# Find all the #8^(th)# roots of #3i-3#?

##
Find all the #8^(th)# roots of

#3i-3#

Also show any one of them in an Argand diagram.

Find all the

Also show any one of them in an Argand diagram.

##### 1 Answer

#### Explanation:

Let

First, we will put the complex number into polar form:

# |omega| = sqrt((-3)^2+3^2) =3sqrt(2)#

# theta =arctan(3/(-3)) = arctan(-1) = -pi/4#

# => arg \ omega = pi/2+pi/4 = (3pi)/4 #

So then in polar form we have:

# omega = 3sqrt(2)(cos((3pi)/4) + isin((3pi)/4)) #

We now want to solve the equation

# z^8 = 3sqrt(2)(cos((3pi)/4) + isin((3pi)/4)) #

Whenever dealing with complex variable equation such as this it is essential to remember that the complex exponential (and therefore the polar representation) has a period of

# z^8 = 3sqrt(2)(cos((3pi)/4+2npi) + isin((3pi)/4+2npi)) \ \ \ n in ZZ #

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

# z = 3sqrt(2)(cos((3pi)/4+2npi) + isin((3pi)/4+2npi))^(1/8) #

# \ \ = (3sqrt(2))^(1/8)(cos((3pi)/4+2npi) + isin((3pi)/4+2npi))^(1/8) #

# \ \ = 3^(1/8)2^(1/16) (cos(((3pi)/4+2npi)/8) + isin(((3pi)/4+2npi)/8))#

# \ \ = 3^(1/8)2^(1/16) (cos theta + isin theta ) #

Where:

# theta =((3pi)/4+2npi)/8 = ((3+8n)pi)/32#

And we will get

After which the pattern continues (due the above mentioned periodicity).

We can plot these solutions on the Argand Diagram: