Show that $|z+1|\le|z+1|^2 +|z|$ for all $z \in \mathbb{C}$

Question

Question: Show that $|z+1|\le|z+1|^2 +|z|$ for all $z \in \mathbb{C}$

So far I have,

Suppose $1\le|z+1|$

$|z+1|\le|z+1|^2$

$|z+1|\le|z+1|^2+|z|$

Now I must show $|z+1|<1$ but this is where I'm stuck

Answer 1

Ok but its a complex numbers question so you should try to understand it in those terms. You have a triangle with vertices the origin, $-1$ and $z$. The side lengths are $1$, $|z|$, and $|z+1|$ so we have $$1 \leq |z+1|+|z|$$ in addition $$2|z+1|\leq |z+1|^2 +1$$ adding we get $$2|z+1|+1\leq |z+1|^2 +1+|z+1|+|z|$$

which gives $$|z+1|\leq |z+1|^2 +|z|$$

(the second inequality follows from ($0 \leq (|z+1|-1)^2$)

Answer 2

Here is a 'less complex' approach:

First, let $w=z+1$, then we want to prove $|w| \le |w|^2+ |w-1|$.

Now let $w=r e^{i \theta}$, and note that $|w-1| = \sqrt{r^2+1-2 r \cos \theta} \ge |1-r|$.

Since $|w| = r$, if we can show $r \le r^2 + |1-r|$, then we are finished.

If $r \ge 1$ this is clearly true because $r^2 \ge r$.

If $r <1$, we have $0 \le r^2 -2r +1 = (r-1)^2$ which is clearly true.

Answer 3

My approach is similar to that of Rene Schipperus, but approaches the problem from a slightly different direction.

We wish to prove that $|z+1| \le |z+1|^2 + |z|$.

Subtracting both sides of the inequality by $|z|$ and multiplying by $|z+1|+|z|$ yields

$$|z+1|^2 - |z|^2 \stackrel?\le |z+1|^2 (|z+1|+|z|)$$

There's just one slightly awkward term here, the $-|z|^2$, so let's isolate that on one side and see what we get:

$$-|z|^2 \stackrel?\le |z+1|^2(|z+1|+|z|-1).$$

Now look at the term $|z+1|+|z|$: that's the sum of the distance between $z$ and $-1$ and the distance between $z$ and $0$. By the triangle inequality, $|z+1| + |z| \ge |-1 - 0| = 1$. Thus the term $|z+1|+|z|-1$ is at least $0$. Multiplying by $|z+1|^2$ doesn't change that, so $$-|z|^2 \le 0 \le |z+1|^2 (|z+1|+|z|-1).$$ All the steps above are reversible (note in particular that dividing by $|z+1|+|z|$ is permissible because we have shown it is strictly greater than $0$).

Note

If you're wondering how I decided to subtract by $|z|$ and multiply by $|z+1|^2 + |z|$, well … I can't quite tell that whole story. I was generally thinking that it's easier to think about squares of complex magnitudes than about the magnitudes themselves, and producing a difference of squares sometimes makes square things without making too many non-square things. The fact that it led to something useful seems to be more luck than anything else.

Answer 4

From the triangle inequality,

$$ \begin{align} |z| + |z+1|^2 - |z+1| &= |z| + |z+1| - 1 + (|z+1|-1)^2 \\ &\geq |1+z-z| - 1 + (|z+1|-1)^2 \\ &= (|z+1|-1)^2 \\ &\geq 0. \end{align} $$