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4u complex numbers (1 Viewer)

jnney

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why does z^n=1 have a modulus of 1?

If w is a complex root of z^5-1=0 with the smallest positive argument, show athat w2, w3 and w4 are other complex roots. prove that


a) 1 + w + w^2 + w^3 + w^4 = 0

I've done this bit up to the part where it becomes a zero.......I have no idea how it becomes 0.

b) Find the quadratic equation whos roots are a=w+w^4 and b=w^2+w^3.
 

thoth1

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well:

take modulus of both sides


therefore modulus of z = 1.
 

jnney

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So if z^n = 2

the modulus would be 2?
 

thoth1

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z^5-1 = 0
(w-1)(1+w+w^2+w^3+w^4) = 0
then u say that since w is a complex root w-1 is not equal to zero. therefore 1+w+....=0.

b) sum of roots = w+w^2+w^3+w^4 = -1 (from part a)
product of roots = = -1 (from part a)
(note: w^5=1)

therefore sum of roots = -1
product of roots = -1

therefore equation: x^2-(sum of roots)x+(product of roots) = 0
therefore: x^2+x-1=0
 
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Hermes1

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z^n = 2
(z^n)=(2)
(z)^n=2
(z)=

the brackets are supposed to be modulus signs.
 

Parvee

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You're starting 4u already? :eek:
 

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