This is incorrect. It should be $\displaystyle z= x^{1/3}W(cos(120)+ i sin(120))$.
If $\displaystyle z= x(cos(\theta)+ i sin(\theta))$ then $\displaystyle |z|= x$ so [tex]|^3|= (9/4)|z| then [tex]x^3= (9/4)x. 4x^3- 9x= x(4x^2- 9)= 0. Can you solve for x?
i am not sure how to do part (iii)
Thanks, Plato. Yes, the "W" was a mistype. It should have been $\displaystyle z= x^{1/3}(cos(120)+ i sin(120))$. The point is that, with |z|= x, you should have had "$\displaystyle x^{1/3}$" not "$\displaystyle x$".
$\displaystyle x^3= (9/4)x$ is equivalent to $\displaystyle 4x^3- 9x= x(4x^2- 9)= 0$ so either $\displaystyle x= 0$ or $\displaystyle 4x^2- 9= (2x- 3)(2x+ 3)= 0$.
Of course, in order to have |z|= x, with z non-zero as shown, x must be a positive real number.