4U Maths Question Thread

[justwannawish]

(Image removed from quote.)



\begin{align*}\therefore z &= 1 \left( \cos \frac\pi3 + i \sin \frac\pi3\right)\\ &= \frac12 + \frac{\sqrt3}{2}i\end{align*}

how would we do the question if the args didn't add up to pi/2 and we couldn't use the circle method

[RuiAce]

how would we do the question if the args didn't add up to pi/2 and we couldn't use the circle method

This is actually a lot harder in general. Whenever you encounter questions like this, usually there's a trick or two behind it all. In general, however, you may need to brute force it and actually find the equation of the ray, before going back to 2U methods for the point of intersection.

[justwannawish]

This is actually a lot harder in general. Whenever you encounter questions like this, usually there's a trick or two behind it all. In general, however, you may need to brute force it and actually find the equation of the ray, before going back to 2U methods for the point of intersection.

:/ That sounds tricky haha. If you don't mind, could you walk me through finding the equation of the ray? e.g. if the question was like arg(z-1)=pi/4, arg(z+1)=pi/3 find arg(z)

Also, what are some of the really common complex number qs that one must now. From looking at past trials, finding the roots of unity and factorising over complex/real fields, finding square roots and then solving a quadratic are often the easy marks at the beginning of the test. Can you think of any others, especially to do with vectors (e.g. finding locus)

Thank you!

[RuiAce]

:/ That sounds tricky haha. If you don't mind, could you walk me through finding the equation of the ray? e.g. if the question was like arg(z-1)=pi/4, arg(z+1)=pi/3 find arg(z)

Also, what are some of the really common complex number qs that one must now. From looking at past trials, finding the roots of unity and factorising over complex/real fields, finding square roots and then solving a quadratic are often the easy marks at the beginning of the test. Can you think of any others, especially to do with vectors (e.g. finding locus)

Thank you!

The method itself is actually not too bad. Take \( \arg (z-1) = \frac\pi4 \). The ray will just be one part of the line, through the point \( (1,0) \), with gradient \( m = \tan \frac\pi4 = 1 \). So using the point gradient form, you just have \( y = 1(x-1) \).

Having said that, whilst that's easy enough, if you did the same thing for \( \arg (z+1) = \frac\pi3 \) you would have \( y = \sqrt{3}(x + 1) \). And then you see the main problem - that \( \sqrt3\) is gonna be hella annoying. It's easy enough to go back to 2U here, and just do simultaneous equations to solve for the point of intersection. But would I expect a 4U student to want to deal with surds for this? Probably not.

Worse, they could move it away from -1 and 1 and give you a random complex number like \( -\sqrt{2} + i\). Or they could swap the angles into something which you'd have to do a bit of work to compute the exact values for.
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Quite a lot of the common ones are all included in my 4U notes book. Vector addition itself is uncommon altogether. For stuff like locus, there are many curves that you'd be expected to know how to draw. On top of that, it should be fairly easy to sketch regions in the Argand plane based off them.

A somewhat common culprit for the trials is to write down the minimum value of \( \arg z \), or \( |z| \), or the real part or whatever, given that you've already sketched its locus. Sometimes they'll swap that out for the maximum value as well. These were covered in my trial lectures last year and you should look at the notes section for a few examples on those.

A lot of the more popular ones tend to overlap into the polynomials topic as well.

[arii]

Hi Rui and other users, I'm not too sure how to do (b) and (c).. I feel like (b) might be related to cyclic quad through constructing another circle but I haven't had much luck doing it.

[RuiAce]

Hi Rui and other users, I'm not too sure how to do (b) and (c).. I feel like (b) might be related to cyclic quad through constructing another circle but I haven't had much luck doing it.

Also with part a), you really should invoke that RASD is a cyclic quadrilateral first (reason: the angles you specified are supplementary), before you draw the circle around the points.


(In this case, to the right of \(DC\)).

Note that the theorem used here is essentially the converse of the "angles subtended to the circumference on the same arc are equal" theorem. It is also one of the reasons why I hate maths in focus, because they don't teach this theorem.

