Question #e8b67

2 Answers
Narad T.
Jul 31, 2017

The solutions are #S={root(4)9(sqrt(2+sqrt2)/2+isqrt(2-sqrt2)/2),z_1, z_2, z_3}#. See the other solutions below

Explanation:

We need

#cos(pi/4)=2cos^2(pi/8)-1#

#2cos^2(pi/8)=1+cos(pi/4)=1+sqrt2/2#

# **cos(pi/8)=sqrt(2+sqrt2)/2** #

#cos(pi/4)=1-2sin^2(pi/8)#

#2sin^2(pi/8)=1-cos(pi/4)=1-sqrt2/2#

# **sin(pi/8)=sqrt(2-sqrt2)/2** #

#cos(5/8pi)=cos(pi/8+pi/2)=cos(pi/8)cos(pi/2)-sin(pi/8)sin(pi/2)#

#=sqrt(2+sqrt2)/2*0-sqrt(2-sqrt2)/2*1=-sqrt(2-sqrt2)/2#

#sin(5/8pi)=sin(pi/8+pi/2)=sin(pi/8)cos(pi/2)+cos(pi/8)sin(pi/2)#

#=sqrt(2-sqrt2)/2*0+sqrt(2+sqrt2)/2*1#

#=sqrt(2+sqrt2)/2#

#cos(9/8pi)=cos(pi/8+pi)=cos(pi/8)cos(pi)-sin(pi/8)sin(pi)=-sqrt(2+sqrt2)/2#

#sin(9/8pi)=sin(pi/8+pi)=sin(pi/8)cos(pi)+cos(pi/8)sin(pi)=-sqrt(2-sqrt2)/2#

#cos(13/8pi)=cos(pi/8+3/2pi)=cos(pi/8)cos(3/2pi)-sin(pi/8)sin(3/2pi)=sqrt(2-sqrt2)/2#

#sin(9/8pi)=sin(pi/8+3/2pi)=sin(pi/8)cos(3/2pi)+cos(pi/8)sin(3/2pi)=-sqrt(2+sqrt2)/2#

We apply Euler's relation

#e^(itheta)=costheta+isintheta#

#e^(ipi/2)=cos(pi/2)+isin(pi/2)=i#

Therefore,

#z^4=9i=9e^(ipi/2+2kpi)#, #AA k in ZZ#

Therefore,

#z=root(4)9e^(ipi/8+2kpi/4)#

When,

#k=0#, #=>#, #z_1=root(4)9e^(i1/8pi)=root(4)9(cos(1/8pi)+isin(1/8pi))=root(4)9(sqrt(2+sqrt2)/2+isqrt(2-sqrt2)/2)#

#k=1#, #=>#, #z_2=root(4)9e^(i5/8pi)=root(4)9(cos(5/8pi)+isin(5/8pi))=root(4)9(-sqrt(2-sqrt2)/2+isqrt(2+sqrt2)/2)#

#k=2#, #=>#, #z_3=root(4)9e^(i9/8pi)=root(4)9(cos(9/8pi)+isin(9/8pi))=root(4)9(-sqrt(2+sqrt2)/2-isqrt(2-sqrt2)/2)#

#k=3#, #=>#, #z_4=root(4)9e^(i13/8pi)=root(4)9(cos(13/8pi)+isin(13/8pi))=root(4)9(sqrt(2-sqrt2)/2-isqrt(2+sqrt2)/2)#

harsh s.
Jul 31, 2017

The four solutions for #z# are
#+-(sqrt(6+3sqrt2)/2+isqrt(6-3sqrt2)/2)#
and
#+-(sqrt(6-3sqrt2)/2-isqrt(6+3sqrt2)/2)#

Explanation:

#z^4=9i#
#rArr# #z=root(4)(9i)#
#rArr# #z=sqrt3*root(4)i#...........#(1)#

Now solve for the fourth roots of #i#

#root(4)i=i^(1/4)=e^(1/4*ln(i))#......................#(2)#

Since,
#ln(i)=....,(-11ipi)/2,color(red)((-7ipi)/2,(-3ipi)/2,(ipi)/2,(5ipi)/2),(9ipi)/2,....#

Plugging in the four highlighted values of #ln(i)# in equation #(2)#,

#root(4)i=e^((-7ipi)/8), e^((-3ipi)/8), e^((ipi)/8), e^((5ipi)/8)#

Plug these four values in equation #(1)# and use Euler's formula to find desired values of #z#.
Note that there are infinite values of #ln(i)# but only four of them are used because others give repeated values for #root(4)i#.
Here I am solving for only one value of #z# but you can try for other values also

#rArr# #z=sqrt3*e^((ipi)/8)=sqrt3*(cos(pi/8)+i*sin(pi/8))#

Since, #cos(pi/8)=sqrt(2+sqrt2)/2, sin(pi/8)=sqrt(2-sqrt2)/2#

#=># #z=(sqrt(6+3sqrt2)/2+isqrt(6-3sqrt2)/2)#

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