# Find the roots of #z^5+1=0#?

##### 2 Answers

Working to 3dp there are 5 solutions:

# z = -1; -0.309+-0.951; z = 0.809+-0.588i #

#### Explanation:

Let

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

# |omega| = 1 #

# arg(omega) = pi #

So then in polar form we have:

# omega = cos(pi) + isin(pi) #

We now want to solve the equation

# z^5 = cos(pi) + isin(pi) #

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^5 = cos(pi+2npi) + isin(pi+2npi) \ \ \ n in ZZ #

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

# z = (cos(pi+2npi) + isin(pi+2npi))^(1/5) #

# \ \ = cos((pi+2npi)/5) + isin((pi+2npi)/5) #

# \ \ = cos theta + isin theta \ \ \ \ # where# theta=((2n+1)pi)/5#

Working to 3dp we get:

Put:

# n=-2 => theta = -(3pi)/5 #

# " " :. z = cos (-(3pi)/5)+ isin (-(3pi)/5) #

# " " :. z = -0.309-0.951i #

# n=-1 => theta = -pi/5 #

# " " :. z = cos (-pi/5)+ isin (-pi/5) #

# " " :. z = 0.809-0.588i #

# n=0 => theta = (pi)/5 #

# " " :. z = cos (pi/5)+ isin (pi/5) #

# " " :. z = 0.809+0.588i #

# n=1 => theta = (3pi)/5 #

# " " :. z = cos ((3pi)/5)+ isin ((3pi)/5) #

# " " :. z = -0.309+0.951 #

# n=2 => theta = pi #

# " " :. z = cos pi+ isin pi #

# " " :. z = -1 #

After which the pattern continues.

We can plot these solutions on the Argand Diagram

#### Explanation:

If we express

so:

We have then:

for any

We have potentially infinite solutions:

but we can see that if

so that:

In conclusion we have five different solutions for

All this points have module equal to