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) = \frac{1 + \cos(2x)}2`

`\sin^2(x) = \frac{1 - \cos(2x)}2`

Then

`\cos\left(-\frac\pi8\right) = \cos\left(\frac\pi8\right) = \sqrt{\frac{1 + \cos\left(\frac\pi4\right)}2} = \sqrt{\frac12 + \frac1{2\sqrt2}}`

`\sin\left(-\frac\pi8\right) = -\sin\left(\frac\pi8\right) = -\sqrt{\frac{1 - \cos\left(\frac\pi4\right)}2} = -\sqrt{\frac12 - \frac1{2\sqrt2}}`

`\cos\left(\frac{3\pi}8\right) = \sin\left(\frac\pi8\right) = \sqrt{\frac12 - \frac1{2\sqrt2}}`

`\sin\left(\frac{3\pi}8\right) = \cos\left(\frac\pi8\right) = \sqrt{\frac12 + \frac1{2\sqrt2}}`

and the roots are equivalently

`\boxed{\pm\sqrt[4]{2}\left(\sqrt{\frac12 + \frac1{2\sqrt2}} - i\sqrt{\frac12 - \frac1{2\sqrt2}}\right) \text{ and } \pm\sqrt[4]{2}\left(\sqrt{\frac12 + \frac1{2\sqrt2}} + i \sqrt{\frac12 - \frac1{2\sqrt2}}\right)}`