Question

Find the fourth roots of 256(cos 280 + I sin 280)

Answer 1

r(cosx+isinx)
r^(1/n)(cos(x/n)+isin(x/n))
256^(1/4)(cos(280/4)+isin(280/4))
4(cos70+isin70)

Answer 2

`\bf \sqrt[{{ n}}]{z}=\sqrt[{{ n}}]{r}\left[ cos\left( \frac{\theta+2\pi k}{{{ n}}} \right) +i\ sin\left( \frac{\theta+2\pi k}{{{ n}}} \right)\right]\quad k\ roots\\\\ -------------------------------\\\\ \sqrt[{{ 4}}]{z}=\sqrt[{{ 4}}]{r}\left[ cos\left( \frac{\theta+2\pi k}{{{ 4}}} \right) +i\ sin\left( \frac{\theta+2\pi k}{{{ 4}}} \right)\right]\quad \begin{array}{llll} k\ roots\\ 0,1,2,3 \end{array}\\\\ -------------------------------\\\\`


`\bf \sqrt[{{ 4}}]{256}\left[ cos\left( \frac{280^o+360^o(0)}{{{ 4}}} \right) +i\ sin\left( \frac{280^o+360^o(0)}{{{ 4}}} \right)\right] \\\\\\ 4[cos(70^o)+i\ sin(70^o)]\implies 4cos(70^o)+i\ 4sin(70^o)\impliedby \begin{array}{llll} 1st\ root\\ k=0 \end{array}`


`\bf -------------------------------\\\\ \sqrt[{{ 4}}]{256}\left[ cos\left( \frac{280^o+360^o(1)}{{{ 4}}} \right) +i\ sin\left( \frac{280^o+360^o(1)}{{{ 4}}} \right)\right] \\\\\\ 4[cos(160^o)+i\ sin(160^o)] \\\\\\ 4cos(160^o)+i\ 4sin(160^o)\impliedby \begin{array}{llll} 2nd\ root\\ k=1 \end{array}`


`\bf -------------------------------\\\\ \sqrt[{{ 4}}]{256}\left[ cos\left( \frac{280^o+360^o(2)}{{{ 4}}} \right) +i\ sin\left( \frac{280^o+360^o(2)}{{{ 4}}} \right)\right] \\\\\\ 4[cos(250^o)+i\ sin(250^o)] \\\\\\ 4cos(250^o)+i\ 4sin(250^o)\impliedby \begin{array}{llll} 3rd\ root\\ k=2 \end{array}`


`\bf -------------------------------\\\\ \sqrt[{{ 4}}]{256}\left[ cos\left( \frac{280^o+360^o(3)}{{{ 4}}} \right) +i\ sin\left( \frac{280^o+360^o(3)}{{{ 4}}} \right)\right] \\\\\\ 4[cos(340^o)+i\ sin(340^o)] \\\\\\ 4cos(340^o)+i\ 4sin(340^o)\impliedby \begin{array}{llll} 4th\ root\\ k=3 \end{array}`



so.. just get their product, for example the first one would be 1.37 + 3.76i rounded up