For #z^5 = 1 + i#, solve the complex number?

1 Answer
Vinícius Ferraz
Jun 14, 2018

#z_0 = root[10]{2}(cos frac{pi}{20} + i sin frac{pi}{20})#
#z_1 = root[10]{2}(cos frac{9pi}{20} + i sin frac{9pi}{20})#
#z_2 = root[10]{2}(cos frac{17pi}{20} + i sin frac{17pi}{20})#
#z_3 = root[10]{2}(cos frac{25pi}{20} + i sin frac{25pi}{20})#
#z_4 = root[10]{2}(cos frac{33pi}{20} + i sin frac{33pi}{20})#

Explanation:

#z^5 = sqrt 2 (sqrt2 /2 + i sqrt 2/2)#

#z^5 = sqrt 2 (cos frac{pi}{4} + i sin frac{pi}{4})#

#z = root[10]{2}(cos frac{pi/4 + 2kpi}{5} + i sin frac{pi/4 + 2kpi}{5}); k = 0, 1, 2, 3, 4#

#z = root[10]{2}(cos frac{pi+8kpi}{20} + i sin frac{pi+8kpi}{20}); k = 0, 1, 2, 3, 4#

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