Answer to Question #144106 in Complex Analysis for Caylin

Question

Let w be a complex number, z a 4th root of w.

1) Show that z(k) = p^1/4[cos((θ + 2kπ)/4) + isin((θ + 12kπ)/4)], k = 0, 1, 2, 3, is a formula for the 4th roots of w, where θ is the argument of w and p its modulus.

2) hence , determine 4th roots of 16.

please assist.

Answer 1

Let's use the trigonometric representation of a complex number :

$w=\rho (cos(\theta)+i\sin(\theta)) = \rho e^{i\theta}$

$z=ae^{i\alpha}$

However we need to remember that the argument is defined up to $2\pi n, n\in\mathbb{Z}$ . We have a pretty simple formula for $z^4$ in this trigonometric form :

$z^4 = a^4 e^{4i\alpha}$

The fact that $z^4=w$ means that $a^4=\rho, e^{4i\alpha}=e^{i\theta}$ . Therefore we find :

$a= \rho^{1/4}, e^{i(4\alpha-\theta)}=1 \Rightarrow 4\alpha-\theta=2\pi k, k\in\mathbb{Z}$ . Finally, for $k=0,1,2,3$ (it is not necessary to study the cases $k<0$ or $k>3$ as these values of arguments would differ by $2\pi$ from the solutions that we will find and so they will not give any new values of $z$) we have:

$z_k = \rho^{1/4}e^{i(\frac{\theta+2\pi k}{4}) } = \rho^{1/4} (\cos(\frac{\theta+2\pi k}{4} )+i\sin(\frac{\theta+2\pi k}{4} ))$

Now let's proceed to calculating the 4th roots of 16:

$16 = 16 \times e^{i\cdot0}$

$z_k=(16)^{1/4} (\cos(\frac{2\pi k}{4})+i\sin(\frac{2\pi k}{4}))$

$z_0=2, z_1=2i, z_2=-2, z_3=-2i$

$z_k = \{2; 2i; -2; -2i\}$