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Let $z=x+iy$, where $x,y\in \mathbb{R}$.

Given $|z|=\sqrt{2}$, so $$|z{|}^2=x^2+y^2=2.$$

Also, $|z^2-i|=1$. Compute $z^2$: $$z^2=(x+iy{)}^2=x^2-y^2+2ixy.$$ So, $$z^2-i=(x^2-y^2)+2ixy-i=(x^2-y^2)+i(2xy-1).$$ Therefore, $$|z^2-i{|}^2=(x^2-y^2{)}^2+(2xy-1{)}^2=1.$$

Now, we have the system: $$\begin{array}{rl}{x}^2+y^2& =2\\ (x^2-y^2{)}^2+(2xy-1{)}^2& =1\end{array}$$

Let us use polar form: $z=r{e}^{i\theta }$, $r=\sqrt{2}$, so $z=\sqrt{2}{e}^{i\theta }$. Then $z^2=2e^{i2\theta }$, so $$z^2-i=2e^{i2\theta }-i.$$

Write $2e^{i2\theta }=2(\cos 2\theta +i\sin 2\theta )$, so $$z^2-i=2\cos 2\theta +i(2\sin 2\theta -1).$$

Therefore, $$|z^2-i{|}^2=(2\cos 2\theta {)}^2+(2\sin 2\theta -1{)}^2=1.$$ Expand: $$\begin{array}{rl}4{\cos }^{2}2\theta +(4{\sin }^{2}2\theta -4\sin 2\theta +1)& =1\\ 4{\cos }^{2}2\theta +4{\sin }^{2}2\theta -4\sin 2\theta +1& =1\\ 4({\cos }^{2}2\theta +{\sin }^{2}2\theta )-4\sin 2\theta +1& =1\\ 4(1)-4\sin 2\theta +1& =1\\ 4-4\sin 2\theta +1=1& \\ 4-4\sin 2\theta =0& \\ 1-\sin 2\theta =0& \\ \sin 2\theta =1& \end{array}$$

So $2\theta =\frac{\pi }{2}+2\pi k$, $k\in \mathbb{Z}$. Thus $\theta =\frac{\pi }{4}+\pi k$.

Therefore, the solutions are $$z=\sqrt{2}{e}^{i\left(\frac{\pi }{4}+\pi k\right)}$$ for $k\in \mathbb{Z}$.

Explicitly, for $k=0$: $$z_1=\sqrt{2}{e}^{i\frac{\pi }{4}}=\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)=\sqrt{2}\left(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)=1+i.$$ For $k=1$: $$z_2=\sqrt{2}{e}^{i\left(\frac{5\pi }{4}\right)}=\sqrt{2}\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)=\sqrt{2}\left(-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right)=-1-i.$$

Thus, the solutions are $z=1+i$ and $z=-1-i$.