Is it possible to have $\|u\|$ = 3, $\|v\|$ = 2 and $\langle u, v \rangle = 5 + 4i$?

Question

True or false? For some inner product space $V$, there exist $u, v \in V$ such that $\|u\|$ = 3, $\|v\|$ = 2 and $\langle u, v \rangle = 5 + 4i$.

I tried focusing on the definition of the inner product and look for a counterexample through the conjugate symmetry property, but I cannot find any. Is this the wrong direction?

Answer 1

Cauchy-Schwarz yields the contradiction

$$6=\sqrt{36}<\sqrt{25+16}=|5+4i|=|\langle u,v\rangle|\le\|u\|\cdot\|v\|=2\cdot3=6.$$

Answer 2

$||u||=3, ||v||=2$ and $\langle u, v \rangle = 5 + 4i$ is not possible, since in this case the Cauchy -Schwarz inequality

$$|\langle u, v \rangle| \le ||u|| \cdot ||v||$$

does not hold.