Complex number question (1 Viewer)

[GaDaMIt]

need help with a question ..

Find the locus of z if

w = (z - i) / (z - 2) and w is purely imaginary


Additionally, my teacher said something about the answer to this question being wrong as there are two exclusions and people forget to look at the top part of the equation when solving questions like these algebraicly, he said when you do it geometrically the two exclusions are evident

anyway the answer is a circle with radius rt 5 / 2, centre (1 , 1/2) excluding A(2 , 0), AB is the diameter B (0 , 1)

but yeah some other exclusion isnt in the answers according to my teacher..

 

[pLuvia]

Let z=x+iy

Rationalise the denominator, then since w is purely imaginary then the real part is equal to zero

Now let the Re(w)=0

And find the exclusions using the denominator, as it cannot equal to zero. And the locus should be found using the numerator

Hope that helps

 

[Wackedupwacko]

w = (z-i)/(z-2), let z = x+iy

w= [x+i(y-1)]/[x-2+iy] = [x+i(y-1)][x-2-iy]/[(x-2)^2 +y^2]

= [x(x-2) +y(y-1) +i{(y-1)(x-2) -yx}]/ [(x-2)^2 + y^2]
Re(w) = 0

=> [x(x-2) +y(y-1)]/ [ (x-2)^2 +y^2] = 0
=> x^2 -2x +y^2-y= 0
=> (x-1)^2 + (y-1/2)^2 = 5/4

circle =]

edit: corrected my mistake its a circle not hyperbola XD

 

[webby234]

w = (z-i)/(z-2), let z = x+iy

w= [x+i(1-y)]/[x-2+iy] = [x+i(1-y)][x-2-iy]/[(x-2)^2 +y^2]

= [x(x-2) +y(1-y) +i{(i-y)(x-2) -yx}]/ [(x-2)^2 + y^2]
Re(w) = 0

=> [x(x-2) +y(1-y)]/ [ (x-2)^2 +y^2] = 0
=> x^2 -2x -y^2 +y = 0

which is a hyperbola

Your second line is wrong - should be y - 1 not 1 - y.

 

[GaDaMIt]

Im more interested in the exclusions which would apply to this particular question. Im only getting (2,0) as (x-2)² + y² cant be 0, but i remmeber my teacher saying that for this particular question there were two exlusions.. one of which arent evident by doing the question in an algebraic manner

 

[Affinity]

I am guessing a bit here, but I think what your teacher is talking about is the fact that w is pure imaginary - this means that Re(w) =0, but also Im(w) can't be zero (i.e., w =/= 0). So, from the top line of the fraction, we can see that z =/= i.

Agreed

but then it depends on how you define an "imaginary number", I don't see why including 0 is not a sensible interpretation.


It's just like the question: what is a natural number?
It was quite fashionable in the past (not so remote either) to exlcude zero, now almost every text I encounter includes 0.

 

[Wackedupwacko]

o yea =( my bad... guess thats what i get for doing itall in my head and a monitor.... need paper for this. thanks for spotting the mistake

editted my answer

 

[GaDaMIt]

Yeah thanks alot guys. Im at school and i just asked my teacher, and yeah the other restriction is the top cant be zero so, (0, 1). He said something about when doing it gemetrically its a ray and yeah, but I understand it from someone elses reply on this post (cbb quoting lol) where they said top part imaginary and real cant both be 0 or something.

Anyway yeah thanks.

Are questions like these common? And if so do people usually forget the second exclusion or what?

 

[airie]

I would do this question geometrically.

Since (z - i) / (z - 2) is purely imaginary, this means that |arg(z-i) - arg(z-2)| is pi/2 ie. the vectors (z-i) and (z-2) are perpendicular to each other. Thus, the lines from a point P(x,y) representing a complex number z=x+iy to the point A(0,1) [represents the number i] and from P to the point B(2,0) [represents the number 2] are perpendicular to each other. So the locus can be determined using coordinate geometry (find gradient of PA and PB and use the fact that their product is -1), or you could directly say that it follows that the locus is a circle with the line joining A(0,1) and B(2,0) as the diameter. But if P is at A or B, one of the lines PA and PB would be a single point, while the other is a straight line, thus both points A and B do not satisfy the conditions. Therefore, the locus is (x-1)² + (y-1/2)² = 5/4, excluding the points (0,1) and (2,0).

 

[Wackedupwacko]

i never liked coordinate geometry...... absolute waste of time to do and use XD!!! id still stick with my algebra... its faster imo just more prone to mistakes... waitits not just a tad faster than coord geometry its alot faster cuz ur just multiplying one pronumeral after another and writing them down... where as coord u needa think about how to do the question....

 

[airie]

i never liked coordinate geometry...... absolute waste of time to do and use XD!!! id still stick with my algebra... its faster imo just more prone to mistakes... waitits not just a tad faster than coord geometry its alot faster cuz ur just multiplying one pronumeral after another and writing them down... where as coord u needa think about how to do the question....

I guess I'm just better with geometry and all. Don't really like fiddling around with algebra too much XP

 

[Affinity]

I'm afraid I disagree with blanket statements like this completely. It depends on the question.

Take this year's 4u paper, Q2 part d - the one involving the ellipse

|z-1-3i| + |z-9-3i| = 10

if you do this question geometrically, you can answer all parts of the question in less than 2 minutes.

But substitute z = x + iy instead and algebra bash and it will take more like 10 minutes - if you actually suceed in getting it out at all.

True.

that said, it's not that long to derive the polar form

let z = 1 + 3i + r[cos(t) + i sin(t)]
then you get

r=9/(5-4cos(t))

 

[Wackedupwacko]

hmm i guess its partially true although in reality you should already know all the easy shapes and their typical eqn and how to solve for the key points from them. those you should instantly recognise and thus spend almost no time trying to work out what it is. its the ones where you dont know what shape it is that i always fall back to algebra.


edit : and also when using coord geometry theres a fair amount of algebra involved as well (coord geomettry not just simple where u know what shape and how to find values from the given eqn) so using algebra straight out is faster.

edit2: also i happen to like algebra.... i tend not to get lost when i have a large amount of algebra (like q7 from ext1 that differentiation question.)

 

[jyu]

hmm i guess its partially true although in reality you should already know all the easy shapes and their typical eqn and how to solve for the key points from them. those you should instantly recognise and thus spend almost no time trying to work out what it is. its the ones where you dont know what shape it is that i always fall back to algebra.


edit : and also when using coord geometry theres a fair amount of algebra involved as well (coord geomettry not just simple where u know what shape and how to find values from the given eqn) so using algebra straight out is faster.

edit2: also i happen to like algebra.... i tend not to get lost when i have a large amount of algebra (like q7 from ext1 that differentiation question.)

What is |z-i| + |z+i| = 1 ?

 

[Wackedupwacko]

What is |z-i| + |z+i| = 1 ?

nah i was thinkn along the lines of z=0

 

[jyu]

nah i was thinkn along the lines of z=0

z = 0 does not satisfy |z-i| + |z+i| = 1.

For this and similar equations, substitution of z = x + iy will lead to a wrong cartesian equation. Using geometry for the interpretation of the equation the solution is { }.