Answer to Question #117213 in Algebra for Nana Opoku Kuffour

Question

#117181
Express the roots of the equation z 3 − α 3 = 0 in terms of α and w, where w is a complex cube root of unity. Use your answer to find the roots of the following equations in the form a + ib. (a) z 3 − 27 = 0 (b) z 3 + 8 = 0 (c) z 3 − i = 0.

Answer 1

For the equation $z^{3}=a^{3}$ , we have that

$z^{3}=a^{3} * 1=a^{3}*e^{(2\pi)i}=a^{3}(cos 2\pi + isin 2\pi)$

Since $sin 2\pi , cos 2\pi$ are periodic functions of the period $2\pi k , k=0,1,2$ , then

$z_{k}^{3}=a^{3}(cos( 2\pi+ 2\pi k)+ isin ( 2\pi+ 2\pi k))$

thus $z_{k}=a(cos( 2\pi+ 2\pi k)+ isin ( 2\pi+ 2\pi k))^{1/3}$

using De Moivre's formula, we have

$z_{k}=a(cos( 2\pi+ 2\pi k)/3+ isin ( 2\pi+ 2\pi k)/3)$ , where $k=0,1,2$

when k=0 , we have that the first root $z_{0}=aw , w=-1/2+(\sqrt3/2)i$

when k=1 , we have that the second root $z_{1}=aw^{2}=a(-1/2-(\sqrt3/2)i)$

when k=2 , we have that the third root $z_{2}=a$

### To solve the problem (a) we put a=3 , then the roots are $-3/2+(3\sqrt3/2)i$ , $-3/2-(3\sqrt3/2)i$ , 3

### To solve the problem (b) , we put $a=-2$ , thus the roots are $1-\sqrt3i , 1+\sqrt3i , -2$

### But the problem (c) is different, so we solve it with different steps

$z^{3}=i=cos \pi/2+ isin \pi/2$ , then $z^{3} = cos( \pi/2+ 2\pi k)+ isin ( \pi/2+ 2\pi k)$

where $, k=0,1,2$

thus the roots are $z_{k} = cos(( \pi/2+ 2\pi k)/3)+ isin ( ( \pi/2+ 2\pi k)/3 )$

when k=0 , $z_{0} = cos( \pi/6)+ isin ( \pi/6)$

when k=1 , $z_{1} = cos( 5\pi/6)+ isin ( 5\pi/6 )$

when k=2 , $z_{2} = cos(9\pi/6)+ i sin ( 9\pi/6 )$