# Thread: HSC 2018-2019 MX2 Marathon

1. ## Re: HSC 2018 MX2 Marathon Originally Posted by mrbunton z= x+iy, w = u+iv; w = z -1/z. , find locus of w if |z|=2
edit:from patel textbook
https://i.imgur.com/DOI8Tanr.jpg

Hopefully im right, correct me if wrong  Reply With Quote

2. ## Re: HSC 2018 MX2 Marathon Originally Posted by si2136 https://i.imgur.com/DOI8Tanr.jpg

Hopefully im right, correct me if wrong
u didnt fully complete it; the hardest part is finding the relationship between v and w.  Reply With Quote

3. ## Re: HSC 2018 MX2 Marathon Originally Posted by mrbunton u didnt fully complete it; the hardest part is finding the relationship between v and w.

Squaring, we have

Replacing variables,

or

Is that right?  Reply With Quote

4. ## Re: HSC 2018 MX2 Marathon

yup  Reply With Quote

5. ## Re: HSC 2018 MX2 Marathon

polynomial/complex number question:
cot(x)= y+1
a,b are roots of z^2-2z+2=0
prove that ( (y+a)^n - (y+b)^n ) / (a-b) = sin(nx) / sinx^n  Reply With Quote

6. ## Re: HSC 2018 MX2 Marathon

This is an interesting question I found in the SGS notes.

If is a complex number, find the locus of if  Reply With Quote

7. ## Re: HSC 2018 MX2 Marathon Originally Posted by fan96 This is an interesting question I found in the SGS notes.

If is a complex number, find the locus of if

Polar bash (mod-arg for the high schoolers) makes this problem trivial...  Reply With Quote

8. ## Re: HSC 2018 MX2 Marathon Originally Posted by fan96 This is an interesting question I found in the SGS notes.

If is a complex number, find the locus of if

=Re(x+iy -(x-iy)/x^2+y^2)=0
x - x/(x^2+y^2) = 0
times by x^2 + y^2 then divide by x;
0= x^2+y^2 -1;
1 = x^2 + y^2  Reply With Quote

9. ## Re: HSC 2018 MX2 Marathon Originally Posted by mrbunton =Re(x+iy -(x-iy)/x^2+y^2)=0
x - x/(x^2+y^2) = 0
times by x^2 + y^2 then divide by x;
0= x^2+y^2 -1;
1 = x^2 + y^2
is not the correct answer.

(Be careful of any restrictions you place on the locus)  Reply With Quote

10. ## Re: HSC 2018 MX2 Marathon Originally Posted by fan96 This is an interesting question I found in the SGS notes.

If is a complex number, find the locus of if  Reply With Quote

11. ## Re: HSC 2018 MX2 Marathon  Reply With Quote

12. ## Re: HSC 2018 MX2 Marathon Originally Posted by altSwift As I said above, that's not the correct answer.

The locus is not .

There are certain values of (specifically ) that need to be considered.

(By the way, is not on the unit circle so it's not necessary to explicitly exclude it if the locus was just the unit circle)

I'll post the solution tomorrow if nobody gets the right answer.  Reply With Quote

13. ## Re: HSC 2018 MX2 Marathon Originally Posted by fan96 As I said above, that's not the correct answer.

The locus is not .

There are certain values of (specifically ) that need to be considered.

(By the way, is not on the unit circle so it's not necessary to explicitly exclude it if the locus was just the unit circle)

I'll post the solution tomorrow if nobody can figure it out.
rip, I'll have another go tonight  Reply With Quote

14. ## Re: HSC 2018 MX2 Marathon Originally Posted by altSwift An exercise in Patel (I believe 4H, Q8) has a similar, but harder question.  Reply With Quote

15. ## Re: HSC 2018 MX2 Marathon

For anyone interested, the Patel question I was talking about is:  Reply With Quote

16. ## Re: HSC 2018 MX2 Marathon Originally Posted by fan96 An exercise in Patel (I believe 4H, Q8) has a similar, but harder question.

Nice! Here is my solution:  Reply With Quote

17. ## Re: HSC 2018 MX2 Marathon

It definitely seems to be something that most people forget. For this type of question it tends to not matter (in terms of marks), but being able to spot when a variable's domain has been restricted has saved me a few marks in exams.

(This might help with the locus problem I posted before...)  Reply With Quote

18. ## Re: HSC 2018 MX2 Marathon Originally Posted by fan96 For anyone interested, the Patel question I was talking about is:  Reply With Quote

19. ## Re: HSC 2018 MX2 Marathon Originally Posted by fan96 It definitely seems to be something that most people forget. For this type of question it tends to not matter (in terms of marks), but being able to spot when a variable's domain has been restricted has saved me a few marks in exams.

(This might help with the locus problem I posted before...)
Ooh these hints  Reply With Quote

20. ## Re: HSC 2018 MX2 Marathon Originally Posted by fan96 This is an interesting question I found in the SGS notes.

If is a complex number, find the locus of if

 Spoiler (rollover to view): let z= rcosθ+isinθ z-1/z = (r-1/r)cosθ + i(r+1/r)sinθ Re(z-1/z) = (r-1/r)cosθ If LHS = 0, then (r-1/r) = 0 or cosθ = 0 hence, r=1 or θ= ±(2n+1)π/2 Hence the locus is x^2+y^2 = 1 and x = 0 (excluding (0,0))  Reply With Quote

21. ## Re: HSC 2018 MX2 Marathon Originally Posted by jathu123 Spoiler (rollover to view): let z= rcosθ+isinθ z-1/z = (r-1/r)cosθ + i(r+1/r)sinθ Re(z-1/z) = (r-1/r)cosθ If LHS = 0, then (r-1/r) = 0 or cosθ = 0 hence, r=1 or θ= ±(2n+1)π/2 Hence the locus is x^2+y^2 = 1 and x = 0 (excluding (0,0))
ahahaha I forgot to consider x = 0 jesus  Reply With Quote

22. ## Re: HSC 2018 MX2 Marathon

A classic one  Reply With Quote

23. ## Re: HSC 2018 MX2 Marathon Originally Posted by jathu123 Spoiler (rollover to view): let z= rcosθ+isinθ z-1/z = (r-1/r)cosθ + i(r+1/r)sinθ Re(z-1/z) = (r-1/r)cosθ If LHS = 0, then (r-1/r) = 0 or cosθ = 0 hence, r=1 or θ= ±(2n+1)π/2 Hence the locus is x^2+y^2 = 1 and x = 0 (excluding (0,0))
Yep.

If you used , you ended up with the locus

Which is the correct answer.

It's tempting to divide both sides by , but if you performed the division you were implicitly assuming that , which lost solutions.

Clearly if , any value of is a solution, except for zero since .

That can also be shown by considering

which is zero for all .

To graph , you could consider two separate cases: , and .

This is the solution from the SGS notes:   Reply With Quote

24. ## Re: HSC 2018 MX2 Marathon Originally Posted by mrbunton polynomial/complex number question:
cot(x)= y+1
a,b are roots of z^2+2z-1=0
prove that ( (y+a)^n - (y+b)^n ) / (a-b) = sin(nx) / sinx^n
I tried taking out a factor of

from the top two brackets but I still cant get to the LHS... any hints?  Reply With Quote

25. ## Re: HSC 2018 MX2 Marathon Originally Posted by altSwift I tried taking out a factor of

from the top two brackets but I still cant get to the LHS... any hints?
glad someone noticed my question
make sure to use all the given equations.
eg: a and b represent their own complex numbers (on the lhs), try finding them;
its easier approaching it from the lhs to the rhs;without trying to take out any common factors immediately  Reply With Quote