Answer to Question #144106 in Complex Analysis for Caylin

Question #144106
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.
1
Expert's answer
2020-11-15T18:04:15-0500

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\\}"



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS