Example 11 The equation about $x$
$$
x^{4}+(k-1) x^{3}+k x^{2}+(k-1) x+1=0
$$

has no real roots. Then the range of real number $k$ is $\qquad$
Solve: Dividing both sides of the original equation by $x^{2}$, we get
$$
x^{2}+\frac{1}{x^{2}}+(k-1)\left(x+\frac{1}{x}\right)+k=0 \text {. }
$$

Let $x+\frac{1}{x}=y \in(-\infty,-2] \cup[2,+\infty)$.
Then $x^{2}+\frac{1}{x^{2}}=y^{2}-2$
$$
\begin{array}{l}
\Rightarrow y^{2}+(k-1) y+k-2=0 \\
\Rightarrow(y+1)[y+(k-2)]=0 \\
\Rightarrow y=-1 \text { (discard) or } y=-k+2 \\
\Rightarrow x+\frac{1}{x}=-k+2 \\
\Rightarrow x^{2}+(k-2) x+1=0 .
\end{array}
$$

Since the original equation has no real roots, we have
$$
\Delta=(k-2)^{2}-4<0 \text {. }
$$

Solving this, we get $0<k<4$.