Question

`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?

Answer 1

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

Explanation:

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}`

Answer 2

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

Explanation:

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