# Complex Numbers

• Jan 13th 2008, 09:53 PM
noreaction
Complex Numbers
I got this question in a complex numbers test and I could not figure it out. Any help would be appreciated.

Question:

R is a positive number and $z_1$ and $z_2$ are complex numbers. Show that the points which represent respectively the numbers $z_1$, $z_2$, $\frac{z_1-iRz_2}{1-iR}$ form the vertices of a right angled triangle
• Jan 13th 2008, 11:06 PM
CaptainBlack
Quote:

Originally Posted by noreaction
I got this question in a complex numbers test and I could not figure it out. Any help would be appreciated.

Question:

R is a positive number and $z_1$ and $z_2$ are complex numbers. Show that the points which represent respectively the numbers $z_1$, $z_2$, $\frac{z_1-iRz_2}{1-iR}$ form the vertices of a right angled triangle

Put $z_3=\frac{z_1-iRz_2}{1-iR}$ then the lengths of the sides of the triangle are $|z_1-z_2|$ , $|z_2-z_3|$ and $|z_1-z_3|.$

Now you need to show that these satisfy Pythagoras's theorem, and then
by the converse of Pythagoras's the points form a right triangle.

RonL
• Jan 13th 2008, 11:10 PM
noreaction
Thank you. I can't believe I didn't think of that.
• Jan 14th 2008, 01:51 PM
Opalg
Quote:

Originally Posted by noreaction
R is a positive number and $z_1$ and $z_2$ are complex numbers. Show that the points which represent respectively the numbers $z_1$, $z_2$, $\frac{z_1-iRz_2}{1-iR}$ form the vertices of a right angled triangle.

Let $z_3 = \frac{z_1-iRz_2}{1-iR}$.

The displacement from z_1 to z_3 is given by $z_3-z_1 = \frac{z_1-iRz_2}{1-iR} - z_1 = \frac{iR(z_1-z_2)}{1-iR}$.

The displacement from z_2 to z_3 is given by $z_3-z_2 = \frac{z_1-iRz_2}{1-iR} - z_2 = \frac{z_1-z_2}{1-iR}$.

Therefore $z_3-z_1 = iR(z_3-z_2)$. So to get from $z_3-z_2$ to $z_3-z_1$ you have to multiply it by R (which changes its length but not its direction) and then multiply it by i (which has the effect of rotating it through a right angle. This shows that the line from z_1 to z_3 is perpendicular to the line from z_2 to z_3.