My first 4u HSC Assessment (1 Viewer)

[Riviet]

Here it is, i posted it up especially for pLuvia, but everyone else feel free to attempt it. My class had to do it in 45 minutes so i suggest you try it with this time limit. I went pretty good for first 3 questions, which is mostly pretty straight forward, Q4 is a bit harder, but i just didnt get enough time to do it.
This paper is good practice for your upcoming 4u assessment too if you haven't done it yet so without further ado...
Good luck!

Q1
a) Solve x²-2x+10=0 , expressing the roots in the form a+ib. (2 marks)
b) Evaluate i/(1+i) and write your answer in the form a+ib and in modulus-argument form. (4 marks)
c) If w=1-i(sqrt3), show that arg(w+w) = arg(w x w). [the underline means conjugate of] (3 marks)

Q2
Find the locus of z if
a) Im(z²)=2 (2 marks)
b) |(z-2)/(z+2)|<1 (3 marks)

Q3
i) Solve x⁴+1=0 and show the four roots on an Argand diagram. (3 marks)
ii) Hence or otherwise exress x⁴+1=0 as a product of real factors. (3 marks)
iii) If z is one of the roots, show that 1+z²+z⁴+z⁶=0 (2 marks)
iv) Evaluate 0->7[SIGMA]zⁿ (2 marks)

Q4
i) On an Argand diagram show the points P and Q, corresponding to the complex numbers p=z1+z2 and q=z1-z2, where z1 and z2 are given complex numbers in the first quadrant. (2 marks)
ii) Show that if ^POQ=pi/2, then |z1|=|z2| (2 marks)
iii) Show that if OP=OQ, then (z2)²/(z1)² is real and negative (3 marks)

I have the worked solutions too if there are any problems.

 

[.ben]

How do you do Q3 iii) and iv)?

edit: is it AP/GP?

 

[Riviet]

Q3 iii) hint: there is a really cool shortcut that i managed to get in like 1-2 mins, use what you have to simplify and show that it equals 0
iv) hint: first write out all the terms in the series added up and simplify, using part i) and iii)

 

[pLuvia]

Hey thanks Riviet, could you post up answers for 1 and 2 and 3(i)
Haven't even done that sigma stuff

Edit: You meant to say for 3(iii) x not z? or I'm wrong

 

[pLuvia]

How do you do 4(iii)
4(ii) I posted that similar question somewhere on BoS

 

[.ben]

Does it have anything to do with summing a geometric series?

 

[Riviet]

No, it is z
z=x, it's just that it's letting z be the variable instead, since z is used as a root alot in complex numbers lol.

These are the final solutions to each question, if you have specific problems with a question, i'll post up that full worked solution for you.

Q1 Answers:
a) 1+3i or 1-3i
b) (1/sqrt2)cis[pi/4]
c) w+w=1-(sqrt3)i + 1+(sqrt3)i
arg(w+w)=arg 2=0
w x w=[1-i(sqrt3)][1+i(sqrt3)]
=1+3
=4
arg(w x w)=arg 4=0
.:arg(w+w)=arg(w x w)

Q2 Answers:
a) hyperbola, xy=1
b) |z+2|>|z-2| distance from -2 to z is greater or equal to the distance from 2 to z, including all the points lying on the y-axis. (use a diagram if you like)

Q3 Answers:
i) |x|=1 [my teacher's solutions had a brief proof for this but it isnt really necessary since we know that mod of each anyway]
x=1(cos@+isin@)
x⁴=(cos@+isin@)⁴
-1=cos4@+isin4@
cos4@=-1 , sin4@=0 [equating real and imaginary parts]
@=pi/4, 3pi/4, 5pi/4, 7pi/4

 

[Riviet]

Does it have anything to do with summing a geometric series?

Nope. XD
Read my hint and work it out, not that hard.

