Same thing with $\varphi_{-Y}$, so I can simplify $\varphi_{X-Y}$ even further so that $\varphi_{X-Y}(t)=\varphi_X(t)\varphi_{Y}(-t)$
Now I've expressed both characteristic functions with $\varphi_X$ and $\varphi_Y$
Did I do it correctly?