# Comple numbers and circles

• Sep 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.

• Sep 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 $\displaystyle \sqqrt{5}-1\leq |z-3|\leq \sqrt{5}+1$

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 $\displaystyle |z-3|$ and transform it in a way that brings $\displaystyle |z-(1+i)|=1$ into play, like this:

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

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

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

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

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

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

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