Q:

Induced map in K-theory by a "trivial" bimodule

Let $R$ be a ring (not necessary commutative) and let $P_{\bullet}$ be a perfect $R$-bimodule (chain complex). I will denote the category of perfect right $R$-chain complexes by $\textbf{Perf}(R)$. The endofunctor $-\otimes_{R}P_{\bullet} :\textbf{Perf}(R)\rightarrow \textbf{Perf}(R)$ induces a map in algebraic $K$-theory given by
$K_{\ast}(-\otimes_{R}P_{\bullet}):K_{\ast}(R)\rightarrow K_{\ast}(R)$. 
If the class $[P_{\bullet}] \in K_{0}(R)$ is trivial $(=0)$ does it mean that 
$K_{\ast}(-\otimes_{R}P_{\bullet})$ is a 0 map ?

A:

No. Let $R=\mathbb{Z}\times\mathbb{Z}$, let $P$ and $Q$ be the projective modules $\mathbb{Z}\times0$ and $0\times\mathbb{Z}$, and let 
$$P_\bullet=\dots\longrightarrow0\longrightarrow P\otimes_\mathbb{Z}P
\stackrel{0}{\longrightarrow}Q\otimes_\mathbb{Z}P\longrightarrow0\longrightarrow\dots$$