 

[pLuvia]

You do Q3 a weird way lol This is how I do it

x⁴+1=0
x⁴=-1
=-1
= cis pi
.: x = (cis pi)^1/4
= cis[(2kpi+pi)/4] where k = 0,+1,2
When k = 0, x = cis pi/4
when k = 1, x = cis 3pi/4
when k = -1, x = cis -pi/4
when k = 2, x = cis 5pi/4

Same thing anyway

Edit: Now I see what you did

 

[SeDaTeD]

You'll kick yourself if you don't see a really easy method for q3 iii and iv.

 

[Riviet]

Ok, problems resolved for pLuvia on msn, anyone else that has problems, feel free to ask any.

 

[pLuvia]

You'll kick yourself if you don't see a really easy method for q3 iii and iv.

I saw it

 

[insert-username]

Our first 4U assessment is tommorrow, and it's open book. I can't be bothered studying because there's only three people in the class and the other two are very likely to drop it. Besides, it's nearly holidays and I'm too burned out to care.


I_F

 

[pLuvia]

Open book maths? Well that just defeats the purpose doesn't it

 

[.ben]

Ok let's lay this open. I'm dumb and most likely to fail 4U. Please tell me. =].......... please?

 

[Riviet]

Haha there's alot more to maths than have all your formulae memorised/written out for a open book text... you guys should all know that by now.

 

[.ben]

Ok let's lay this open. I'm dumb and most likely to fail 4U. Please tell me. =].......... please?

lol you guys.

 

[insert-username]

Just because it's open book doesn't mean it's worth nothing. If you're flicking through a book, you're wasting time that could go to answering and checking questions.


I_F

 

[pLuvia]

Are you sure you really want to know, just if you change your mind

Spoiler

z⁴=-1

3(ii)
1+z²+z⁴+z⁶
Since x⁴ = -1
= 1+z²+(-1)+(z²)(-1)
= 0

3(iii)
0->7[SIGMA]zⁿ
= z⁰+z¹+z²+z³+z⁴+z⁵+z⁶+z⁷

Since z⁴=-1 and (1+z²+z⁴+z⁶) = 0
= (1+z²+z⁴+z⁶)+z¹+z³+z⁵+z⁷
= 0+z¹+z³+z⁵+z⁷
= z¹+z³+(z)(-1)+z³(-1)
= z¹+z³-z-z³
= 0

 

[Yip]

imo 3(ii) is a little pointless seeing as it is a quartic unity q lol...
x^4+1=(x^2+1)^2-2x^2
=(x^2+root2x+1)(x^2-root2x+1)

Alternate 3.(iv) method
0->7[SIGMA]zn
= 1+z+z^2+z^3+z^4+z^5+z^6+z^7
=[z^8-1]/z-1
=[(z^4)^2-1]/z-1
=0/z-1
=0


im not sure about my ans for q4...

4.(ii)
p=z1+z2 and q=z1-z2 form the diagonals of a parallelogram
From diagram, OQ is parallel to shorter diagonal of parallelogram (equal vectors have equal argument)
Therefore p and q are perpendicular (complementary cointerior angles on parallel lines)
Thus the parallelogram is a rhombus, and adjacent sides of rhombi are equal
|z1|=|z2|
(iii)
Now, let point B on the diagram be (1,0), let complex number z2 be D and complex number z1 be A and select an arbritrary point C such that triangle OAB|||triangle ODC

Using equal ratio corresponding sides in similar triangles,
1/|OA|=|OC|/|OD|
|OC|=|OD|/|OA|=|z2|/|z1|

arg[OD]=angleAOB-arg[z2] (due to similar triangles having corresponding angles)
=arg[z1]-arg[z2]
Thus OD represents the complex number |z2|/|z1|

If OP=OQ, the rhombus becomes a square because squares have equal perpendicular diagonals
Thus angleAOB-arg[z2]=90
arg[(z2)^2/(z1)^2]=2arg[z2/z1]=180
complex numbers with arguments of 180 lie on the negative side of the x-axis
Thus (z2)^2/(z1)^2 is negative and real

And yes, my diagrams suck badly