Find the value of alpha and theta?

#Z_1=2(cos(18^@)+isin(18^@))#
#Z_2=alpha(cos(40^@)+isin(40^@))#
#Z_3=32(cos(theta)+isin(theta))#
Find value of #alpha# and #theta# if
#(Z_1^10)/(Z_2-Z_3)=4i#

1 Answer
Jan 19, 2018

Given:

#Z_1=2(cos(18^@)+isin(18^@))" [1]"#

#Z_2=alpha(cos(40^@)+isin(40^@))" [2]"#

#Z_3=32(cos(theta)+isin(theta))" [3]"#

#(Z_1^10)/(Z_2-Z_3)=4i" [4]"#

Substitute equation [1] into equation [4]:

#((2(cos(18^@)+isin(18^@)))^10)/(Z_2-Z_3)=4i#

Use De Moivre's formula:

#(2^10(cos(18^@xx10)+isin(18^@xx10)))/(Z_2-Z_3)=4i#

#(1024(cos(180^@)+isin(180^@)))/(Z_2-Z_3)=4i#

#(1024(-1+i0))/(Z_2-Z_3)=4i#

#(-1024+i0)/(Z_2-Z_3)=4i" [4.1]"#

Multiply both sides of equation [4.1] by #Z_3-Z_2#:

#1024+i0=4i(Z_3-Z_2)" [4.2]"#

Substitute equations [2] and [3] into equation [4.2]:

#1024+i0=4i(32(cos(theta)+isin(theta))-alpha(cos(40^@)+isin(40^@)))#

Perform the multiplication:

#1024+i0=4i(32cos(theta)+i32sin(theta))-alphacos(40^@)-ialphasin(40^@))#

#1024+i0=i128cos(theta)-128sin(theta)-i4alphacos(40^@)+4alphasin(40^@))#

Separating the real and imaginary parts into two equations:

#1024 = 4alphasin(40^@)-128sin(theta)" [5]"#

#i0 = i128cos(theta)-i4alphacos(40^@)" [6]"#

Solve equation [6] for #alpha# in terms of #theta#:

#0 = 32cos(theta)-alphacos(40^@)#

#alpha = 32cos(theta)/cos(40^@)" [6.1]"#

Substitute equation [6.1] into equation [5]:

#1024 = 4(32cos(theta)/cos(40^@))sin(40^@)-128sin(theta)#

Multiply both sides by #cos(40^@)/128#:

#8cos(40^@) = cos(theta)sin(40^@)-sin(theta)cos(40^@)#

Multiply both sides by -1:

#-8cos(40^@) = sin(theta)cos(40^@)-cos(theta)sin(40^@)#

The right side is the identity for #sin(A-B)# where #A=theta# and #B = 40^@#:

#sin(theta-40^@) = -8cos(40^@)#

The above equation does not have a solution, because the right side is outside of the the range of the sine function, #-1 <= sin(x) <= 1#. This means that there are not real values for #theta# and #alpha#