Question:

What are the complex 4th roots of z = -72 + i*72*sqrt(3)?

Answers:

A-Best: First convert -72 + 72i√3 to polar form. |-72 + 72i√3| = √[(-72)^2 + (72√3)^2] = √(5184 + 15552) = 144 The angle between the real axis and -72 + 72i√3 is: tan^-1(-72√3 / 72) = -π/3 + 2π = 5π/3 Adding π to this gives the angle in the correct quadrant (8π/3). So -72 + 72i√3 = 144(cos 8π/3 + i sin 8π/3) From De Movire's Theorem, we have: Root 1 = 144^(1/4)[cos (8π/3)/4 + i sin (8π/3)/4] = 144^(1/4)(cos 2π/3 + i sin 2π/3) = 144^(1/4)[(-1 + i√3)/2] = -√3 + 3i Root 2 = 144^(1/4)[cos (8π/3 + 2π)/4 + i sin (8π/3 + 2π)/4] = 144^(1/4)[cos (8π/3 + 6π/3)/4 + i sin (8π/3 + 6π/3)/4] = 144^(1/4)[cos (14π/3)/4 + i sin (14π/3)/4] = 144^(1/4)(cos 7π/6 + i sin 7π/6) = 144^(1/4)[(-√3 - i)/2] = -3 - i√3 Root 3 = 144^(1/4)[cos (8π/3 + 4π)/4 + i sin (8π/3 + 4π)/4] = 144^(1/4)[cos (8π/3 + 12π/3)/4 + i sin (8π/3 + 6π/3)/4] = 144^(1/4)[cos (20π/3)/4 + i sin (14π/3)/4] = 144^(1/4)(cos 5π/3 + i sin 5π/3) = 144^(1/4)[(1 - i√3)/2] = √3 - 3i Root 4 = 144^(1/4)[cos (8π/3 + 6π)/4 + i sin (8π/3 + 6π)/4] = 144^(1/4)[cos (8π/3 + 18π/3)/4 + i sin (8π/3 + 18π/3)/4] = 144^(1/4)[cos (26π/3)/4 + i sin (26π/3)/4] = 144^(1/4)(cos 13π/6 + i sin 13π/6) = 144^(1/4)[(√3 + i)/2] = 3 + i√3 I hope that helps!  

A: This is a general way to find the nth roots of any complex number. Let the complex number have a magnitude r and an angle t. The nth roots will all have a magnitude r ^(1/n) and angles of t, (t + 2π/n), (t + 4π/n), ....... (t + 2(n - 2)π/n), (t + 2(n - 1)π/n). Note that t + 2π is the same as t, so the angles repeat after this. Hope this helps.