45 If the conjugate of the complex number $z$ is $\bar{z}$, and $|z|=1$. Also, $A=(-1,0), B(0,-1)$ are fixed points, then the function $f(z)=|(z+1)(\bar{z}-i)|$ takes its maximum value when, in the complex plane, the figure formed by $Z$ and $A, B$ is
A. Isosceles right triangle
B. Equilateral triangle
C. Right triangle
D. Isosceles triangle
45 D. Let $z=\cos \theta+i \sin \theta$, then $f(z)=\sqrt{2\left[1+\sqrt{2} \sin \left(\theta+\frac{\pi}{4}\right)\right]^{2}}$. When $\theta=2 k \pi+\frac{\pi}{4}$ ($k \in \mathbf{Z}$), $f(z)$ reaches its maximum value $2+\sqrt{2}$. At this time, $z=\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}$. The figure formed by points $Z, A, B$ is an isosceles triangle.