Thread: Complex Variables - Upper Bound

1. Complex Variables - Upper Bound

I'm starting to work on this one question, but I'm not sure how to apply this inequality in it:

$\left|\left|z_1\right|-\left|z_2\right|\right|\leq\left|z_1+z_2\right|\le q\left|z_1\right|+\left|z_2\right|$

The question is : Find an upper bound for $\left|\frac{-4}{z^3-5z+1}\right|$ if $\left|z\right|=2$.

Now, I managed (in a previous question) to show that the lower and upper bounds on $\left|z^3-5z+1\right|$ when $\left|z\right|=2$ and I got $3\leq\left|z^3-5z+1\right|\leq 19$....I have a feeling I need to incorporate this, but I'm not sure how.

Any input would be appreciated!!

2. Originally Posted by Chris L T521
I'm starting to work on this one question, but I'm not sure how to apply this inequality in it:

$\left|\left|z_1\right|-\left|z_2\right|\right|\leq\left|z_1+z_2\right|\le q\left|z_1\right|+\left|z_2\right|$

The question is : Find an upper bound for $\left|\frac{-4}{z^3-5z+1}\right|$ if $\left|z\right|=2$.

Now, I managed (in a previous question) to show that the lower and upper bounds on $\left|z^3-5z+1\right|$ when $\left|z\right|=2$ and I got $3\leq\left|z^3-5z+1\right|\leq 19$....I have a feeling I need to incorporate this, but I'm not sure how.

Any input would be appreciated!!
I wonder how you got the bounds on $|z^3-5z+1|$: consider $z=2$... The best lower bound is $|z^3-5z+1|\geq |5z|-|z^3+1|\geq 5|z|-|z|^3-1=1$.

Then all the work is done: $\left|\frac{-4}{z^3-5z+1}\right|=\frac{4}{|z^3-5z+1|}\leq \frac{4}{1}=4$ by the previous lower bound...

3. Originally Posted by Laurent
I wonder how you got the bounds on $|z^3-5z+1|$: consider $z=2$... The best lower bound is $|z^3-5z+1|\geq |5z|-|z^3+1|\geq 5|z|-|z|^3-1=1$.
To satisfy your curiosity, this is how I got the bounds:

L.B: $\left|z^3-5z+1\right|=\left|z^3-\left(5z-1\right)\right|\geqslant\left|\left|z^3\right|-\left|5z-1\right|\right|\geqslant\left|\left|z\right|^3-\left(5\left|z\right|+1\right)\right|=\left|-3\right|=3$

U.B: $\left|z^3-5z+1\right|=\left|\left(z^3-5z\right)+1\right|\leqslant\left|\left|z^3-5z\right|+1\right|\leqslant\left|\left|z\right|^3+ 5\left|z\right|+1\right|=19$

Then all the work is done: $\left|\frac{-4}{z^3-5z+1}\right|=\frac{4}{|z^3-5z+1|}\leq \frac{4}{1}=4$ by the previous lower bound...
That makes sense. I figured that I had to do something like this.

Thank you!

4. Originally Posted by Chris L T521
To satisfy your curiosity, this is how I got the bounds:

L.B: $\left|z^3-5z+1\right|=\left|z^3-\left(5z-1\right)\right|\geqslant\left|\left|z^3\right|-\left|5z-1\right|\right|{\color{red}\geqslant}\left|\left|z \right|^3-\left(5\left|z\right|+1\right)\right|=\left|-3\right|=3$
I feel like my previous remark about $z=2$ was too allusive. I was pointing at a mistake: it lies in the inequality in red. Indeed, the function $x\mapsto|x|$ is not increasing, so you can't write what you did: $\left|z^3\right|-\left|5z-1\right|\leq |z|^3-5|z|+1$ does not imply $\left|\left|z^3\right|-\left|5z-1\right|\right|\leq \left||z|^3-5|z|+1\right|$.