Find the complex number z^9. (Enter your answer in a + bi form.)

Question

9th root of z = 4 + 4 root3 i. In a+bi form. 

 

I am stuck! 

 

 

Answer 1

Euler's formula is not necessary:

 

z = 4 + 4i³   r = 8, verify with diagram and Pythagoras. θ = π/3.

 

z⁹ = 8⁹(cos π/3 + i sin π/3)⁹

z⁹ = 134217728 (cos 9(π/3) + i sin 9(π/3))

z⁹ = 134217728 (cos 3π + i sin 3π)

z⁹ = 134217728 (-1 + i(0))

z⁹ = -134217728 + 0i

Answer 2

 

Hi Sydney,

 

     z = 4 + i4√3

 

        = 4(1 + i√3)

 

        = 4·2[(1/2) + i(√3/2)]

 

        = 8[cos(π/3) + isin(π/3)]

 

        = 8e^i(π/3)   (Euler's formula, cosθ + isinθ = e^iθ)

 

     z⁹ = (8e^i(π/3))⁹

 

          = 8⁹·(e^i(π/3))⁹

 

          = 8⁹·e^i3π

 

          = 8⁹·[cos(3π) + isin(3π)]

 

          = -8⁹

 

Hope this helps.