$\begin{aligned} z &=x+i y \\ &=\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)^{-25} \\ &=\left[\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right]^{-25} \end{aligned}$
$\because(\operatorname{Cos} \pi-\mathrm{i} \sin \pi)^{\mathrm{n}}=\operatorname{Cos} \mathrm{n} \pi-\mathrm{i} \operatorname{Sin} \mathrm{n} \pi$
$\left[\begin{array}{l}\cos (-\theta)=\cos \theta \\ \sin (-\theta)=-\sin \theta\end{array}\right]$
$\begin{array}{l}\mathrm{z}=\left[\cos \left(\frac{25 \pi}{4}\right)+i \sin \left(\frac{25 \pi}{4}\right)\right] \\ = & {\left[\cos \left(6 \pi+\frac{\pi}{4}\right)+i \sin \left(6 \pi+\frac{\pi}{4}\right)\right]} \\ =\cos \frac{\pi}{4}+i \sin \frac{\pi}{4} \\ =\frac{1+i}{\sqrt{2}} \\ \frac{z-\sqrt{2}}{z-i \sqrt{2}}=\frac{(1+i-2) \sqrt{2}}{\sqrt{2}(1+i-2 i)} \\ =\frac{-1+i}{1-i}=-1 \\ \text { Amplitude }=\tan ^{-1} \frac{b}{a}=\tan ^{-1}\left(\frac{0}{-1}\right)=\tan ^{-1} 0 \\ =\tan ^{-1}(\tan \pi)=\pi\end{array}$
Hence fundamental amplitude of $\left(\frac{z-\sqrt{2}}{z-i \sqrt{2}}\right)$ is $\pi$