# locus of complex no.

• May 19th 2008, 10:22 AM
afeasfaerw23231233
locus of complex no.
p293 q15c
question: Suppose that $arg \frac {z-1}{z+1} =\frac {\pi}4$where z is a complex no.
(a) let z = x + yi , where x and y are real. Show that the locus of z is an arc of a circle.
(b) Find the radius and centre of the circle found in (a)
(c) Find the farthest point from the origin on the locus

i've done (a)(b) correctly but don't know how to do (c).
My working:
(a) $\frac {x-1+yi}{x+1+yi}$
$= \frac {x^2 +y^2-1+2yi} {x^2+y^2 +2x+1}$
$\frac{2y}{x^2+y^2-1} = tan \frac{\pi} 4 = 1$
$x^2+y ^2-2y-1=0$
b) centre=(0,-1) radius = $\sqrt 2$

don't know how to do (c). Thanks.
• May 20th 2008, 12:06 AM
CaptainBlack
Quote:

Originally Posted by afeasfaerw23231233
p293 q15c
question: Suppose that $arg \frac {z-1}{z+1} =\frac {\pi}4$where z is a complex no.
(a) let z = x + yi , where x and y are real. Show that the locus of z is an arc of a circle.
(b) Find the radius and centre of the circle found in (a)
(c) Find the farthest point from the origin on the locus

i've done (a)(b) correctly but don't know how to do (c).
My working:
(a) $\frac {x-1+yi}{x+1+yi}$
$= \frac {x^2 +y^2-1+2yi} {x^2+y^2 +2x+1}$
$\frac{2y}{x^2+y^2-1} = tan \frac{\pi} 4 = 1$
$x^2+y ^2-2y-1=0$
b) centre=(0,-1) radius = $\sqrt 2$

don't know how to do (c). Thanks.

You will have noticed that the question mentioned an "arc". Now the equation:

$
\frac{2y}{x^2+y^2-1} = \tan\left( \frac{\pi} 4\right) = 1
$

for points on circle of radius $\sqrt{2}$ implies that $y \ge 0$, so the arc in question is that part of the circle not below the real axis, and the furtherest points from the origin on this arc are the ends where $y=0$.

(A sketch will help)

RonL