Complex Analysis - Proving an inequality

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
8 replies · 3K views
NewtonianAlch
Messages
453
Reaction score
0

Homework Statement


Show that if |z| = 10 then 497 ≤ |z[itex]^{3}[/itex] + 5iz[itex]^{2}[/itex] − 3| ≤ 1503.



The Attempt at a Solution



I'm not an entirely sure how to begin this one, or if what I'm doing is correct.

If I sub in |z| = 10 into the equation; |1000 + 500i - 3| = 997 +500i

Then the modulus of that is sqrt(997[itex]^{2}[/itex]+500[itex]^{2}[/itex]) = 1115.35

497 ≤ 1115.35 ≤ 1503
 
Physics news on Phys.org
You subbed in z=10, but there are many other values of z with |z|=10 (for example, 10i, 5sqrt(2)+5sqrt(2)i, etc.)

You just have to apply the triangle inequality (in a different form for each bound)
 
I took a look at triangle inequalities for Complex Analysis, and there are essentially two types?

One for the upper bound and one for the lower:

|z1 + z2| ≤ |z1| + |z2|

|z1 + z2| ≥ ||z1| − |z2||

But I'm not too sure how that would fit in here.

Would I need to find the roots of z as a starting point?
 
It's really not clicking still.

So I'm meant to show 497 ≤ |z1| + |z2| + |z3|, so would z1 = z^3, z2 = 5iz^2, and z3 = -3 ?
 
NewtonianAlch said:
It's really not clicking still.

So I'm meant to show 497 ≤ |z1| + |z2| + |z3|, so would z1 = z^3, z2 = 5iz^2, and z3 = -3 ?

That wouldn't prove anything, since 497 ≤ |z1| + |z2| + |z3| does not imply that 497 ≤ |z1 + z2+ z3|

However, the other direction might be more fruitful
 
OK.

|z1 + z2 + z3| [itex]\geq[/itex] 497

z1 = z^3
z2 = 5iz^2
z3 = -3

z^3 + 5iz^2 [itex]\geq[/itex] 500
|z| = 10, so |z|^2 = 100

Simplifying equation to 1 + 5i [itex]\geq[/itex] 1/2

Am I even remotely in the right direction?

Really appreciate your help by the way. Thanks for that.
 
I'm still stuck on this problem if anyone can help that'd be great =)
 
Nvm, it is solved now! Can't believe it was actually very easy, once the tutor did it in class, I was like doh!