By de Moivre's theorem,
[tex]1 - i = \sqrt2\,e^{-i\pi/4} \implies (1-i)^2 = 2\,e^{-i\pi/2}[/tex]
[tex]\implies \sqrt[4]{(1 - i)^2} = \sqrt[4]{2}\,e^{i(2\pi k-\pi/2)/4} = \sqrt[4]{2}\,e^{i(4k-1)\pi/8}[/tex]
where [tex]k\in\{0,1,2,3\}[/tex]. The fourth roots of [tex](1-i)^2[/tex] are then
[tex]k = 0 \implies \sqrt[4]{2}\,e^{-i\pi/8}[/tex]
[tex]k = 1 \implies \sqrt[4]{2}\,e^{i3\pi/8}[/tex]
[tex]k = 2 \implies \sqrt[4]{2}\,e^{i7\pi/8}[/tex]
[tex]k = 3 \implies \sqrt[4]{2}\,e^{i11\pi/8}[/tex]
or more simply
[tex]\boxed{\pm\sqrt[4]{2}\,e^{-i\pi/8} \text{ and } \pm\sqrt[4]{2}\,e^{i3\pi/8}}[/tex]
We can go on to put these in rectangular form. Recall
[tex]\cos^2(x) = \dfrac{1 + \cos(2x)}2[/tex]
[tex]\sin^2(x) = \dfrac{1 - \cos(2x)}2[/tex]
Then
[tex]\cos\left(-\dfrac\pi8\right) = \cos\left(\dfrac\pi8\right) = \sqrt{\dfrac{1 + \cos\left(\frac\pi4\right)}2} = \sqrt{\dfrac12 + \dfrac1{2\sqrt2}}[/tex]
[tex]\sin\left(-\dfrac\pi8\right) = -\sin\left(\dfrac\pi8\right) = -\sqrt{\dfrac{1 - \cos\left(\frac\pi4\right)}2} = -\sqrt{\dfrac12 - \dfrac1{2\sqrt2}}[/tex]
[tex]\cos\left(\dfrac{3\pi}8\right) = \sin\left(\dfrac\pi8\right) = \sqrt{\dfrac12 - \dfrac1{2\sqrt2}}[/tex]
[tex]\sin\left(\dfrac{3\pi}8\right) = \cos\left(\dfrac\pi8\right) = \sqrt{\dfrac12 + \dfrac1{2\sqrt2}}[/tex]
and the roots are equivalently
[tex]\boxed{\pm\sqrt[4]{2}\left(\sqrt{\dfrac12 + \dfrac1{2\sqrt2}} - i\sqrt{\dfrac12 - \dfrac1{2\sqrt2}}\right) \text{ and } \pm\sqrt[4]{2}\left(\sqrt{\dfrac12 + \dfrac1{2\sqrt2}} + i \sqrt{\dfrac12 - \dfrac1{2\sqrt2}}\right)}[/tex]
