Answer to Question #96508 in Complex Analysis for Olajide Olaitan

Question

Find the fifth root of -1

Answer 1

1) Let $a<0$ and $a$ is not a complex number.

If $n$ is an odd number, then there is one and only one real number $x$ such that $x^n = a$. This number is $x = \sqrt[n]{a}$ and is called the root of an odd degree $n$ from a negative number $a$.

So, $\sqrt[5]{-1} = \sqrt[5]{(-1)*(-1)*(-1)*(-1)*(-1)}=\sqrt[5]{(-1)^5} = -1.$

2) Let $a \in\Complex$ (complex number) and $a=i^2$

Find the trigonometric form of a complex number

$x=Re(a)=-1, y=Im(a)=0$.

$x<0,y\geq0\implies arg(a)=\phi=\pi-\arctan(\cfrac{y}{\vert x \vert})=\pi -0=\pi$.

Thus, the trigonometric form of the complex number $a=i^2$ is $a=\cos(\pi)+i\sin(\pi)$.

The fifth roots are $a_k = \sqrt[5]{a}=\sqrt[5]{\vert a \vert}(\cos\cfrac{\phi+2\pi k}{5}+i\sin\cfrac{\phi+2\pi k}{5} ), k=0,1,2,3,4$.

$k=0\implies a_0 = \sqrt[5]{\vert a \vert}(\cos\cfrac{\phi+2\pi *0}{5}+i\sin\cfrac{\phi+2\pi *0}{5} )=\cos\cfrac{\pi}{5}+i\sin\cfrac{\pi}{5}=cos36\degree+i\sin36\degree = 0.809017+0.587785i$ ,

$k=1\implies a_1 = \sqrt[5]{\vert a \vert}(\cos\cfrac{\phi+2\pi *1}{5}+i\sin\cfrac{\phi+2\pi *1}{5} )=\cos\cfrac{3\pi}{5}+i\sin\cfrac{3\pi}{5}=cos108\degree+i\sin108\degree=-0.309017+0.951057i$ ,

$k=2\implies a_2 = \sqrt[5]{\vert a \vert}(\cos\cfrac{\phi+2\pi *2}{5}+i\sin\cfrac{\phi+2\pi *2}{5} )=\cos\cfrac{5\pi}{5}+i\sin\cfrac{5\pi}{5}=\cos\pi +i\sin\pi = -1$ ,

$k=3\implies a_3 = \sqrt[5]{\vert a \vert}(\cos\cfrac{\phi+2\pi *3}{5}+i\sin\cfrac{\phi+2\pi *3}{5} )=\cos\cfrac{7\pi}{5}+i\sin\cfrac{7\pi}{5}=cos252\degree+i\sin252\degree=-0.309017-0.951057i$ ,

$k=4\implies a_4 = \sqrt[5]{\vert a \vert}(\cos\cfrac{\phi+2\pi *4}{5}+i\sin\cfrac{\phi+2\pi *4}{5} )=\cos\cfrac{9\pi}{5}+i\sin\cfrac{9\pi}{5}=cos324\degree+i\sin324\degree=0.809017-0.587785i$

For evaluation was used cosine and sine tables (https://onlinemschool.com/math/formula/cosine_table/ , https://onlinemschool.com/math/formula/sine_table/ ).

Answer: $0.809017\pm0.587785i, -0.309017\pm0.951057i, -1.$