Question

One of the solutions of equation
${z}^{3}$
= w is z1 = -5
$\sqrt{3}$
+ 5i. What is one of other two solutions if w is a complex number?

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Answer 1

Answer:

shdhwgaodveu3eueiej3iw0i

Answer 2

Given:

• $Z^{3} = w$ which has a solution $Z_1 = -5\sqrt{3}+5i$

To Find:

• The other two solutions($Z_2, Z_3$) when w is a complex number.

Solution:

• We already know the roots of unity satisfy for $w \neq 1$ and $w^{3}$ = 1.
• Consider, $w^3 = 1$
• $w^{3} - 1 = 0$ further solving we get,
• $(w-1)(w^{2}+w+1) = 0$  since we need two solutions.
• On solving we get, $w = \frac{1}{2}(-1+/-i\sqrt{3} )$
• We get two complex numbers here.
• $w = \frac{1}{2}(-1+i\sqrt{3} )$ and other is $w* = w^{2}$
• We can say the other two can be written as $Z_1w$ and $Z_1w*$
• So, $(Z_1w)^{3} = (Z_1)^{3}w^{3} = a(1) = a$   since, $(Z^{3} = a)$
• Same goes for $Z_1w*$
• $Z_1 = 5(-\sqrt{3}+i)$  
• $Z_2 = Z_1w = 5(-\sqrt{3}+i)\frac{1}{2} (-1+i\sqrt{3}) = \frac{5}{2}(\sqrt{3}-\sqrt{3}-1-3i) = -10i$  
• $Z_3 = Z_1w* = \frac{5}{2}(-\sqrt{3}+i)(-1-\sqrt{3}) = \frac{5}{2}(\sqrt{3}+\sqrt{3}+i+3i) = 5(\sqrt{3}+i)$  
• Here we get additional information saying a = $(-10i)^{3} = 1000i$ because  $(Z^{3} = a)$.

The two solutions are:

• $Z_2 = Z_1w = 5(-\sqrt{3}+i)\frac{1}{2} (-1+i\sqrt{3}) = \frac{5}{2}(\sqrt{3}-\sqrt{3}-1-3i) = -10i$
• $Z_3 = Z_1w* = \frac{5}{2}(-\sqrt{3}+i)(-1-\sqrt{3}) = \frac{5}{2}(\sqrt{3}+\sqrt{3}+i+3i) = 5(\sqrt{3}+i)$