Thread: Imaginary Part

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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 .

Originally Posted by scubasteve123

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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 .

I think you are misinterpreting the question. It asks for NOT

Originally Posted by Drexel28

I think you are misinterpreting the question. It asks for NOT

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okay but I have another practice question where |Re(2+z +z^3)|<=4 and thats the method that was used there so I thought it would be similar but i suppose not. Are there any ideas that would help me?

Originally Posted by scubasteve123

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okay but I have another practice question where |Re(2+z +z^3)|<=4 and thats the method that was used there so I thought it would be similar but i suppose not. Are there any ideas that would help me?

but that doesn't help here, does it?

Originally Posted by Drexel28

but that doesn't help here, does it?

No I found that property also and cant find much use unfortunately

Originally Posted by scubasteve123

No I found that property also and cant find much use unfortunately

So how do you think we should go about doing this?

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

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