# [SOLVED]Triangle inequality question

#### dwsmith

##### Well-known member
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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#### Plato

##### Well-known member
MHB Math Helper
I am trying to show $|(n+z)^2|\leq n^2 -|z|^2$ where is complex
I don't think that you really mean that.
$|(2+i)^2|=|3+4i|=5$
$(2)^2-|i|^2=3$

• dwsmith

#### dwsmith

##### Well-known member
I don't think that you really mean that.
$|(2+i)^2|=|3+4i|=5$
$(2)^2-|i|^2=3$
This may help (below) then but I figure the above was just a straight application.
$$\sum_{n = N+1}^{ \infty}\frac{1}{(z+n)^2}$$
$R>0$ 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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#### Prove It

##### Well-known member
MHB Math Helper
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.
Are n and z both complex? Or is only z complex?

#### dwsmith

##### Well-known member
Are n and z both complex? Or is only z complex?
z and n is a natural