#x^3+i =0# Find all complex number solutions. Write in trigonometric form. I know I need to find r and the angle. Need help with the steps?

We want to solve #x^3+i=0# or #x^3=-i#

Let #k in ZZ#

#-i=cos(3/2pi+2kpi)+sin(3/2pi+2kpi)i#

and

#x=r(cos(theta)+sin(theta)i)#

So

#r(cos(theta)+sin(theta)i)=(cos(3/2pi+2kpi)+sin(3/2pi+2kpi)i)^(1/3)=cos(1/2pi+2/3kpi)+sin(1/2pi+2/3kpi)#

We obtain the above using De Moivre's formula:

#(cos(theta)+sin(theta)i)^n=cos(ntheta)+sin(ntheta)i#

Letting #k=0,1,-1#, we get values of #x# as

#x_1=cos(1/2pi)+sin(1/2pi)i=i#

#x_2=cos(1/2pi+2/3pi)+sin(1/2pi+2/3pi)=sqrt3/2-1/2i#

#x_3=cos(1/2pi-2/3pi)+sin(1/2pi-2/3pi)=sqrt3/2+1/2i#

So, we have our original #x# satisfying

#x in{i,sqrt3/2-1/2i,-sqrt3/2-1/2i}#

We seek solutions to:

# x^3+i = 0 #

Let # omega=-i #, and then #x^3 = omega#

And we will put the complex number into polar form (visually):

# |omega| = 1 #
# arg(omega) = -pi/2 #

So then in polar form we have:

# omega = cos(-pi/2) + isin(-pi/2) #

We now want to solve the equation #x^3=omega# for #x# (to gain #3# solutions):

# x^3 = cos(-pi/2) + isin(-pi/2) #

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^3 = cos(-pi/2+2npi) + isin(-pi/2+2npi) \ \ \ n in ZZ #

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

# z = (cos(-pi/2+2npi) + isin(-pi/2+2npi))^(1/3) #
# \ \ = cos((-pi/2+2npi)/3) + isin((-pi/2+2npi)/3) #
# \ \ = cos theta + isin theta \ \ \ \ #

Where:

# theta= (-pi/2+2npi)/2 = ((4n-1)pi)/6 #

Put:

# n=-1 => (-5pi)/6 #
# " " :. z = cos (-(5pi)/6)+ isin (-(5pi)/6) #
# " " :. z = -sqrt(3)/2-1/2i #

# n=1 => (3pi)/6 #
# " " :. z = cos ((3pi)/6)+ isin ((3pi)/6) #
# " " :. z = 0+i #

# n=0 => theta = (-pi)/6 #
# " " :. z = cos ((-pi)/6)+ isin ((-pi)/6) #
# " " :. z = sqrt(3)/2-1/2 #

After which the pattern continues.

We can plot these solutions on the Argand Diagram