How do you find the fifth root of #32(cos((5pi)/6)+isin((5pi)/6))#?

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Answer 1 · 2017-02-23T09:23:36

Fifth root of #32(cos((5pi)/6)+isin((5pi)/6))# is #2(sqrt3/2+i1/2)#. Other four roots can also be obtained. See below for details.

According to DeMoivre's theorem if #z=r(costheta+isintheta)#

then #z^n=r^n(cosntheta+isinntheta)#

It has also been proved that it is true for all #ninQ#

Hence, if #z=32(cos((5pi)/6)+isin((5pi)/6))#

#root(5)z=z^(1/5)=32^(1/5)(cos((5pi)/6xx1/5)+isin((5pi)/6xx1/5)#

= #2(cos(pi/6)+isin(pi/6))#

= #2(sqrt3/2+i1/2)#

In fact you should find five fifth roots, using

#z=32(cos(2npi+(5pi)/6)+isin(2npi+(5pi)/6))#, where #n# is an integer and then

#root(5)z=2(cos((2npi)/5+pi/6)+isin((2npi)/5+pi/6))#

As may be seen the above root is obtained by putting #n=0# and other roots can be obtained by putting #n=1,2,3" or "4#