Vectors in complex numbers. (1 Viewer)

[x.Exhaust.x]

Do they occasionally pop up? (e.g. addition of vectors, subtraction of vectors, multiplication of vectors). What style of questions do they ask on these?

For locus problems e.g. locus of z being a circle, perpendicular bisector etc, I usually substitute x+iy and algebraically solve. Does anyone else approach this method? Or do you memorise the formula/style and draw it? Substituting in x+iy takes too much time methinks.

Has anybody heard of "PACMAN" questions? How frequent are "PACMAN" questions? For example:

Suppose arg(z-2/z+2) = pi/3. Show that the locus of z is a circular arc and provide a neat sketch of it.

Are those type of questions still examined? It's in the syllabus right? Because I can't seem to find it in there .

And finally, we covered subsets of the complex plane. Does the HSC examine subsets within complex locus problems? It gives me a headache...

Thanks.

 

[untouchablecuz]

Do they occasionally pop up? (e.g. addition of vectors, subtraction of vectors, multiplication of vectors). What style of questions do they ask on these?

geometric addition+subtraction: yes, theres usually a question or two asked in an exam
geometric multiplication: not so much

in terms of the styles of questions, often they concern transformations of certain complex numbers to another

e.g. on the complex plane, point O(0,0) A(a,a') B(b,b') [a>0, a'>0, b<0, b'>0] form an equilateral triangle

if the point A represents the point Z then what does B represent? answer in terms of Z

because its an equilateral triangle |Z|=|B| and angle AOB=pi/3

so B is a rotation of A clockwise through pi/3

this is equivalent to multiplying Z by cis(pi/3)

so, B=Zcis(pi/3)

ALSO, take note of the fact that multiplying a complex number by i is a rotation through pi/2, multiplying by i^2 is a rotation through pi etc very important

For locus problems e.g. locus of z being a circle, perpendicular bisector etc, I usually substitute x+iy and algebraically solve. Does anyone else approach this method? Or do you memorise the formula/style and draw it? Substituting in x+iy takes too much time methinks.

usually, the cartesian equation of a locus can be easily derived geometrically

i.e. locus of z such that |z-(a+ib)|=r is a circle with equation (x-a)^2+(x-b)^=r^2

locus of z such that |z-a|=|z-b| is the perpendicular bisector of the line joining A and B (equation can easily be found)

if you get something like,

find the locuz of z such that 2Im(z)+|z|Re(z)=3, i think your much safer substituting x+iy

Has anybody heard of "PACMAN" questions? How frequent are "PACMAN" questions? For example:

Suppose arg(z-2/z+2) = pi/3. Show that the locus of z is a circular arc and provide a neat sketch of it.

Are those type of questions still examined? It's in the syllabus right? Because I can't seem to find it in there .

i've only seen this type of Q only once before in a HSC exam

and the locus is generally pretty easy to find geometrically; consult cambridge for a good explanation

And finally, we covered subsets of the complex plane. Does the HSC examine subsets within complex locus problems? It gives me a headache...

can you elaborate (with an example or something)?

 

[addikaye03]

Suppose arg(z-2/z+2) = pi/3. Show that the locus of z is a circular arc and provide a neat sketch of it.

Those kinda Q are alright though.

So arg(z-2)-arg(z+2)=pi/3

arg(z-2)=pi/3+arg(z+2)

So then you just sketch it on an argand diagram where A(0,2) represents arg(z-2) crossing the real axis, similar for arg(z+2). The angle between there intersection is pi/3 and therefore pi/3 must be added to arg(z+2) to make there vectors parrallel (because of alternate angles). So now we determine that the locus of z if above the Real axis on the major arc.

Make another sketech, pi/3 be the angle between arg(z+2) and arg(z-2) again, there angle at centre= 2pi/3. Find Centre and length of radius.
Find an expression in terms of (x-a)^2+(y-b)^2=r^2.

-When i do past papers etc, i usually feel that if a type of Q hasn't popped up then its more likely this year.

 

[x.Exhaust.x]

Thanks again untouchable and addikaye.

can you elaborate (with an example or something)?

Alright here's are some examples (I know how to do them - so don't bother completing the solution for them, unless you're bored):

Sketch and describe the following subsets of the complex plane:

A = {z : |z-4| < 4}
B = {z: Arg(z) > 0}


Etc.

 

[untouchablecuz]

ah ok

identical to graphing regions on the xy-plane

 

[addikaye03]

Thanks again untouchable and addikaye.



Alright here's are some examples (I know how to do them - so don't bother completing the solution for them, unless you're bored):

Sketch and describe the following subsets of the complex plane:

A = {z : |z-4| < 4}
B = {z: Arg(z) > 0}


Etc.

|z-4|<4

Let z=x+iy

|(x-4)+iy|<4

Since rt(a^2)=|a|

(x-4)^2+y^2<16

Therefore Circle Centre (4,0) and radius <4

Therefore region is within the Circle

Add to Argand Sketch, Arg(z)>0 ie. Shade Quardant one and tw, dotted line extending alone Re(z) axis.

THen Find the region that satifies both.

I was bored lol