Question

asked Oct 7, 2021 91.1k views

1 Answer

Carmelle · Answer 1 · 2021-10-13T13:13:41+0000

Answer:

For K=0

$cos((0+2\pi 0)/(5))+isin((0+2\pi 0)/(5))\\cos(0)+isin(0)=1+i0$

For K=1:

$cos((0+2\pi 1)/(5))+isin((0+2\pi 1)/(5))\\cos(2\pi/5)+isin(2\pi/5)=0.3090+i0.9510$

For K=2:

$cos((0+2\pi 2)/(5) )+isin((0+2\pi 2)/(5) )\\cos(4\pi/5)+isin(4\pi/5)=-0.809+i0.587$

For K=3:

$cos((0+2\pi 3)/(5) )+isin((0+2\pi 3)/(5) )\\cos(6\pi/5)+isin(6\pi/5)=-0.809-i0.587$

For K=4:

$cos((0+2\pi 4)/(5) )+isin((0+2\pi 4)/(5) )\\cos(8\pi/5)+isin(8\pi/5)=0.3091-i0.9510$

Explanation:

Fifth Root is given by:

$\sqrt[5]{z}=1+0i$

The above equation will become:

z=(1+0i)^5

It can be written as:

z=[cos(0)+isin(0)]^5

|z|=1,

According to De-moivre's Theorem:

$z=cos((0)/(5))+isin((0)/(5))\\ z=cos(0)+isin(0)$

Now, Fifth Roots of unity in standard form a + bi :

$\sqrt[5]{z}=[{cos(0+2\pi k)+isin(0+2\pi k)}]^(1/5)$

k=0,1,2,3,4