Complex Numbers (1 Viewer)

wagig

Member
Joined
Jan 29, 2013
Messages
152
Gender
Male
HSC
2015
Hey guys, i have an exam coming up this thursday and i'm not sure about an answer to a questions:
"On the Argand diagram, clearly illustrate the locus of z if w = (z+2)/(z-4i) is purely imaginary"
 

rumbleroar

Survivor of the HSC
Joined
Nov 30, 2011
Messages
2,271
Gender
Female
HSC
2014
First, I would let z=x+iy
Then I would solve w (realise your fraction,etc)
Then let re(w)=0 since it's purely imaginary
Would probably get a locus and sketch

Hoped it helped!


Sent from my iPhone using Tapatalk
 

SpiralFlex

Well-Known Member
Joined
Dec 18, 2010
Messages
6,960
Gender
Female
HSC
N/A
You're missing the "y ≠ 0" condition. If Re(w) = 0, y = 0, then it is Real not Imaginary. 0 + 0i = Real.
That's a condition. I was talking about another condition.

The condition was when z = 0 + 4i. i.e (to draw a 'hole' at (0, 4), since it is not part of the locus of z).
 
Last edited:

ChillTime

Member
Joined
Dec 9, 2013
Messages
40
Gender
Male
HSC
2005
This unassuming question is turning out quite interesting.

You're missing the "y ≠ 0" condition. If Re(w) = 0, y = 0, then it is Real not Imaginary. 0 + 0i = Real.
Did you mean there was a discontinuity at y = 0 or Im(w) = Re(w) = 0?

Because w = that huge expression in my working above.

That's a condition. I was talking about another condition.

The condition was when z = 0 + 4i. i.e (to draw a 'hole' at (0, 4), since it is not part of the locus of z).
Right you are.
 
Last edited:

anomalousdecay

Premium Member
Joined
Jan 26, 2013
Messages
5,766
Gender
Male
HSC
2013
That's a condition. I was talking about another condition.

The condition was when z = 0 + 4i. i.e (to draw a 'hole' at (0, 4), since it is not part of the locus of z).
Yes this. Remember that the denominator is NOT equal to zero, hence the condition.
 

ilikecake

New Member
Joined
Nov 19, 2013
Messages
5
Gender
Male
HSC
2014
An easier way would be to do this geometrically.

Because the result is purely imaginary, we can simply say that arg[(z+2)/(z-4i)] = +/- (pi/2).
Subsequently, arg (z+2) - arg (z-4i) = +/- (pi/2).
The shape itself becomes a circle, with z+2 and z-4i as two points on the circle that, when you draw a line from each to another point on the circle, the angle formed is pi/2.
 

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

Top