really hard complex number question (1 Viewer)

[andy.sanders]

Z^4+1=0 factorise into quadratic factors and deduce this

cos2Θ=2(cosΘ-cos45)(cosΘ-cos135)

 

[andy.sanders]

i figured it out just then however if you know a way which could be easier that would be great

 

[bored of sc]

Z^4+1=0 factorise into quadratic factors and deduce this

cos2Θ=2(cosΘ-cos45)(cosΘ-cos135)

Step 1: Write in mod-arg form and add 2kpi to argument
i.e. -1 = cis(pi+2kpi) = cis[(1+2k)pi]

z = (-1)^1/4
= [cis(1+2k)pi]^1/4

Step 2: De Moivre's theorem
z = cis[(1+2k)pi]/4 where k = -2, -1, 0, 1

z = cis(-3pi/4), cis(-1pi/4), cis(pi/4), cis(3pi/4) --- as you can see they are conjugate pairs

Step 3: Factorise --> [z-cis(-3pi/4)][z-cis(3pi/4)][z-cis(pi/4)][z-cis(-pi/4)] = 0

Now make conjugates into quadratics.

(z²-cis(3pi/4)z-cis(-3pi/4)z+cis0)(z²-cis(-pi/4)z-cis(pi/4)z+cis0)

Now as a side bit take the middle bit by themselves,
-cos3pi/4-isin3pi/4-cos-3pi/4-isin-3pi/4
= -cos3pi/4-isin3pi/4-cos3pi/4+isin3pi/4 --- since cosA = cos-A and sinA = -sinA
= -2cos3pi/4
-cos-pi/4-isin-pi/4-cospi/4-isinpi/4
= -cospi/4+isinpi/4-cospi/4-isinpi/4
= -2cospi/4

therefore factored it is:
(z²-2cos3pi/4.z+1)(z²-2cospi/4.z+1) = 0

cos2Θ = 2(cosΘ-cos45)(cosΘ-cos135)
Let Θ = pi/4
RHS = 2(cospi/4-cospi/4)(cospi/4-cos3pi/4)
= 2.0.2cospi/4
= 0
= cos90
= cos2.pi/4
But Θ = pi/4
thus
= cos2Θ
= LHS

That second part is intense.

 

[bored of sc]

The second bit I did from cos2theta onwards is absolute garbage. Actually the first might also be garbage too.

 

[tommykins]

bosc - the question states 'deduce', which I believe is more likely to be a prove that LHS = RHS rahter than RHS = LHS.

 

[youngminii]

What. The. Fuck.
Did not understand jackall

 

[andy.sanders]

Step 1: Write in mod-arg form and add 2kpi to argument
i.e. -1 = cis(pi+2kpi) = cis[(1+2k)pi]

z = (-1)^1/4
= [cis(1+2k)pi]^1/4

Step 2: De Moivre's theorem
z = cis[(1+2k)pi]/4 where k = -2, -1, 0, 1

z = cis(-3pi/4), cis(-1pi/4), cis(pi/4), cis(3pi/4) --- as you can see they are conjugate pairs

Step 3: Factorise --> [z-cis(-3pi/4)][z-cis(3pi/4)][z-cis(pi/4)][z-cis(-pi/4)] = 0

Now make conjugates into quadratics.

(z²-cis(3pi/4)z-cis(-3pi/4)z+cis0)(z²-cis(-pi/4)z-cis(pi/4)z+cis0)

Now as a side bit take the middle bit by themselves,
-cos3pi/4-isin3pi/4-cos-3pi/4-isin-3pi/4
= -cos3pi/4-isin3pi/4-cos3pi/4+isin3pi/4 --- since cosA = cos-A and sinA = -sinA
= -2cos3pi/4
-cos-pi/4-isin-pi/4-cospi/4-isinpi/4
= -cospi/4+isinpi/4-cospi/4-isinpi/4
= -2cospi/4

therefore factored it is:
(z²-2cos3pi/4.z+1)(z²-2cospi/4.z+1) = 0

cos2Θ = 2(cosΘ-cos45)(cosΘ-cos135)
Let Θ = pi/4
RHS = 2(cospi/4-cospi/4)(cospi/4-cos3pi/4)
= 2.0.2cospi/4
= 0
= cos90
= cos2.pi/4
But Θ = pi/4
thus
= cos2Θ
= LHS

That second part is intense.

ye i did that the same way but now i think that question is quite bad seeing as all the text books dont have anything similar

 

[andy.sanders]

Step 1: Write in mod-arg form and add 2kpi to argument
i.e. -1 = cis(pi+2kpi) = cis[(1+2k)pi]

z = (-1)^1/4
= [cis(1+2k)pi]^1/4

Step 2: De Moivre's theorem
z = cis[(1+2k)pi]/4 where k = -2, -1, 0, 1

z = cis(-3pi/4), cis(-1pi/4), cis(pi/4), cis(3pi/4) --- as you can see they are conjugate pairs

Step 3: Factorise --> [z-cis(-3pi/4)][z-cis(3pi/4)][z-cis(pi/4)][z-cis(-pi/4)] = 0

Now make conjugates into quadratics.

(z²-cis(3pi/4)z-cis(-3pi/4)z+cis0)(z²-cis(-pi/4)z-cis(pi/4)z+cis0)

Now as a side bit take the middle bit by themselves,
-cos3pi/4-isin3pi/4-cos-3pi/4-isin-3pi/4
= -cos3pi/4-isin3pi/4-cos3pi/4+isin3pi/4 --- since cosA = cos-A and sinA = -sinA
= -2cos3pi/4
-cos-pi/4-isin-pi/4-cospi/4-isinpi/4
= -cospi/4+isinpi/4-cospi/4-isinpi/4
= -2cospi/4

therefore factored it is:
(z²-2cos3pi/4.z+1)(z²-2cospi/4.z+1) = 0

cos2Θ = 2(cosΘ-cos45)(cosΘ-cos135)
Let Θ = pi/4
RHS = 2(cospi/4-cospi/4)(cospi/4-cos3pi/4)
= 2.0.2cospi/4
= 0
= cos90
= cos2.pi/4
But Θ = pi/4
thus
= cos2Θ
= LHS

That second part is intense.

wait a sec cosΘ has 4 solutions you missed out two i think but anyway it doesnt matter.

 

[OLDMAN]

Just concerned that the second part has not been answered properly.
Z^4+1=(z2-2cos3pi/4.z+1)(z2-2cospi/4.z+1)
Divide both sides by z^2, to get the z+1/z =2cosΘ structure which would be useful.
Seen a similar problem like this in the Blue Patel.