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Show that Im(2 + z +3z^4) <= 4 when |z|<=1

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The regular way is to go |2 + z + 3z^4| <= 2 + |z| + 3|z|^4

then we know |z|=1 so 2+1+3 =6 and we want <=4, so im missing some sort of identity or substitution .

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- Jan 11th 2010, 04:16 PMscubasteve123Imaginary Part
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Show that Im(2 + z +3z^4) <= 4 when |z|<=1

_

The regular way is to go |2 + z + 3z^4| <= 2 + |z| + 3|z|^4

then we know |z|=1 so 2+1+3 =6 and we want <=4, so im missing some sort of identity or substitution . - Jan 11th 2010, 04:22 PMDrexel28
- Jan 11th 2010, 04:42 PMscubasteve123
- Jan 11th 2010, 04:43 PMDrexel28
- Jan 11th 2010, 04:55 PMscubasteve123
- Jan 11th 2010, 06:58 PMDrexel28
- Jan 12th 2010, 02:15 AMdifferentiate
It is given that mod z = 1

that means that

x^2 + y^2 = 1

Let z = X + iy

subtitute this into the given expression, and then take the imaginary part. Substitute x^2 + y^2 =1 into the imaginary part wherever possible - Jan 12th 2010, 02:16 AMtsunamijon
Hopefully this is right....

Im(2 + z +3z^4)

let, z= rcosX+isinX

z^n = r^n {cos(nX)+isin(nX)} by de moivre

Im(2 + z +3z^4) = Im(2 + r{cosX+isinX} + 3r^4 cos(4X)+isin(4X))

= rsinX + 3r^4(sin{4X})

but |z|<=1, so r<=1 for all X

so Im(2 + z +3z^4) <= 1 + 3 = 4

QED