(Complex analysis). Show that the inequality holds

[richyw]

Homework Statement

Show that the inequality$$\left|\frac{z^2-2z+4}{3x+10}\right|\leq3$$holds for all $z\in\mathbb{C}$ such that $|z|=2$

Homework Equations

Triangle inequality

The Attempt at a Solution

I'm not really sure how to go about this. the x is throwing me off. Should I write it out with $z=x+iy$?

 

[Mark44]

Homework Statement

Show that the inequality$$\left|\frac{z^2-2z+4}{3x+10}\right|\leq3$$holds for all $z\in\mathbb{C}$ such that $|z|=2$

Homework Equations

Triangle inequality

The Attempt at a Solution

I'm not really sure how to go about this. the x is throwing me off. Should I write it out with $z=x+iy$?

Could it be a typo and they really intended x to be z? That x really looks out of place to me.

 

[richyw]

I don't know if it is a typo. If it is not then I could say that

$$3x+10=Re(3z+10)$$$$|Re(3z+10)|\leq | 3z+10 | = 3|z|+10=16$$

 

[richyw]

on the top I say that $$|z^2-2z+4|\leq |z|^2+2|z|+4 = 16$$

 

[richyw]

not sure how that would help me...

 

[dirk_mec1]

Must be a type: suppose x => -10/3 and y => sqrt(4-(10/3)^2)* i with z=x+yi then the denominator heads towards zero and the fraction goes to inf.

 

[davidmoore63@y]

An outline of a solution, assuming the d should be a z.

Think about what the numerator and the denominator do to the circle mod z=2.

The denominator converts it to a circle of radius 6, 10 units shifted to the right. Thus the distance from the origin ranges from 4 to 12.

The numerator is a bit more tricky. Thinking of it as (z-1)^2 +3 you we that it first shifts the circle to the left one unit (making the distance to origin range from 1 to 3). Then you square it, which meAns the distance ranges from 1 to 9, with these extrema being on the real axis. Then you add 3 to get a distance range of 4 to 12.

Thus both numerator and denominator have modulus between 4 and 12, so their ratio cannot be greater than 3