# #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?

##### 2 Answers

#### Explanation:

We want to solve

Let

and

So

We obtain the above using De Moivre's formula:

Letting

So, we have our original

# x = i, (+-sqrt(3)-i)/2 #

#### Explanation:

We seek solutions to:

# x^3+i = 0 #

Let

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 = 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

# 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