## Find the fifth roots of 32(cos 280° + i sin 280°).

Question

Find the fifth roots of 32(cos 280° + i sin 280°).

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1. I) Let z = 32 (cos 280 degrees + i sin 280 degrees)
= 32 (cos (k* 360 + 280) + isin (k*360 + 280)),where k = integer
II) z^ (1/5) = (32 (cos (k*360 + 280) + i sin (k*360 + 280))) ^ (1/5)
then,
z^ (1/5) = 2(cos ((k*360 + 280)/5) + i sin((k*360 + 280)/5))
We apply De Moivre’s theorem:
z^ (1/5) = 2(cos(72k + 56) + i sin (72k +56))
We can get the five roots by assigning k = 0, 1, 2, 3, and 4
When K = 0, 1st root = 2 + i sin (cos 56 + i sin 56)k = 1, 2nd root = 2 (cos 128 + i sin 128)k = 2, 3rd root = 2 (cos 200 + i sin 200)k = 3, 4th root = 2 (cos 272 + i sin 272)k = 4, 5th root = 2 (cos 344 + i sin 344)

The fifth root is 5th root = 2 (cos 344 + i sin 344)

z1= 2(cos 56 deg + i sin 56 deg)n

z2= 2(cos 128 deg + i sin 128 deg)n

z3 = 2(cos 200 deg + i sin 200 deg)n

z4 = 2(cos 272 deg+ i sin 272 deg)

z5 = 2(cos 344 deg+ i sin 344 deg)

Step-by-step explanation:

De Moivre’s Theorem–> e^i theta = cos theta + i sin thetan

The fifth root of 32 is 2 and 280/5 is 56n

z1= 2(cos 56 deg + i sin 56 deg)n

360deg/ 5 = 72 deg

(If it was asking for cubed root u would do 360/3. whatever the root is you divide that by 360)

z2= 2(cos 56 + 72) + (i sin 56 + 72)n

z2= 2(cos 128 deg + i sin 128 deg)n

z3 = 2(cos 128 + 72) + ( i sin 128 + 72)n

z3 = 2(cos 200 deg + i sin 200 deg)n

z4 = 2(cos 200 + 72) + (i sin 200 + 72)n

z4 = 2(cos 272 deg + i sin 272 deg)

z5 = 2(cos 272 + 72) + (i sin 272 + 72)n

z5 = 2(cos 344 deg + i sin 344 deg)

deg means degrees