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- Thread starter dwsmith
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I don't think that you really mean that.I am trying to show $|(n+z)^2|\leq n^2 -|z|^2$ where is complex

$|(2+i)^2|=|3+4i|=5$

$(2)^2-|i|^2=3$

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This may help (below) then but I figure the above was just a straight application.I don't think that you really mean that.

$|(2+i)^2|=|3+4i|=5$

$(2)^2-|i|^2=3$

$$

\sum_{n = N+1}^{ \infty}\frac{1}{(z+n)^2}

$$

$R>0$ and $N>2R$. For the inequality, we are assuming $|z|<R$. Additionally, $n>N$.

So I want to show

$$

\left|\frac{1}{(z+n)^2}\right|\leq\frac{1}{(n-|z|^)2}

$$

So if that is true, then $|(z+n)^2|\leq (n -|z|)^2$.

---------- Post added at 04:22 PM ---------- Previous post was at 03:25 PM ----------

$|(n + z)^2|\leq n^2 + 2n|z| + |z|^2\leq n^2 - 2n|z| + |z|^2\leq (n - |z|)^2$

Is this correct?

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Are n and z both complex? Or is only z complex?I am trying to show $|(n+z)^2|\leq (n -|z|)^2$ where is complex

$|(n+z)^2| = |n^2 + 2nz + z^2| \leq n^2 + 2n|z| + |z|^2$ But I can't figure out the connection for the final piece.

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z and n is a naturalAre n and z both complex? Or is only z complex?