4U Maths Question Thread

[RuiAce]

Yewww I can say I posted in 4u maths 

Helloooo, this is such a dumb question that has no relevance to any mathematical problems whatsoever, but I am so confused about the structure of 4u maths so I thought I might as well ask here since the exam is over  Sooo I know you guys sit the exam the same time as Mathematics, meaning you don't do that exam, but do you do the Mathematics course still? Im 10/10 confused and always wanted to know the answer to this

- from your fellow 4 unit (English) student 

Perhaps the easiest way to put it

The 4U student is taught the 2U content as well, but then it doesn't get examined. It just becomes assumed knowledge for the 4U student

[paigek3]

Perhaps the easiest way to put it

The 4U student is taught the 2U content as well, but then it doesn't get examined. It just becomes assumed knowledge for the 4U student

Riiiight that makes sense!!! Ahhh thank you for answering this I have been trying to figure it out for so long

[itssona]

solve x^2cos^2 theta + x sin2theta +1 =0

[RuiAce]

solve x^2cos^2 theta + x sin2theta +1 =0

Assuming that theta is not a typo, what are we trying to solve for here? Because if we're trying to solve for x, the easiest way out is really just the quadratic formula

[RuiAce]

Please note that these were demonstrated in the lectures. So I will not go into much depth with the solution.

Would someone please do this. Solving quadratic equations with complex coefficients: finding the complex number equations.

4x^2-4(1+2i)x-(3-4i)=0

\begin{align*}0&=4x^2-4(1+2i)x-(3-4i)\\ x&= \frac{4(1+2i)\pm \sqrt{16(1+2i)^2 +16(3-4i)}}{8}\tag{quadratic formula}\\ &= \frac{4+8i\pm \sqrt{16(1+4i-4)+16(3-4i)}}{8}\\ &= \frac{4+8i \pm \sqrt{0}}{8}\\ &= \frac{1}{2}+ i\end{align*}

Hi, guys, do you use De Moivre's Thereom to solve:
z^5= -32
Thank you

Handwriting as "cis" is ok



So our solutions are 2cis(pi/5), 2cis(3pi/5), ...

Remark: You could've taken k=-2,-1,0,1,2

[itssona]

lil question
say i have 2 complex roots which are conjugates of eachother. One is alpha and the other is beta.
how do we prove alpha^2=beta and beta^2=alpha

thank you

[RuiAce]

lil question
say i have 2 complex roots which are conjugates of eachother. One is alpha and the other is beta.
how do we prove alpha^2=beta and beta^2=alpha

thank you

That usually isn't true. Do they give more information?

[itssona]

That usually isn't tru6e. Do they give more information?

well alpha=(-1+root 3 i)/2
beta=(-1-root3 i)/2

verify that a^2=b, b^2=a, a^3=b^3=1, 1+a+b=0

maybe it's related to roots of unity? idk :/

[RuiAce]

well alpha=(-1+root 3 i)/2
beta=(-1-root3 i)/2

verify that a^2=b, b^2=a, a^3=b^3=1, 1+a+b=0

maybe it's related to roots of unity? idk :/

Lol.

Yes, I bet you those are the cube roots of unity

[itssona]

Lol.

Yes, I bet you those are the cube roots of unity

oh lmao how do I verify all that stuff tho

[RuiAce]

Tbh, whilst there might be a crafty solution, at a time like that I would just be lazy and actually do the multiplication by hand

[itssona]

Tbh, whilst there might be a crafty solution, at a time like that I would just be lazy and actually do the multiplication by hand

ooh okay thank you Rui

[itssona]

Heya, I tried doing this by subbing in x1+iy1 for w1 and x2+iy2 for w2 but im still not getting it :/

solve for w1, w2
2w1 + 3iw2 =0
(1-i)w1 + 2w2 = i-7

(when i write w1, i mean w subscript 1)

[RuiAce]

Heya, I tried doing this by subbing in x1+iy1 for w1 and x2+iy2 for w2 but im still not getting it :/

solve for w1, w2
2w1 + 3iw2 =0
(1-i)w1 + 2w2 = i-7

(when i write w1, i mean w subscript 1)

Note that you will need to divide complex numbers.

[itssona]

Thank you so much Rui

Alsoo, kinda not getting anywhere with this despite using trig formulas and whatnot:
if z is cistheta
prove that
2/(1+z) = (1+costheta+isintheta)/(1+costheta) = (2cos^2theta - 2isin(theta/2) cos(theta/2))/ 2cos^2(theta/2) = 1-itan(theta/2)

or if you can point me in the general direction? thank you!!!