Solve the equation #ix^3 + 1 - isqrt3 = 0#. Express your answer in trigonometric form?

Question

Answer is #2^(1/3)(cosalpha + isinalpha)# #"for"# #alpha = 50^@, 122^@, 194^@, 266^@, 338^@#.

Thanks in advance.

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Answer 1 · 2018-05-28T22:03:00

Only three cubic roots for angles 10, 150 and 250. See below

From given equation

#x^3=(-1+sqrt3i)/i=((-1+sqrt3i)i)/(i·i)=(-i-sqrt3)/-1=sqrt3+i#

Lets pass to polar form #sqrt3+i#

#rho=sqrt(3+1)=2#

#theta=arctan(1/sqrt3)=30º#

Our complex number is #2_(30)#

three cubic roots

#z_0=root(3)2_(30/3)=root(3)2_10=root(3)2(cos10+isin10)#
#z_1=root(3)2_(130)=root(3)2(cos130+isin130)#
#z_2=root(3)2_250=root(3)2(cos250+isin250)#