Given the complex number $z$ has a modulus of 1. Find
$$
u=\frac{(z+4)^{2}-(\bar{z}+4)^{2}}{4 i} \text {. }
$$

the maximum value.
$$
\begin{array}{l}
\text { Given }|z|=1 \text {, we can set } z=\cos x+i \sin x \text {. Then, } \\
u=(4+\cos x) \sin x \text {. } \\
\text { By } u^{2}=(4+\cos x)^{2} \sin ^{2} x \\
=\frac{(4+\cos x)(4+\cos x)(\sqrt{6}+1)(1+\cos x) \cdot(\sqrt{6}+3)(1-\cos x)}{(\sqrt{6}+1)(\sqrt{6}+3)} \\
\leqslant \frac{1}{9+4 \sqrt{6}} \\
\cdot\left[\frac{(1+\cos x)+(1+\cos x)+(\sqrt{6}+1)(1+\cos x)+(\sqrt{6}+3)(1-\cos x)}{4}\right]^{4} \\
=\frac{1}{9+4 \sqrt{6}}\left(\frac{6+\sqrt{6}}{2}\right)^{4} \\
=\frac{9+24 \sqrt{6}}{4} \text {. } \\
\end{array}
$$

When $4+\cos x=$ ( $\sqrt{6}+1)(1+\cos x)$
$$
=(\sqrt{6}+3)(1-\cos x) \text {, }
$$

i.e., $\cos x=\frac{\sqrt{6}-2}{2}$, the maximum value is
$$
u_{\max }=\frac{\sqrt{9+24 \sqrt{6}}}{2}
$$

The coefficients $\sqrt{6}+1, \sqrt{6}+3$ are determined by the method of undetermined coefficients.