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.

1 Answer
Aug 23, 2017

Steve M

Explanation:

Let # omega=-3+3i #, and let #z^8 = omega#

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)) #

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We now want to solve the equation #z^8=omega# for #z# (to gain #8# solutions):

# 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 #2pi#, so we can equivalently write (incorporating the periodicity):

# 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 #8# unique solutions by choosing appropriate values of #n#. Working to 3dp, and using excel to assist, we get:
Steve M

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

We can plot these solutions on the Argand Diagram:

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