# How do you find the cube root of #64 (cos (pi/5) + isin (pi/5)) #?

##### 1 Answer

# \ \ \ \ \ \ 4 (cos (pi/15) + isin (pi/15)) #

# \ \ \ \ \ \ 4 (cos ((11pi)/15) + isin ((11pi)/15)) #

# \ \ \ \ \ \ 4 (cos ((7pi)/5) + isin ((7pi)/5)) #

#### Explanation:

Let

Whenever dealing with complex variable equation such as this it is essential to remember that the complex exponential has a period of

# z^3=64 (cos (pi/5+2npi) + isin (pi/5+2npi)) \ \ \ n in NN#

By De Moivre's Theorem we can write this as:

# z = {64 (cos (pi/5+2npi) + isin (pi/5+2npi))}^(1/3) #

# \ \= 64^(1/3) (cos (pi/5+2npi) + isin (pi/5+2npi))}^(1/3 #

# \ \= 64^(1/3) (cos ((pi/5+2npi)/3) + isin ((pi/5+2npi)/3)) #

# \ \= 4 (cos ((pi/5+2npi)/3) + isin ((pi/5+2npi)/3)) #

Put:

# n=0 => z = 4 (cos ((pi/5)/3) + isin ((pi/5)/3)) #

# " " = 4 (cos (pi/15) + isin (pi/15)) #

# n=1 => z = 4 (cos ((pi/5+2pi)/3) + isin ((pi/5+2pi)/3)) #

# " " = 4 (cos ((11pi)/15) + isin ((11pi)/15)) #

# n=2 => 4 (cos ((pi/5+4pi)/3) + isin ((pi/5+4pi)/3)) #

# " " = 4 (cos ((7pi)/15) + isin ((7pi)/15)) #

After which the pattern continues

So the three roots are:

# z = 4 (cos (pi/15) + isin (pi/15)) #

# \ \ \ \ \ \ 4 (cos ((11pi)/15) + isin ((11pi)/15)) #

# \ \ \ \ \ \ 4 (cos ((7pi)/5) + isin ((7pi)/5)) #

We can view these roots on the Argand diagram: