Solve the equation ix^3 + 1 - isqrt3 = 0ix3+1−i√3=0. Express your answer in trigonometric form?

Answer is 2^(1/3)(cosalpha + isinalpha)213(cosα+isinα) "for"for alpha = 50^@, 122^@, 194^@, 266^@, 338^@α=50∘,122∘,194∘,266∘,338∘.

Thanks in advance.

1 Answer
F. Javier B.
May 28, 2018

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

Explanation:

From given equation

x^3=(-1+sqrt3i)/i=((-1+sqrt3i)i)/(i·i)=(-i-sqrt3)/-1=sqrt3+ix3=−1+√3ii=(−1+√3i)ii⋅i=−i−√3−1=√3+i

Lets pass to polar form sqrt3+i√3+i

rho=sqrt(3+1)=2ρ=√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)

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