Question

Z3=w is Z1=-5√3+5i what is one of other two solution if w is a complex numbers​

Answer 1

Given :  Z₁ = -5√3+5i  

Z³ = w

To Find : one of other two solution

Solution:

Z³ = w

=> (Z₁)³ = w  Eq1

Z₁ = -5√3+5i  

= 10 ( -√3/2  + i/2)

= 10 ( cos (5π/6)  + iSin(5π/6))

 Cubing both sides

=>  (Z₁)³ = 10³ (cos(15π/6) + iSin(15π/6) )  Eq2

Equate Eq1 and Eq2

=> 10³ (cos(15π/6) + iSin(15π/6) ) = w  

=> 10³(cos(3π/6) +iSin(3π/6))  =w

=> 10³(cos( π/2) +iSin(π/2))  =w

Generalized solution

=> 10³(cos( (4n + 1)π/2) +iSin((4n+1)π/2))  =w

z³ = w  =>  z = ∛w

Taking cube root

=> 10 (cos( (4n + 1)π/6) +iSin((4n+1)π/6)) =  ∛w

Substituting

n = 0

= 10 cos π/6 + i sin π/6  = 5√3 + 5i  

n = 1

=  10 ( cos (5π/6)  + iSin(5π/6)) = -5√3+5i     already given

n = 2

10 (cos( (9π/6) +iSin(9π/6)) = -10i

Hence  two other solutions

are   5√3 + 5i  and -10i

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Answer 2

Answer:

w=1,2,3

Step-by-step explanation:

Solution

Whenever you're dealing with complex roots, remember complex roots of unity. The answer comes in a similar way. So immediately convert into polar coordinates.

$z_{1}=-5 \sqrt{3}+5 i$

$z_{1}=100 \cdot e^{-\frac{\pi}{6} i}$

Since that is one of the solutions, all of the solutions can be expressed with the formula

$z=100 \cdot e^{\left(-\frac{\pi}{6}+k \cdot \frac{2}{3} \pi\right) i}$,

where w=0,1,2

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