Answer to Question #100458 in Algebra for Rahul Bisht

Question

find all the 5th roots of 5-i

Answer 1

What is actually asked here is to solve the equation

$$z^5=5-i.$$

In other words, the modulus here is

$$m=\sqrt{5^2+(-1)^2}=\sqrt{26},$$

and the argument is

$$\phi=\text{atan}\frac{-1}{5}=-0.197\text{ rad}.$$

Therefore, we can rewrite the equation into the following polar form:

$$z^5=\sqrt{26}\cdot e^{-0.197i+2k\pi i}.$$

Simplify:

$$z^5=\sqrt{26}\cdot e^{(-0.197+2k\pi)i}.$$

Let

$$z=re^{i\theta}.$$

Hence:

$$\big(re^{i\theta}\big)^5=\sqrt{26}\cdot e^{(-0.197+2k\pi)i},\\ \space\\ r^5e^{5i\theta}=\sqrt{26}\cdot e^{(-0.197+2k\pi)i},$$

equate the modulus and the power of $e$:

$$r^5=\sqrt{26},\space\space\space\space\space\space\space5\theta=-0.197+2k\pi;\\ \space\\ r=\sqrt[10]{26},\space\space\space\space\space\space\space\theta=\frac{2k\pi-0.197}{5}.\\$$

Now the question is: what values of $k$ must we choose in order to get five different and unique roots? The answer is the values starting from zero to $n-1$, that is,

$$k=0,1,2,3,4.$$

Substitute these values and get 5 roots:

$$z_1=\sqrt[10]{26}\cdot e^\frac{-0.197}{5},\\ z_2=\sqrt[10]{26}\cdot e^\frac{2\pi-0.197}{5},\\ z_3=\sqrt[10]{26}\cdot e^\frac{4\pi-0.197}{5},\\ z_4=\sqrt[10]{26}\cdot e^\frac{6\pi-0.197}{5},\\ z_5=\sqrt[10]{26}\cdot e^\frac{8\pi-0.197}{5}.$$