Complex No. Q (Make that two) (1 Viewer)

[Sparcod]

Two Questions today.
1.Solve 4z^5 +z=0 over complex field.

2.Prove cos 4@ in terms of cos @ is
8 cos^4 +8cos^2@+1)
(argghhh the link)
Hence solve 8x^4- 8x^2 +1=0
Hence find cos n/8 and cos 5n/8 (n is pi)


The teacher gave us these questions from a past paper.

 

[Riviet]

For 1, take the z out:
z(4z⁴+1)=0
.'. z=0 or z⁴=-1/4
Use De Moivre's Theorem or any other convenient method to solve the second equation.

 

[gyponese]

for the first one, take Z as a common factor, then Z = 0 , and Z^4 = -1.. you need to use general solution so solve that (i think for that one its (2kpie + pie)/4 )

for the second part.. expand (Cos@ + iSin@)^4 using the binomial theorem, and also using demovire's theorem.. equate the two, then equate real and imaginary, and there u go..

 

[Sparcod]

I don't get the equating real parts bit. Thanks anyway. I'll try this myself.

For 1, take the z out:
z(4z⁴+1)=0
.'. z=0 or z⁴=-1/4
Use De Moivre's Theorem or any other convenient method to solve the second equation.

z^4 =-1/4
= 1/4 cis n

z= 1/ v2 cis n/4 where n is pi (I can't see where the 4 solutions are)

am i right there?

 

[Riviet]

z⁴=1/4.cis(pi+2k.pi), where k is an integer [the 2k.pi is just a rotation of the root about 2pi, or in other words the general solution to this equation]
z=2^-1/2.cis[(pi+2k.pi)/4], where k=0, +1, 2
Substitute each value of k to obtain your 4 solutions to this equation.

 

[icycloud]

z^4 = -1/4
Let z = r cis K
r^4 cis 4K = 1/4 cis Pi
r = 1/Sqrt[2]
4K = Pi + 2kPi

Thus, z_k = 1/Sqrt[2] Cis[ (2k+1) Pi/4] k = -2,-1,0,1