Question

(1-i)^2 find 4 th root​

Answer 1

By de Moivre's theorem,

$1 - i = \sqrt2\,e^{-i\pi/4} \implies (1-i)^2 = 2\,e^{-i\pi/2}$

$\implies \sqrt[4]{(1 - i)^2} = \sqrt[4]{2}\,e^{i(2\pi k-\pi/2)/4} = \sqrt[4]{2}\,e^{i(4k-1)\pi/8}$

where $k\in\{0,1,2,3\}$. The fourth roots of $(1-i)^2$ are then

$k = 0 \implies \sqrt[4]{2}\,e^{-i\pi/8}$

$k = 1 \implies \sqrt[4]{2}\,e^{i3\pi/8}$

$k = 2 \implies \sqrt[4]{2}\,e^{i7\pi/8}$

$k = 3 \implies \sqrt[4]{2}\,e^{i11\pi/8}$

or more simply

$\boxed{\pm\sqrt[4]{2}\,e^{-i\pi/8} \text{ and } \pm\sqrt[4]{2}\,e^{i3\pi/8}}$

We can go on to put these in rectangular form. Recall

$\cos^2(x) = \dfrac{1 + \cos(2x)}2$

$\sin^2(x) = \dfrac{1 - \cos(2x)}2$

Then

$\cos\left(-\dfrac\pi8\right) = \cos\left(\dfrac\pi8\right) = \sqrt{\dfrac{1 + \cos\left(\frac\pi4\right)}2} = \sqrt{\dfrac12 + \dfrac1{2\sqrt2}}$

$\sin\left(-\dfrac\pi8\right) = -\sin\left(\dfrac\pi8\right) = -\sqrt{\dfrac{1 - \cos\left(\frac\pi4\right)}2} = -\sqrt{\dfrac12 - \dfrac1{2\sqrt2}}$

$\cos\left(\dfrac{3\pi}8\right) = \sin\left(\dfrac\pi8\right) = \sqrt{\dfrac12 - \dfrac1{2\sqrt2}}$

$\sin\left(\dfrac{3\pi}8\right) = \cos\left(\dfrac\pi8\right) = \sqrt{\dfrac12 + \dfrac1{2\sqrt2}}$

and the roots are equivalently

$\boxed{\pm\sqrt[4]{2}\left(\sqrt{\dfrac12 + \dfrac1{2\sqrt2}} - i\sqrt{\dfrac12 - \dfrac1{2\sqrt2}}\right) \text{ and } \pm\sqrt[4]{2}\left(\sqrt{\dfrac12 + \dfrac1{2\sqrt2}} + i \sqrt{\dfrac12 - \dfrac1{2\sqrt2}}\right)}$