locus (1 Viewer)

=)(= · Dec 10, 2022

for part b is it just -0.5≤x≤0.5 cause this lies inside of the condition of a?

Drongoski · Dec 10, 2022

For b): I think x>= 0.5. So locus is the arc of the unit circle, centre O, with $-\frac {\pi}{3} \leq arg z \leq \frac {\pi}{3}$

=)(= · Dec 10, 2022

ohh i see but how did you find the argument region

Drongoski · Dec 10, 2022

=)(= said:
ohh i see but how did you find the argument region

I draw a unit circle, centre the origin.

$z = cis \theta $ is just a point on the circle and $ \frac {1}{z} = \overline z $ is just another point directly below z. $ z + \frac {1}{z} $ form a diagonal on the x-axis of the rhombus, formed by z and $\frac {1}{z} $ as adjacent sides. You can then work things out from there. Of course a related diagram would be so much more helpful! $.$

locus (1 Viewer)

=)(=

Active Member

Drongoski

Well-Known Member

=)(=

Active Member

Drongoski

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)