Question

3. Find the cube roots of (-2 + 2i) and express them in the form a + bi.

Answer 1

Answer:

there fore -2 = a

2 = b

Step-by-step explanation:

We know that , -2 + 2i = ( -2 , 2 )

= -2( 1 , 0 ) + 2( 1 , 0 )

Answer 2

Given : (-2 + 2i)

To Find :  cube roots of (-2 + 2i)

Solution:

a + bi  = ∛(-2 + 2i)

Taking cube on both sides

=> (a + bi)³ = -2  + 2i

=> a³ + (bi)³ + 3abi(a + bi)  =  -2 + 2i

=> a³ - ib³ + 3a²bi  - 3ab² = - 2 + 2i

=> (a³ - 3ab²)+ i(-b³ + 3a²b)  = - 2 + 2i

a³ - 3ab²  = - 2

-b³ + 3a²b =   2

Adding both

(a³ - b³) + 3ab(a - b)  = 0

=> (a - b) (a² + b²  + ab) + 3ab(a - b)  = 0

=>  (a - b)  (a² + b² + 4ab) = 0

=> (a - b) = 0

=> a = b  

a = b

a³ - 3ab²  = - 2

=>  a³ - 3a³ = -2

=> -2a³ = - 2

=> a³ = 1

=> a = 1

=> b = 1

a + bi  = 1 + i  

cube roots of (-2 + 2i)  = 1 + i

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