Suppose $\mathbb F_q$ is the finite field of order $q$ and $\mathbb{F}_q^d$ is $d$-dimensional vector space over finite field $\mathbb{F}_q$. Given $f:\mathbb{F}_q^d\to \mathbb{C}$, define the Fourier transform $\hat{f}:\mathbb{F}_q^d\to \mathbb{C}$ by the formula $$\hat{f}(m)=q^{-d}\sum \limits_{x\in \mathbb{F}_q^d} e^{-\dfrac{2\pi i}{q} x\cdot m}$$ I was trying to prove Fourier inversion formula which states that $$f(x)=\sum \limits_{m\in \mathbb{F}_q^d}e^{\dfrac{2\pi i}{q} x\cdot m}\hat{f}(m)$$
In the RHS of the above equality I plugged in the value of $\hat{f}(m)$ in order to derive that it is actually $f(x)$. However, after some manipulations I ran into the following sum:
$$q^{-d}\sum \limits_{y\in \mathbb F_q^d} \left(\sum \limits_{m\in \mathbb F_q^d} e^{\dfrac{2\pi i}{q}(x-y)\cdot m} \right)f(y)$$ and this sum can be splitten into two sums where the first one corresponds to $y=x$ and the second over $y\neq x$.
The sum which corresponds to $y=x$ is equal to $q^{-d}f(x)|\mathbb F_q^d|=q^{-d}f(x)q^d=f(x).$
How to show that the second sum is equal to zero? I have tried but I failed.
Would be thankful to see the solution, please.
Remark: So we basically need to show that for $z\in \mathbb{F}_q^d, z\neq 0$ we have $$\sum \limits_{m\in \mathbb F_q^d} e^{\dfrac{2\pi i}{q} z\cdot m}=0.$$ Am I right?
EDIT: Here is the original version of the post.
Let $\mathbb{F}_q$ be a finite field of order $q$ and $\mathbb{F}_q^d$ be a $d$-dimensional vector space over $\mathbb{F}_q$, where $d\geq 2$.
Given $f:\mathbb{F}_q^d\to \mathbb{C}$ we define Fourier transform $\widehat{f}:\mathbb{F}_q^d\to \mathbb{C}$ by formula $$\widehat{f}(m)=q^{-d}\sum \limits_{x\in \mathbb{F}_q^d} \chi(-x\cdot m) f(x),$$ where $\chi$ is a non-trivial principal character on $\mathbb{F}_q$.
Lemma: With the notation above, $$f(x)=\sum \limits_{m\in \mathbb{F}_q^d}\chi(x\cdot m) \widehat{f}(m)$$ and $$\sum \limits_{m\in \mathbb{F}_q^d}|\widehat{f}(m)|^2=q^{-d}\sum \limits_{x\in \mathbb{F}_q^d}|f(x)|^2.$$ These identities are called Fourier inversion and Plancherel.