15. Let $z$ be an imaginary number, $w=z+\frac{1}{z}$, and $-1<w<2$. Then the range of the real part of $z$ is $\qquad$ .
15. $\left(-\frac{1}{2}, 1\right)$.

Let $z=a+b$ i. Then
$$
\begin{array}{l}
-1<a+b \mathrm{i}+\frac{a-b \mathrm{i}}{a^{2}+b^{2}}<2 \\
\Rightarrow b-\frac{b}{a^{2}+b^{2}}=0 \\
\Rightarrow b=0 \text { or } a^{2}+b^{2}=1 .
\end{array}
$$

When $b=0$; no solution;
When $a^{2}+b^{2}=1 \Rightarrow-\frac{1}{2}<a<1$, which meets the condition.