# Comple numbers and circles

• September 12th 2010, 05:23 AM
berachia
Comple numbers and circles
Let S be the interior of the circle |z 1 i| = 1. Show, by using suitable inequalities for |z1 ± z2|, that if z S then 5 1 < |z 3| < 5 + 1.

Obtain the same result geometrically by considering the line containing the centre of the circle and the point 3.

• September 12th 2010, 06:42 AM
Failure
Quote:

Originally Posted by berachia
Let S be the interior of the circle |z 1 i| = 1. Show, by using suitable inequalities for |z1 ± z2|, that if z S then 5 1 < |z 3| < 5 + 1.

I don't think that this is exactly right, it should be $\sqqrt{5}-1\leq |z-3|\leq \sqrt{5}+1$

Quote:

Obtain the same result geometrically by considering the line containing the centre of the circle and the point 3.

So for the algebraic part you start with $|z-3|$ and transform it in a way that brings $|z-(1+i)|=1$ into play, like this:

$(*)\qquad |z-3|=|(z-(1+i))+(-2+i)|$

Now, you should remember that $|z_1+z_2|\leq |z_1|+|z_2|$ (triangle inequality) and, similarly, $||z_1|-|z_2||\leq |z_1+z_2|$ (a consequence of the triangle inequality).

Now set $z_1 := z-(1+i)$ and $z_2 := -2+i$ in (*) to get

$|z-3|\leq |z-(1+i)|+|-2+i|=1+\sqrt{5}$,

because we are allowed to assume that $|z-(1+i)|=1$, and

$\sqrt{5}-1=||z-(1+i)|-|-2+i||\leq |z-3|$

For the geometrical part: the values of z that satisfy $|z-(1+i)|=1$ lie on a circle with center 1+i and radius 1, the values of z that satisfy $|z-3|=\sqrt{5}\pm 1$ lie on one of the two circles with center 3 and radii $\sqrt{5}\pm 1$.