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\begin{align*}\pi - \angle DST &= \angle DCT \tag{part ii}\\ &= \angle DAR \tag{ext. angle of cyclic quad}\\ &= \angle DSR \tag{part i}\end{align*}

[arii]

Also with part a), you really should invoke that RASD is a cyclic quadrilateral first (reason: the angles you specified are supplementary), before you draw the circle around the points.


(In this case, to the right of \(DC\)).

Note that the theorem used here is essentially the converse of the "angles subtended to the circumference on the same arc are equal" theorem. It is also one of the reasons why I hate maths in focus, because they don't teach this theorem.

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\begin{align*}\pi - angle DST &= \angle DST \tag{part ii}\\ &= \angle DAR \tag{ext. angle of cyclic quad}\\ &= \angle DSR \tag{part i}\end{align*}

Thank you for your reply. I wanted to ask about the third-to-second-last line for the solution to part (c). I don't understand how Angle DST = Angle DAR through the external angle of cyclic quad. DAR is the external angle to the cyclic quad, but DST is not one of the four corners in cyclic quad ABCD, it's kinda just floating in the middle... I think the external angle of cyclic quad means Angle DAR = Angle DCB?

[RuiAce]

Thank you for your reply. I wanted to ask about the third-to-second-last line for the solution to part (c). I don't understand how Angle DST = Angle DAR through the external angle of cyclic quad. DAR is the external angle to the cyclic quad, but DST is not one of the four corners in cyclic quad ABCD, it's kinda just floating in the middle... I think the external angle of cyclic quad means Angle DAR = Angle DCB?

Typo's from doing it before I went to sleep. Fixed

[arii]

Hi Rui (and other users), so this one is a bit of a challenge.. I'm having some trouble actually visualising what is descriptively told to us so if someone can include a diagram, that'd be amazing.

[jazzycab]

Hi Rui (and other users), so this one is a bit of a challenge.. I'm having some trouble actually visualising what is descriptively told to us so if someone can include a diagram, that'd be amazing.

This is what the image looks like:

[kaustubh.patel]

Hey there guys need some help with polynomials, in the first que i get that 2 roots are complex conugates as all coefficient are real and to have 2 turning points there must be a negative gradient (thus c<0) but the rest is tough to get.

[RuiAce]

Hey there guys need some help with polynomials, in the first que i get that 2 roots are complex conugates as all coefficient are real and to have 2 turning points there must be a negative gradient (thus c<0) but the rest is tough to get.

What are the questions?

[kaustubh.patel]

Ohhhh so sorry i completely forgot the ques my bad. Here they are.

And please a bit more help with another hard conics question im having trouble with.
NP is the ordinate of a point P(x1,y1) on the hyperbola x^2/a^2 - y^2/b^2=1. The tangent at p meets the x axis at T. Prove thata (OT)(ON)=a^2, where O is the origin.

[RuiAce]

Ohhhh so sorry i completely forgot the ques my bad. Here they are.

And please a bit more help with another hard conics question im having trouble with.
NP is the ordinate of a point P(x1,y1) on the hyperbola x^2/a^2 - y^2/b^2=1. The tangent at p meets the x axis at T. Prove thata (OT)(ON)=a^2, where O is the origin.

PDF attached. As stated, will get back to part a) of Q15 later. (I wasn't really able to use the internet at the time of writing up a response.)

I'm also not sure what the wording of your conics question suggests. The 'ordinate' is just the y-coordinate of a point, yet it states \(NP\), which is a line (more accurately a segment). The ordinate of the point \(P\) is just \(y_1\).
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EDIT




GeoGebra simulation



Note that this is guaranteed, because \( x^3-3x+k \) is guaranteed to never have a triple root.

[kaustubh.patel]

oh i see what the k does it shifts the graph up and that restricts the the cubic to intersect the x axis more than once. Thank you heaps Rui deeply appreciate it even if you had no internet. The other conics que i asked is from sk patel (6E q8)  textbook, take your time and thank you for helping us out.