Question from half yearly... (1 Viewer)

[dawso]

I so reckon i have been fuked over by my maths department this year, they really dont like me, here is another question i swear i got right but they hav marked me wrong....

(i) in the diagram, POQ is a right angled, isosceles triangle. P represents the complex number z. Find the complex number represented by Q.

(ii) what is the complex number represented by the vector PQ?

(diagram in attatchment)

also this one:

Desicribe and sketch the locus of those points z=x+iy such that:

arg ((z-i)/(z+i)) = pi/2

for this, how much working do u think u shud hav 2 show, cause i just did it in my head and drew it and they didnt like that....

 

[KFunk]

i) If P represents z then Q is a 90° rotation of z clockwise which is equivalent to multiplying z by '-i'. Hence Q can be expressed as:
'-iz'

 

[dawso]

mmhmm, thank you kfunk (thats wat i got and guess wat...no marks!!) the rest please peoples??

btw kfunk, i hav 2 say that ur famous picture of 'conics is the devil' is now my 4unit book cover , lol

 

[KFunk]

mmhmm, thank you kfunk (thats wat i got and guess wat...no marks!!) the rest please peoples??

btw kfunk, i hav 2 say that ur famous picture of 'conics is the devil' is now my 4unit book cover , lol

You don't know how proud I am of that . Spread the word!!!

P.S. For ii) was there anything to indicate that PQ was perpindicular to the x axis. Also, how did they justify gving you no marks? What did they want?

 

[dawso]

nah, pg isnt necessarily perp 2 axis, its quite easy i think but i wont giv any hints, see wat others get, and they gave no excuse, just didnt mark it right or wrong just put a total of 1 mark 4 all of wat i posted (out of 4)

 

[KFunk]

In which case I'm thinking it would just be vector Q - vector P.
P = z = x + iy
Q = -iz = y - ix

PQ = Q(y - ix) - P(x +iy) = y - x - i(y + x)

I don't know if that's the way to go about it, I'm starting to feel rather rusty with the complex stuff (quite prematurely, might I add).

 

[Slidey]

btw kfunk, i hav 2 say that ur famous picture of 'conics is the devil' is now my 4unit book cover , lol

You're an ideas man.

(ii) what is the complex number represented by the vector PQ?

You can use vector components:

OP_vector = (x,y)
OQ_vector = (y,-x)
OP+PQ=OQ, so PQ=OQ-OP:
If we take Q to be -iz, then:
PQ=(y,-x)-(x,y)=(y-x,-x-y)=-(x-y)-(x+y)i

"Describe and sketch the locus"

Note the describe. If you only drew a sketch you will not get full marks, but you should get some marks.

 

[dawso]

ok guys, now me and steele hav been havin a little intellectual discovery session here, and i know, it is quite sad that im spendin my holidays argueing bout maths, but its to cold and there is no swell, so wat else is there 4 a guy 2 do?

now in simple terms someone answer me,

arg(n/m) = arg(n) - arg(m) true/false??

 

[KFunk]

True . . . . . .

Arg(z₁z₂) = argz₁ + arg z₂

Arg (z₁/z₂) = argz₁ - arg z₂ or so I have been led to believe.

since if z₁= r₁cisθ₁ and z₂ = r₂cisθ₂

z₁/z₂ = r₁cisθ₁/ r₂cisθ₂
= (r₁cisθ₁/r₂cisθ₂) x (r₂cis(-θ₂)/r₂cis(-θ₂))
= r₁r₂cis(θ₁ - θ₂)/ (r₂)²cis0

i.e z₁/z₂ = r₁/r₂.cis(θ₁ - θ₂) which would appear to make the above true.

 

[Slidey]

That's good. Which then leads to the question: what type of conic section is the locus of arg ((z-i)/(z+i)) = pi/2?

I owe Matt a cookie if he's correct.

 

[FinalFantasy]

Desicribe and sketch the locus of those points z=x+iy such that:

arg ((z-i)/(z+i)) = pi/2


i would say Locus is half circle left of the Im(z) axis, such that the angle inscribed in it is pi\2, the locus exluces the points i and -i

 

[Slidey]

Yes that's what Matt got. Do you have any working?

 

[FinalFantasy]

Yes that's what Matt got. Do you have any working?

o..i don't think i can produce any working to show dat
but we had such questions in exams before and what i had there was sufficient to acquire full marks...

 

[KFunk]

There's probably a way but I havn't seen a conclusive algebraic proof for these types of question. You just need logic and a bit of geometry.

You observe that the case of arg(z-i) = arg(z+i) + π/2 holds true to the left of the im-axis and that since the difference in args is 90° you've got a semi circle (from circle geo). Hence the centre is at (0,0) with a radius of half the distance between (0,1) and (0,-1) i.e radius = 1 unit.

therefore arg(z-i) = arg(z+i) + π/2 when x² + y² = 1 where -1&le;x<0

 

[Slidey]

Hmm. That makes sense. Thanks.

Know where the fallacy lies here?

arg((z-i)/(z+i)) = π/2
.'. the complex number (z-i)/(z+i) is purely imaginary, so:
Re[(z-i)/(z+i)]=0
(x+(y-1)i)(x-(y+1)i)/|z+i|^2
Re[(x+(y-1)i)(x-(y+1)i)]=0 (denominator real)
x^2+(y^2-1)=0
x^2+y^2=1

Uh. Nevermind. Matt gets his cookie.

 

[KFunk]

That's pretty solid, nice work .