Complex Analysis - Proving an inequality

[NewtonianAlch]

Homework Statement

Show that if |z| = 10 then 497 ≤ |z^{3} + 5iz^{2} − 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^{2}+500^{2}) = 1115.35

497 ≤ 1115.35 ≤ 1503

 

[Office_Shredder]

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)

 

[NewtonianAlch]

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?

 

[Office_Shredder]

|z1 + z2| ≤ |z1| + |z2|

For example, from this you can also show
| z_1 + z_2 + z_3 | \leq |z_1| + |z_2| + |z_3|

 

[NewtonianAlch]

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 ?

 

[Office_Shredder]

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

 

[NewtonianAlch]

OK.

|z1 + z2 + z3| \geq 497

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

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

Simplifying equation to 1 + 5i \geq 1/2

Am I even remotely in the right direction?

Really appreciate your help by the way. Thanks for that.

 

[NewtonianAlch]

I'm still stuck on this problem if anyone can help that'd be great =)

 

[NewtonianAlch]

Nvm, it is solved now! Can't believe it was actually very easy, once the tutor did it in class, I was like doh!