The polar form of 1 − i = 2 ( cos ( − π 4 + i sin ( − π 4 ) ) . 1−i=\sqrt{2}(\cos(-\dfrac{\pi}{4}+i\sin(-\dfrac{\pi}{4})). 1 − i = 2 ( cos ( − 4 π + i sin ( − 4 π )) .
According to the De Moivre's Formula, all 3-th roots of a complex number
2 ( cos ( − π 4 + i sin ( − π 4 ) ) \sqrt{2}(\cos(-\dfrac{\pi}{4}+i\sin(-\dfrac{\pi}{4})) 2 ( cos ( − 4 π + i sin ( − 4 π )) are given by
( 2 ) 1 / 3 ( cos ( − π / 4 + 2 π k 3 ) + i sin ( − π / 4 + 2 π k 3 ) ) , (\sqrt{2})^{1/3}(\cos(\dfrac{-\pi/4+2\pi k}{3})+i\sin(\dfrac{-\pi/4+2\pi k}{3})), ( 2 ) 1/3 ( cos ( 3 − π /4 + 2 πk ) + i sin ( 3 − π /4 + 2 πk )) , k = 0 , 1 , 2 k=0,1,2 k = 0 , 1 , 2 k = 0 : k=0: k = 0 :
( 2 ) 1 / 6 ( cos ( − π / 4 + 2 π ( 0 ) 3 ) + i sin ( − π / 4 + 2 π ( 0 ) 3 ) ) (2)^{1/6}(\cos(\dfrac{-\pi/4+2\pi (0)}{3})+i\sin(\dfrac{-\pi/4+2\pi (0)}{3})) ( 2 ) 1/6 ( cos ( 3 − π /4 + 2 π ( 0 ) ) + i sin ( 3 − π /4 + 2 π ( 0 ) ))
= ( 2 ) 1 / 6 ( cos ( π 12 ) − i sin ( π 12 ) ) =(2)^{1/6}(\cos(\dfrac{\pi}{12})-i\sin(\dfrac{\pi}{12})) = ( 2 ) 1/6 ( cos ( 12 π ) − i sin ( 12 π ))
k = 1 : k=1: k = 1 :
( 2 ) 1 / 6 ( cos ( − π / 4 + 2 π ( 1 ) 3 ) + i sin ( − π / 4 + 2 π ( 1 ) 3 ) ) (2)^{1/6}(\cos(\dfrac{-\pi/4+2\pi (1)}{3})+i\sin(\dfrac{-\pi/4+2\pi (1)}{3})) ( 2 ) 1/6 ( cos ( 3 − π /4 + 2 π ( 1 ) ) + i sin ( 3 − π /4 + 2 π ( 1 ) ))
= ( 2 ) 1 / 6 ( cos ( 7 π 12 ) + i sin ( 7 π 12 ) ) =(2)^{1/6}(\cos(\dfrac{7\pi}{12})+i\sin(\dfrac{7\pi}{12})) = ( 2 ) 1/6 ( cos ( 12 7 π ) + i sin ( 12 7 π ))
k = 2 : k=2: k = 2 :
( 2 ) 1 / 6 ( cos ( − π / 4 + 2 π ( 2 ) 3 ) + i sin ( − π / 4 + 2 π ( 2 ) 3 ) ) (2)^{1/6}(\cos(\dfrac{-\pi/4+2\pi (2)}{3})+i\sin(\dfrac{-\pi/4+2\pi (2)}{3})) ( 2 ) 1/6 ( cos ( 3 − π /4 + 2 π ( 2 ) ) + i sin ( 3 − π /4 + 2 π ( 2 ) ))
= ( 2 ) 1 / 6 ( cos ( 5 π 4 ) + i sin ( 5 π 4 ) ) =(2)^{1/6}(\cos(\dfrac{5\pi}{4})+i\sin(\dfrac{5\pi}{4})) = ( 2 ) 1/6 ( cos ( 4 5 π ) + i sin ( 4 5 π ))
= ( 2 ) − 1 / 3 ( 1 + i ) =(2)^{-1/3}(1+i) = ( 2 ) − 1/3 ( 1 + i )
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