# How do you find all solutions to #x^3+64i=0#?

##### 1 Answer

#### Explanation:

Let

First let us plot the point

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

# |omega| = 64 #

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

So then in polar form we have:

# omega = 64(cos(-(pi)/2) + isin(-(pi)/2)) #

We now want to solve the equation for

# x^3 = 64(cos(-(pi)/2) + isin(-(pi)/2)) #

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

# x^3 = 64(cos(2npi-(pi)/2) + isin(2npi-(pi)/2)) \ \ \ n in ZZ #

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

# x = 64(cos(2npi-(pi)/2) + isin(2npi-(pi)/2))^(1/3)#

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

# \ \ = 4{cos((2npi-(pi)/2)/3) + isin((2npi-(pi)/2)/3)}#

# \ \ = 4(cos theta + isin theta) \ \ \ \ # where# theta=((4n-1)pi)/6#

Put:

# n=-1 => theta = (-5pi)/6 #

# " " :. x = 4(cos ((-5pi)/6)+ isin ((-5pi)/6)) #

# " " :. x = 4(-sqrt(3)/2-1/2i) #

# " " :. x = -2sqrt(3)-2i #

# n=0 => theta = (-pi)/6 #

# " " :. x = 4(cos ((-pi)/6)+ isin ((-pi)/6)) #

# " " :. x = 4(sqrt(3)/2-1/2i) #

# " " :. x = 2sqrt(3)-2i #

# n=1 => theta = (pi)/2 #

# " " :. x = 4(cos ((pi)/2)+ isin ((pi)/2)) #

# " " :. x = 4(0+i) #

# " " :. x = 4i #

After which the pattern continues.

We can plot these solutions on the Argand Diagram