Triangle inequality

1. Oct 2, 2009

sara_87

1. The problem statement, all variables and given/known data

Use the triangle inequality to show:
$$\left|z^2+3\right|$$ $$\leq(12)$$ for $$\left|z\right|$$=3
where z is a complex number

2. Relevant equations

triangle inequality: $$\left|z_1+z_2\right|$$$$\leq$$ $$\left|z_1\right|$$+$$\left|z_2\right|$$

3. The attempt at a solution

I understand the triangle inequality but i cant seem to do the question.

2. Oct 2, 2009

defunc

Simply substitute z1 with z^2 and z2 with 3.

3. Oct 2, 2009

sara_87

but how do i use the information: 'on abs(z)=3' ?

4. Oct 2, 2009

sara_87

oh i see:
like this:
$$\left|z^2+3\right|$$$$\leq$$$$\left|z^2\right|$$+$$\left|3\right|$$
$$\left|z^2+3\right|$$$$\leq$$9+3=12
am i missing something?

5. Oct 2, 2009

defunc

Thats right. And note that |z^2|= |z|^2

6. Oct 2, 2009

thank you.

7. Oct 3, 2009

sara_87

I'm working on a similar question but this time, i must show that:
$$\left|z^2(2+i)+1\right|$$ $$\geq$$ 1 for $$\left|z\right|$$=1

this time we can use the triangle inequality:
$$\left|z_1-z_2\right|$$ $$\geq$$ $$\left|abs(z_1)-abs(z_2)\right|$$

do i substitute: z1 with z2(2+i) and z2 with -1 ?
because when i do this, i get:
$$\left|z^2(2+i)+1\right|$$ $$\geq$$ $$\left|z^2(2+i)-1\right|\right|\left|$$
I think im making a mistake

8. Oct 3, 2009

defunc

Your substitution is correct, but evaluate the absolute values on the right hand side

9. Oct 3, 2009

sara_87

after evaluating the absolute values, the right hand side would look like:
2z^2 +iz^2-1
so somehow i must show that this is 1
?

10. Oct 3, 2009

defunc

The absolute value a complex number a+ib is
Sqrt(a^2+b^2), and thus a real number. Further you know abs(z)=1. And further abs(xy) = abs(x)abs(y). See what you can do with that

11. Oct 3, 2009

sara_87

oh right, thanks, so i will get:
right hand side:
abs(1*abs(2+i)-1)=abs(4+i^2-1)=abs(sqrt(3)-1)=sqrt(3+1)=2
?? not 1 ?