Triangle Inequality: Solving |z^2+3| ≤ (12) for |z|=3

[sara_87]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

I understand the triangle inequality but i can't seem to do the question.

 

[defunc]

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

 

[sara_87]

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

 

[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|\leq9+3=12
am i missing something?

 

[defunc]

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

 

[sara_87]

thank you.

 

[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 I am making a mistake

 

[defunc]

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

 

[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
?

 

[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

 

[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 ?