Question

Express the fifth roots of unity in standard form a + bi. with 1 + 0i

Answer 1

Answer:

For K=0

$cos(\frac{0+2\pi 0}{5})+isin(\frac{0+2\pi 0}{5})\\cos(0)+isin(0)=1+i0$

For K=1:

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

For K=2:

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

For K=3:

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

For K=4:

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

Step-by-step 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:

$z=cos(\frac{0}{5})+isin(\frac{0}{5})\\ z=cos(0)+isin(0)$

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

For K=0

$cos(\frac{0+2\pi 0}{5})+isin(\frac{0+2\pi 0}{5})\\cos(0)+isin(0)=1+i0$

For K=1:

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

For K=2:

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

For K=3:

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

For K=4:

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