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I'm trying to get some practice using the Maximum Modulus theorem, and want to use it to conclude that the maximum of $(x^2-y^2-1)^2+4x^2y^2$ occurs at $x=0$, $y=\pm 1$, supposing $x^2+y^2\leq 1$.

My thinking is I want to find some suitalbe complex function $f(z)$ such that $|f(z)|^2=(x^2-y^2-1)^2+4x^2y^2$, and then apply the Maximum Modulus principle, since I know the maximum will occur somewhere on the boundary $x^2+y^2=1$. However, I can't find such a function, so maybe I"m approaching it incorrectly.

I did manage to find that $$ z^2-1=(x+iy)^2-1=(x^2-y^2-1)+2xyi $$ which looked somewhat close, but no cigar. How can this be done better? Thanks.

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    $(x+iy)^2 - 1 = (x^2 - y^2 - 1) + 2ixy$. So why no cigar?2012-03-31
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    @Aryabhata My mistake, I just noticed I wrote that wrong. I'm going to rethink this.2012-03-31

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Take $f(z) = z^2 - 1$. If $z = x + iy$, then $|f(z)|^2 =$ the expression you have.

EDIT: Strangely, I missed that you already found $z^2-1$. You have a mistake in your working:

$$(x+iy)^2 - 1 = (x^2 - y^2 - 1) + 2ixy$$

and

$$|(x+iy)^2 - 1|^2 = (x^2 - y^2 - 1)^2 + 4x^2y^2$$

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    Actually $|f(z)|^2$.2012-03-31
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    @RobertIsrael: Right. Thanks!2012-03-31
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    @anon: Thanks for the edit.2012-03-31
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    Ok, so I now have the expression, and I substitute $1-x^2$ for $y^2$, which simplifies to $-4x^4+4$, so it obtains its maximum when $x=0$. Thanks!2012-03-31
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    @Cowbell: I don't get exactly that (the $x^4$ terms seems to cancel out), but yes, something like that would work.2012-03-31
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$x^2+y^2=1$

$(x^2-y^2-1)^2+4x^2y^2=(x^2+y^2)^2-2(x^2-y^2)+1=4-4x^2\leq 4$

$x=0,y=±1$