Find all the fourth roots of the complex number #-8+8i(sqrt 3)# ?

3 Answers
P dilip_k
Dec 9, 2017

Square roots of the given quantity are

#=pmsqrt(-8+8i(sqrt 3)#

#=pmsqrt(4(-2+2(sqrt 3i))#

#=pmsqrt(4(1^2+(sqrt3i)^2+2*1*(sqrt 3i))#

#=pmsqrt(2^2(1+(sqrt3i)^2)#

#=pm2(1+(sqrt3i)#

Now square roots of #=+2(1+(sqrt3i)#

#=pmsqrt(2(1+(sqrt3i))#

#=pmsqrt(2+2sqrt3i)#

#=pmsqrt((sqrt3)^2+i^2+2sqrt3i)#

#=pmsqrt((sqrt3+i)^2#

#=pm(sqrt3+i)#

Again square roots of #=-2(1+(sqrt3i)#

#=pmsqrt(-2(1+(sqrt3i))#

#=pmisqrt(2+2sqrt3i)#

#=pmisqrt((sqrt3)^2+i^2+2sqrt3i)#

#=pmisqrt((sqrt3+i)^2#

#=pmi(sqrt3+i)#

#=pm(sqrt3i+i^2)#

#=pm(-1+sqrt3i)#

So four fourth roots of the complex number #(-8+8i(sqrt 3)# are

  • #=pm(sqrt3+i)#
    and
  • #=pm(-1+sqrt3i)#
Narad T.
Dec 9, 2017

The solution is #S={(sqrt3+i),(-1+isqrt3),(-sqrt3-i),(1-isqrt3)}#

Explanation:

Apply Euler's Identity

#costheta+isintheta=e^(itheta)#

Let #z=-8+8sqrt3i#

#z=8(-1+sqrt3i)#

#=16(-1/2+sqrt3/2i)#

#=16(cos(2/3pi)+isin(2/3pi))#

#=16e^(2/3ipi+2kpi)#

Therefore, the fourth root is

#z^(1/4)=16^(1/4)e^(2/12pi+kpi/2)#, #AA k in ZZ#

#=2e^(pi/6+kpi/2)#

Let #k=0#, #=>#, #z_1=2(cos(pi/6)+isin(pi/6))=2(sqrt3/2+i/2)=(sqrt3+i)#

Let #k=1#, #=>#,
#z_2=2(cos(pi/6+pi/2)+isin(pi/6+pi/2))=2(cos(2/3pi)+isin(2/3pi))=2(-1/2+sqrt3/2i)=(-1+isqrt3)#

Let #k=2#, #=>#, #z_3=2(cos(pi/6+pi)+isin(pi/6+pi))=2(cos(7/6pi)+isin(7/6pi))=2(-sqrt3/2-1/2i)=(-sqrt3-i)#

Let #k=3#, #=>#, #z_4=2(cos(pi/6+3/2pi)+isin(pi/6+3/2pi))=2(cos(5/3pi)+isin(5/3pi))=2(1/2-sqrt3/2i)=(1-isqrt3)#

NickTheTurtle
Dec 9, 2017

See below

Explanation:

We need to find a certain complex number #z# such that #z^4=-8+8isqrt(3)#.

First of all, all complex numbers can be written in polar form. Thus, we can try to rewrite the right-hand side from Cartesian form to polar form #a+bi=r(cos(theta)+i sin(theta))#, where #r=sqrt(a^2+b^2)# and #theta=arctan(b/a)#. Note that, since #-8+8isqrt(3)# is in the third quadrant of the Argand plane, #theta# must be in the third quadrant of the unit circle.

Then, #r=sqrt((-8)^2+(8sqrt(3))^2)=16# and #theta=arctan((8sqrt(3))/-8)=(2pi)/3+2kpi,k in ZZ# (the reason for #k# will be clear soon, but realize that adding an integer multiple of #2pi# to an angle does not change its value).

So we have #-8+8isqrt(3)=16(cos((2pi)/3+2kpi)+i sin((2pi)/3+2kpi))#.

Then, #z^4=16(cos((2pi)/3+2kpi)+i sin((2pi)/3+2kpi)#, or #z=(16(cos((2pi)/3+2kpi)+i sin((2pi)/3+2kpi))^(1/4)#

Using de Moivre's formula, we know that #(r(cos(theta)+i sin(theta)))^n=r^n(cos(ntheta)+i sin(ntheta))#.

Thus, #z=(2(cos(((2pi)/3+2kpi)/4)+i sin(((2pi)/3+2kpi)/4))),k in ZZ#.

Since #cos(theta)# and #sin(theta)# are periodic every #2pi#, we only need to substitute #0≤k<4# (substituting other integers will reveal the same solution with one in this range).

The solutions are the following:
#z=2(cos(pi/6)+i sin(pi/6))=sqrt(3)+1/2i#
#z=2(cos((2pi)/3)+i sin((2pi)/3))=-1+isqrt(3)#
#z=2(cos((7pi)/6)+i sin((7pi)/6))=-sqrt(3)-1/2i#
#z=2(cos((5pi)/3)+i sin((5pi)/3))=1-isqrt(3)#

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