Mathematics, 23.01.2020 22:31 rosyjesus297
Find the fourth roots of 256(cos 280 + i sin 280)
Answers: 1
![\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}\\\\ -------------------------------\\\\](/tpl/images/0467/9497/8a7e3.png)
![\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}](/tpl/images/0467/9497/b5b8e.png)
![\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}](/tpl/images/0467/9497/874ba.png)
![\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}](/tpl/images/0467/9497/eb22b.png)
![\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}](/tpl/images/0467/9497/bfe64.png)
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