When is $f$ a normal operator? Given that $f$ is defined as $f(\mathbf{x})=\mathbf{x}+c \mathbf{u}(\mathbf{v},\mathbf{x})$ being the notation $(,)$ the inner hermitian product between two vectors, $c\in \mathbb{C}$ a constant, and $(\mathbf{u},\mathbf{v})\neq 0$.
I successfully found the expression of $f^* (\mathbf{x})$, which is $f^* (\mathbf{x})=\mathbf{x}+\overline{c}\mathbf{v}(\mathbf{u},\mathbf{x})$ so with this I should be able to solve the question since normal operators satisfy that $ff^* =f^*f$ so then by doing the multiplication I get: $$ff^* =(\mathbf{x},\mathbf{x})+\mathbf{x}\overline{c}\mathbf{v}(\mathbf{u},\mathbf{x})+\mathbf{x}c\mathbf{u}(\mathbf{v},\mathbf{x})+c\overline{c}(\mathbf{u},\mathbf{v})(\mathbf{v},\mathbf{x})(\mathbf{u},\mathbf{x})$$ But how am I supposed to even operate this? It seems like something out my touch
EDIT: Example of a problem that uses $ff^*$ as a product of $f$ and $f^*$: Under what conditions of the self-adjoint operators $a$ and $b$ is $f$ a normal operator? Given that $f=a+ib$
Solution from the book: To prove $f$ is normal, we can use the definition $ff^*=f^*f$ so we have $$ff^*=(a+ib)(a-ib)=(a-ib)(a+ib)=f^* f$$ $$a^2+b^2+i(ba+ab)=a^2+b^2+i(ab+ba)$$ which concludes that $ab=ba$ for $f$ to be normal
First of all, note that this is the identity operator in the case that $c = 0$, which means that $f$ is normal.
For the remaining cases, one thing that you can do to make the problem easier is note that $f$ is normal if and only if $g = \frac 1c(f - \operatorname{id})$ is normal. That is, it suffices to determine when the operator $$ g(\mathbf x) = (\mathbf v,\mathbf x) \mathbf u $$ is normal. As you have correctly found, $g^*$ is the map given by $$ g^*(\mathbf x) = (\mathbf u, \mathbf x) \mathbf v. $$ Now, it's easy to look at the two compositions of these two operators. We have $$ (g \circ g^*)(\mathbf x) = (\mathbf v,(\mathbf u,\mathbf x) \mathbf v)\mathbf u = (\mathbf v,\mathbf v)(\mathbf u, \mathbf x) \mathbf u\\ (g^* \circ g)(\mathbf x) = (\mathbf u,(\mathbf v,\mathbf x) \mathbf u)\mathbf v = (\mathbf u,\mathbf u)(\mathbf v, \mathbf x) \mathbf v $$ So, when are these two ouputs identical? Note that the the outputs of $g \circ g^*$ (i.e. the range of $g \circ g^*$) consist of multiples of $\mathbf u$, while those of $g^* \circ g$ are multiplies of $\mathbf v$. It follows that $\mathbf u$ and $\mathbf v$ must be multiples of each other.
Indeed, $\mathbf u,\mathbf v$ being multiples of one another is also a sufficient condition: for the trivial case that $\mathbf u$ or $\mathbf v$ is zero, $g$ is the zero-map and hence normal. Otherwise, if we take $\mathbf u = \alpha \mathbf v$ for some $\alpha \in \Bbb C$, then plugging into the above formulas for $g \circ g^*$ and $g^* \circ g$ yields $$ (g \circ g^*)(\mathbf x) = \bar \alpha \alpha (\mathbf v, \mathbf v)(\mathbf v, \mathbf x)\mathbf v = (g^* \circ g)(\mathbf x). $$ Thus, $g$ is normal if and only if the set $\{\mathbf u, \mathbf v\}$ is linearly dependent.