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

Let # omega=-1 #, and then #z^5 = omega#

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=omega# for #z# (to gain #5# solutions):

# 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 #2pi#, so we can equivalently write (incorporating the periodicity):

# 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

If we express #z# in polar form, #z= rho e^(i theta)# we have that:

#z^5 = rho^5 e^(i 5theta)#

so:

#z^5 = -1 => rho^5 e^(i 5theta) = e^(ipi) => {(rho^5 = 1),(5theta =pi+2kpi):}#

We have then:

#{(rho = 1),(theta = pi/5+(2k)/5pi):}#

for any #k in ZZ#

We have potentially infinite solutions:

#z_k=e^(i(pi/5+(2k)/5pi))#

but we can see that if #j-=kmod5#, then #j-k=5n# with #n in ZZ# and we have:

#pi/5+(2jpi)/5 = pi/5+(2(k+5n)pi)/5 = pi/5+(2kpi)/5+2npi #

so that:

#z_j = e^(i(pi/5+(2j)/5pi)) = e^(i(pi/5+(2k)/5pi+2npi)) = e^(i(pi/5+(2k)/5pi))e^(i2npi) = z_k#

In conclusion we have five different solutions for #k=0,1,..,4#:

#z_0 = e^(i(pi)/5) = 1/4((1+sqrt5)+isqrt(10+sqrt5)))#

#z_1 = e^(i(3pi)/5)#

#z_2 = e^(ipi) = -1#

#z_3 = e^(i(7pi)/5)#

#z_4 = e^(i(9pi)/5)#

All this points have module equal to #1#, so they lie on the complex plan on the circle with center the origin and radius #rho = 1#, and they are the vertices of a regular pentagon inscribed in such circle.